This is a problem that deals with the center of mass. The center of MASH equation can generally be written something like this, where the numerator has the center of mass of each individual object times, its mass divided by the denominator, which is just the sum of all of the individual masses. This equation looks a lot more complicated than it actually is. So let me explain this by drawing the picture first and foremost, so we have a table here. We have a leg, there's the top of the table and we have a second leg. And on top of the table there are three blocks. They're kind of stacked off of the table. The first one has one quarter of its length off of the table. The second one has an additional quarter of its length off the table, and the third one has once again an additional quarter of its length off the table. So all of these blocks have a total length of L because they're the exact same block, and they also have the same mass of em. So it looks kind of confusing, but it's not actually that bad. The first thing we need to do is set our X equals zero point and we're going to set that right here at the edge of the table because that just makes the most sense with geometry. Now what we need to do is find the individual center of mass is for each of these objects because it's just literally a block. The center of mass is just going to be the very center of the object. So I'm drawing the X here to kind of show where they are. Now that we have that we just need all of these values. So what I'm going to be looking for is the distance between the center of Mass and the X equals zero point for every block. And that's what we plug in for the X I right here. So if we just start plugging things into the center of mass equation, we have in the denominator the some of the three Masses, which is just M plus M plus m. You can also write that as three m. But I wanted to be a bit clearer here, so if we start with the red block because this distance here it's l over four off of the edge of the table. That means that from the edge of the table to the center of Mass is just going to be once again another L over four. And that's because the center of mass for each of these blocks is just going to be at half of its length because that's how center of mass works. So we can write up here that we have The center of mass of the red block is l over four. We multiply that by its mass. We will now add the center of mass of the green block. And this one is actually easy because l half of the green block, the very center of the green block, is on the edge of the table. So we have zero times m, which is kind of irrelevant, but we'll write it out anyway. And then we do the same thing for this last block. Now, my drawing doesn't really indicate this, but if this length here is l fourth, this length here is another l fourth and we have a third l fourth here. That means we have three l over four until we get to the very edge of the table. And what that means is that from the edge of the table to the center of mass is just a distance of l divided by four. The main difference here, though, compared to our example in red, is that now we're on the left side of the X equals zero axis. And what that means is we're not actually going to have a positive value of a center of mass. This is going to be a negative center of mass, so it will be negative. L divided by four, multiplied by the mass. Let's just make this denominator a little bit larger. So when you plug all of this in your you're obviously going to have 3 a.m. in the denominator, the numerator, the zero times M is irrelevant. But we also have that the red number l four times m is exactly equal, but opposite to the blue number, which is just the minus. L divided by four times m. So if you have the same number subtracting itself, that just becomes zero, which is the final answer. The center of mass of the final configuration is just zero that if the center of mass is zero right on the edge of the table. This is a stable configuration and it won't fall. It's close to it. If we were a little bit further to the left, if we had a negative X value for the center of mass, then it wouldn't be stable. But here it is.