Chapter nine problem 11. 75 kilogram adult sits at one end of a nine meter long board. His 25 kilogram child sits on the other end. Aye, Where should a pivot be placed so that the Board of Balanced ignoring the board's mass and B where should replaced if it has a mess of 15 kilograms and uniforms. So here is our free body diagram. And if we look, we can see that No, we have the 75 kilogram adult sitting at the left end with called Big Temps every five K G's your little and 25 G's your arms. All right. And then we also have the mass of the board. Right? But this is all this is changing. Yes, I'm gonna put that somewhere else just for you, m b. So I could be zero or it could be 15. This is basically the difference between part and part B, right? So let's talk about the free body diagram real quick. So the full length here, this is the full length of the of the longboard, a nine meter long board. So l is going to be 9.0 meters and the distance between. Well, we don't really know where the pivot point is quite yet, right? Because that's the question. Where should they be placed? If it's balanced, that's where we have this thing here that I put the variable pivot point. So because we don't know what it is, we kind of have to guess a little bit first and then with our guessing and with our educated guess this weekend, find no where exactly this place is depending on the masses. And so this is a very important there's a very important feature. So I want you guys to note this is the variable pivot point this X. It depends on the mass. Well, it depends on everybody, all the every mass in the problem, and you'll see that coming scene. But we have a variable, Matt, we have a variable pivot point. And what that means is, we could still we can talk about torques around this pivot point and then we can use these descriptions of the distance like X and then l minus exercise will never left from ex happy Subject X. And so once we do that, since we know all the masses and every since we know the masses on one of the length we should be able to find X with all of our givens. And we can't forget the pivot point itself. So I'm just I just drew this right under where my perfect point was. He really could be anywhere. Um, but this is going to be the forced due to the pivot point. And then you also have on board itself. So M b G's the weight of the board pulling down. But two MGs are the kid in the adults and F. P s, the pivot force and the distance between the pivot and the huh And the center of gravity is going to be l half minus x right, because the center of gravity is going to be in the center of the board. Halfway through the board, it's 4.5 meters, but then it's also going to be subtracted by well, wherever excess will defined his distance. Right? So we still don't know what. Actually, that's the best we're searching for. So set up the pivot with counterclockwise is positive so that there's no rotation. In other words, this means that or I eat. It means that the sum of the talks of zero and they were allowed to do Cem. Very nice stuff. So we have we set up the pivot right there. Right? Who sent this pivot right here than F P actually goes to zero. Right? What does an f p itself doesn't go to here, but the torque caused by any the torque caused by the pivot itself is not existent. Because to have a talk, he must have a distance and a force acting at that distance. Which the only thing we have in the first case in part a is just these two, right? You attempted to include this one, but in the first in a part, a part be part pay M B zero. So we're actually going to ignore it for now. Come back to later. But we can't ignore it from We just have big mg, little mg. And that's going to give us big mg times X. So this is exactly where we had big mg and ex distance away. Right from the pivot from the pivot. Too big Angie from the period. Also too little mgs l minus hex. So just gonna be a little empty times l'm ice picks. And then when you do all the outer before this, you have rights and equal zero, of course. And then you just if we put all the exits on one side and put all the put everything else on the other side and you should get about 2.3 meters 2.25 to be exact for part B, we have to just include the mass of the part B. If you include the mass, and that just means one more factor, right? So if we go back to our be arguing here from the pivot points to the ER to the center of mass of the board, it's just gonna be l half minus x. And so when we include this one, she's gonna be NBG multiplied by el half minus six. Which is exactly what I have here. So same process is a little owns a little bit longer just because it has more terms after you distribute it. But I promise it's not that hard. Um, you just bye. Distribute all your terms and then separate X only on this, like just like 1/2 here. X x things only on the left and then everything else on the right. Then you can factor out ex divide and you should end up with the next pages result which we do here you can pull out. You can also factor g out of both the top in the bottom of the nominator denominator. And if you just plug in those masses, big M Beghin was Ah, what was that? 75? Little in was 25 M B is now 15 kilograms. Way have to include that in there. And this should basically change the position to about 2.5, so it just moves 0.2 after.