Here's a popular statics problem. You have a board on two scales and you have a mass on the board and you ask yourself, what do the scales read as you move the mass around? How does the weight get distributed between the two scales? Let's start by drawing it. We've got our board here, and it's sitting on a scale here, and it's sitting on a scale here and then you've got the mass up here. Let's give the board mass little m and the weight, the big heavyweight mass, big M and this is the left scale and this is the right scale. Let's put it on an axis, on the x-axis with one end and the end, we'll just say we're the scale and the end of the board, it's all one place. I'll put it like right here. One end is at the origin, the other end is at L. Now we just treat it like a statics problem. Let's do a free body diagram for the board. We're really going to think about the board itself. So free body diagram of the board. Well, we know it has its own weight. We know that it has a little mg always right at its center of mass, which were uniform board is in the center. We know the way of scale works is the board pushes on the scale and the scale pushes on the board is an action-reaction force. The scale reads that push force, but it also pushes back with the same force. Whenever the scale reads is also being pushed up, and I'll call it a normal force from left scale and there's a normal force from the right scale. Those are the two were curious if they're going to change as we move the weight around. Then the weight, the mass sits somewhere big M and it applies a force to the board, big M_g, wherever it's sitting, so it can move around to different positions along the x-axis. That's the setup. Now we're going to do statics. We only really care about one direction for translation and that's the sum of the forces in the y meeting up and down are zero because it's not moving in the y. So for that, that equation is pretty straight forward. You have the normal force left is up plus the normal force right is up. Minus mg is down for the board, minus big M g is down for the weight and those all sum to zero and nothing changes as you move the mass around. Doesn't really matter, if this moves left and right, doesn't matter for the force. We also care about the torque. We know it's not rotating. We know the sum of the torques around any axis you want to pick is zero and we're going to do the torque around the origin. So around here, we're going to torque this thing around, right around one side. Let's see what are all the torques we could think about? Well, the normal force on the left is at the point of rotation. So that one is zero. I'll go and write it down, zero for that force. The normal force on the right does create a torque and it is positive because it's counterclockwise and it's being applied perpendicular to the length of the board, which is along this axis, so it's just in our times L. I think the order I wrote that was yeah, N_R times L. Then the 2m_g forces, let's see, the weight of the board is applied in the center. That's an L over 2 and it's negative because it's trying to turn it counterclockwise. Minus L over 2 little m_g. Then the one that gives us a variable here is the big mass. It's a big M_g is at a position x, it's a distance x from the point of rotation from the origin. Also trying to make it go counterclockwise. It's minus x, big M_g, and those are going to be equal to zero. It's not rotating. So there can be no network. The question was, what are the two scales read? It's all going to do is solve for the right normal force and the lift normal force. So actually the torque by itself, since we put the rotation axis here, NL doesn't show up. We only have the right normal force. We can actually get the first answer just by solving this. The whole thing is right here, if we take these two terms over to that side and divide through by NL, then we've got the right normal force and therefore the reading of the right scale and let's see, it must be one half little m_g because that L cancel that L, plus x over L Big M_g. Let's think about that one for a second. What does that mean? That means that the right scale always registers half the weight of the board and that makes sense because as you move the big mass around the board, nothing changes about the board. The boards gravitational force is always dead center between the two scales. Yeah, they're probably going to share in that part of the weight. But it's reading due to the big mass does vary. If x equals zero, if it's all the way over here, then you get no contribution to the right scale. It's all the way over here. Then L over L is 1 and the entire weight of the right scale includes big M_g. Let's solve then for in L. Let's see. The left one NL, yeah, it equals, let's see, the negative of N_R. I'll see some modest one half m_g minus x over L big M_g, plus little m_g plus big M_g. You can see what that turns into is this and this, gives you one half m_g again. That makes sense because we said the two scales probably share the mass or the weight of the board and then these become, let's see, 1 minus x over L big M_g. That's the left scale reading. That actually makes perfect sense. They share the weight of the board and for the big mass, now instead of x over L, we have one minus x over L. It's all the way over here, x over L is zero and it's just big M_g. So it's all the way over here on the left scale reads big M. If it's at the other side, L over L, 1 minus 1 is 0 and the right scale has all agreed. It implies that the share of the force, and it's just linear with position as you move it back and forth. Now we're going to see if that is true. We are going to now read the scales and move this mass around left and right. But actually this is boring, I mean, it's only eight kilograms, 16 pounds. I think we can do better than this. I think I should get out. Let me get on the board myself. I think that'll make it a little more exciting. I'm going to get on over on this side. Here we go and you can see the reading on the scale. Looks like it's about a 170 pounds. So the board's really heavy, its finest mahogany. But what I'm going to do now, so you have about a 170 here and there's probably some reading over there due to the board not quite zero. I'm not going to start moving over. Here we go. If I go partway over. Now that one's gone up a little bit and the left scales gone down a little bit because I'm farther away from it and now let me get right in the middle. I think if I get right in the middle, they're both going to say about 85 pounds clips. Now as I move closer to this side, it's going to get closer to 170 on the right and go less than less on the left. Let's go a little further over almost all the way. Now I'll go all the way over and one on this side, it all works. The moving load.
