Alright, everyone hopefully got a chance to solve this one. We have these three weights and we're given the coordinates of where they are on the Y access and we want to find the center of mass of these three weights. Remember the center of mass is just if you could collapse all of these into a single object, where would that be located? So what I'm gonna do is I'm going to draw a Y axis, that's my Y axis and this is the X. Like this. And basically I'm gonna draw out where each of these things are. So the first one is a one kg block er waits at two m positive. So I'm gonna say this is the plus Y axis. So I'm just gonna draw this out. It doesn't have to be perfect. So this is going to be um my one kg And my wife value is two. The next one is a 1.5 kg at the origin. So in other words at the origin here, we have 1.5 kg and finally we have a 7.5 kg at negative 1.5 m. In other words, if this is negative one, this is negative two, it's gonna be somewhere in the middle here and I'm gonna draw a really, really, really big dot. Alright, so this is gonna be 7.5 kg. Alright, so if you could collapse all of these into a single object, where would the center of mass B before we even start calculating anything. Remember the center of mass tends to sort of skewed towards the higher mass. So in other words, this one here which has a higher mass than anything else, is probably going to have the center of mass to be very close to it. Right, rather than the other two. So how do we calculate this? Remember we just have an equation with this for X. The center of mass. Where you just do em one X one, M two, X two. As many objects as you have. The only thing that's different is that we're just doing in the Y axis now. So we have Y. X. For our Y axis, center of mass. We're going to do M. One, Y. One, M. two, Y 2 & M. three, Y 3 And then divide by the total mass. So m. one M. Two, M. three. All right. So basically this M. One here, we're gonna have to just assign which one is M. One which is M. Two. Which one's M. Three? It really doesn't matter the order in which you do it. So, this one could be M. One or it could be M. Three? It doesn't matter. You still get the right answer. I'm just gonna call this M. One. What does matter is that you keep the positions consistent? So, if this is M. One, this has to be Y one. If this is your M. Two, this is going to be your Y. Two which is zero And then your M. three And then your Y three is going to be negative 1.5. All right. So that has to be consistent. So just plugging this stuff in, we usually have one times two plus and then we have 1.5 and if it's the origin then the Y. Apple Y. Value is just zero and that whole term goes away and then we have 7.5 times negative 1.5. The negative does matter in this case because that's the position. Right? So now we divide by the total mass which is one plus 1.5 plus 7.5. All right. So when you work this out, what you're gonna get is you're gonna get let's see two plus. And this is gonna be negative 11.25 divided by a grand total of 10. And when you work this out, what you're gonna get is a negative 0.93 m. That is your final answer. So, that's where the center of mass is located. And if you were to draw this out, this would be somewhere around here, somewhere like right under right above the negative one mark. This is where your center of mass is. And that makes some sense. Again, the center of mass should be sort of in the middle of all of them, but it's gonna be skewed towards the higher mass on the bottom. Alright folks, so that's it for this one

3

example

Coordinates of Center of Mass of 4 objects

Video duration:

5m

Play a video:

Hey everyone. So let's work this problem out together, we're gonna calculate the X and y coordinates of the center of mass of this system of objects here in centimeters. So here's the idea up until now, all of the problems that we've looked at have had objects on either the X or the y axis. But in this problem it's sort of arranged in a square. So we have both X and y. It turns out that we can still use the same exact center of mass equations for the X and the Y axis. All we really are doing here is we're just trying to figure out if you take this sort of system of objects, where's the one center where you could basically just replace all of the mass as if we were located at that point, is it gonna be over here is gonna be over here or over here or over here or something like that? Well, a good guess is it's going to be that it's going to be closer towards the heavier objects. So probably it's going to be somewhere over here. And basically, what you need to do is figure out what is the X coordinates. And then what is the Y coordinate, some of the words, the X center of mass and the y center of mass. So really, that's all we're doing here. And this is basically that's that's gonna be the location of that center of mass and to do that. We just need to go ahead and use our X and Y equations remember with a two dimensional sort of problem, you can always break it down into one dimensional problems. So basically we just need to figure out the extent of mass and that's going to be M A X A plus MB XB plus EMC XC plus M D X D. Divided by the total mass. Right? This is this is the big M basically remember this M one plus M two. All of this really is just big M. So just to make the math a little bit simpler, we can actually just calculate that right then. So we don't have to plug in all these numbers. If you work this out you're gonna get is 0.25 plus 0.4 plus 0. plus 0.6. Just all the masses added up together. And you're gonna get 1.65 kg. So in other words, the X coordinate here is really just going to be 0.25. And then what's the X coordinate for A Well actually the X coordinate is just gonna be 000 because it's at the origin. And in fact I'm going to go ahead and just write the coordinates for each one of the points. So this is going to be 008. Um This is going to be 8208. This is gonna be eight comma zero. Remember this is the X coordinate and that's the y coordinate for each one of the respective points. So then to get back to this, uh the X coordinate for A. Is zero. So, what about the mass for B. It's 0.4. What about the X coordinate for B. It's still zero, right? Because it's still basically just along the Y axis. It has no X coordinates. So that goes away as well. What about this one? This is going to be 0.6 times this is going to be eight and this is gonna be 0.4 times eight again. All right, so the only two that really are going to be non zero are gonna be these two. These are the only two that actually have an X coordinate. That makes sense. Now, we're just going to divide by 1 61.65. What you should get here is an X coordinate of 4. cm. All right. So that is your X coordinate. And not going to make some sense here. If this side length here is 8cm right? Because it's over here, then the center of mass is going to be not halfway. Now, this would be like four. So, this would be four. It's actually gonna be a little bit beyond that because remember this big mass over here at that corner is sort of pulling or skewing all of the center of mass towards it. So it's not gonna be before it's actually gonna be a little bit more than that. Okay, so what about the Y center of mass? We actually just do the exact same process, but we're just going to do it with Wise instead. So M. A. Y. A plus M B, Y B plus M C Y C plus M. D. Y. D. All divided by 1.65. So in other words this is going to be 0.25. And then what's the Y coordinate for A the Y coordinate is zero because it's still at the origin. What about the Y coordinate for B. It's gonna be eight. Right, That's that corner plus the Y coordinate for C. So that's gonna be eight as well. And then finally four D. It's going to be just zero. Right? That's the white corner for zero. So then if you work this out, what happens is these two terms will cancel out, divided by 1.65. And when you work this out you're gonna get 4.85 cm again. Now this isn't sort of this isn't just a coincidence, it's actually just because if you think about this, these two blocks over here, these four the 0.4 kg blocks are sort of symmetrical about this axis of the square. So really what happens is the center of mass is really just determined by these two blocks over here, it's these two. So what happens here is that this is the smaller one and this is the larger one. So the center of mass sort of gets shifted towards it, and it's going to end up over here. So, because of the symmetry over here, that's why you ended up with the same coordinates for the X and Y axis. That's it for this one. Guys let me know if you have any questions.

Do you want more practice?

We have more practice problems on Intro to Center of Mass