Problem 11.87. So there's a bar that we're trying to measure the mass of. So capital M is the mass. This bar is four meters long and we're told that it has holes and stuff in it. So it's ah, its center of mass is not in the middle. In fact, the center of mass of the bar, we're told, is 1.88 beers from its left edge. So we then are told that we place the bar when we determine this by by placing it on a knife edge. And then we're told that we placed the bar, uh, slightly further to the right and then suspend a mask from one end and then another mass that we move along and measure where the center of mass of the whole system moves, too. And we have this table here, and there are a few things we need to do. So the first thing is to draw free body diagram, and this will help illustrate what it is exactly that we're talking about. So we have our bar. Mass one is suspended 0.2 meters from the left hand side. The knife of it is here one and 1/2 meters from, uh, from the left hand side. Recall the normal force it exerts F p. The center of mass of the bar is over here at M G. On this is, um, 1.5 meters from the air. It is It's 1.5 meters from left hand side. This is 1.88 meters and then, uh, this is the second mass and the distance from the left hand side to the second mass is X and removed. We pick different masses for this and find out where we have to position it to balance the bar on the So here's a free body diagram. Next we want the plot or not. We want to use the equilibrium condition to find the our X. Here is a function of the second mass. So this way over here has a mess of two kilograms and we're using the access for the access were using the pivot. This acts 1.3 meters from the pivot. We have the mass of the bar and its weight on the acceleration of gravity to make this its weight. This is one point. It's eight meters right now 1.88 meters from the left hand side. Uh, so it's 0.38 meters from the pivot. This will have second masters attracted to from it and then X minus 1.5 meters because exes from the left hand side. But we're measuring cross. If it this all needs to equal zero. And so we just solve this for X as a function of em too. But it will be a function of all these things, but we the rest of the constant, um and we get one over M too. I actually this is, uh, make a little more room for this. X is going to be sorry. One over M too. Times two kilograms. Times 1.3 meters. Um, hunts 0.38 meters plus 1.5 meters. So now I want to plot. Uh, X is a function of one over em too. Um, this is an inverse kilograms. This is the meters, so point to interest kilograms. It's two meters and then it one inverse kilogram, 3.5 meters here would be our best fit line. Obviously, you should plot and your plot all of the values that were given. But this is just a sketch of what it and looking like on then. However you like to do these things, you, uh, make a best fit lion. And from there, uh, you will find that m the mass of the bar is 1.59 I feel like art scene. So now, um, from doing the best fit line, we also find that the Y intercepts. So this right here is at 1.5 meters. And so this makes sense because as one over m goes to zero or one over him, to be specific goes to zero. This is the same is saying that M two goes to infinity. And so, as as this becomes larger than all of the other masses involved, you're going to have to recall this is the location of the night EJ. And so you're going to have to, you know, basically put this giant mass as close as possible to the pivot as it gets larger and larger so that the, uh, the center of mass of the system will be over the pivot. So this makes sense because it just gets bigger. It has to get closer to the pivot so that it's the moment arm of the torque that it produces a smaller because the force is increasing.