 In this video, I'm going to try to solve for the force in a left support of a beam that I have here at some construction site. The beam has two supports, one on the left, one on the right, and I'm going to try to figure out the force on the left side. Now, I'm going to start as usual, very innocently, and then I'm going to just do the free body diagram on the left side. So if I do the free body diagram, I would have, if I simplify it to a point, I have a force down, which is gravity. Then I have a force from the left, and I have a force from the right, which are up. Then I would choose my coordinate system, so x, y, and z direction, and then I would do my sum of all forces must be zero. This is clearly static equilibrium. We don't want the beam that is currently addressed to not be addressed anymore. So I would do in y direction sum of all forces must be zero. Therefore, I have my left force, my right force plus fg. If I attribute the signs, my left force is up, my right force is up, and gravity minus mg is down. Now, the problem I'm having with this is I can calculate mg 300 kilograms times 9.8 ish, but I don't have the right force and I don't have the left support force. Therefore, I can actually not solve this. I need an additional equation. Now, where do we get additional equations from? Well, if this beam is addressed and supposed to remain addressed, not only do the sum of all forces in x and y direction have to be zero, but also the sum of all torque around any point on the beam have to be zero. So I can say, okay, in addition, the sum of all torque has to be zero. If I want to calculate the torque of the various forces, the free body diagram, as I have it on the left, is actually quite useless. So in problems like this, you're not going to do a free body diagram like this anymore because you actually need the dimensions of the object to know where the forces are acting. So instead, I'm going to draw my free body diagram with the entire beam as it is, and then I draw my forces where they're acting. The force of gravity acts in the center of gravity, which is, let's assume it's uniform distributed mass in the middle, then I have my left force here and I have my right force here. Now, as I said before, if the beam is addressed and supposed to remain addressed, then the torque around any point on the beam must be zero. So for the pivot point, I can choose this point, this point, this point. I can choose any point. But as you're interested in the left force and not in the right force, I'm going to make my life easier. I'm going to put the pivot right here. And by doing so, the torque caused by the right force will be zero. Therefore, my math will get much simpler. So what forces are causing torque here? I can have a torque from my left force. I can have a torque of gravity. And I could have this torque from the right support, which we know will be zero at its acting on the pivot. Now, when we go for directions, we have to assign directions. Usually, everything that is counterclockwise is considered positive. Everything that's clockwise is going to be considered negative. So let's have a look. My left force, if my beam would be hinged here, and this was allowed to push it up, there's no other force, will turn it in clockwise direction. Therefore, I'm getting minus the torque from the left. My gravity, if it was the only force acting, would push it downwards like this, so it would turn in counterclockwise direction. Therefore, plus the torque caused by gravity. And here we say the torque from the right one, the right force would be zero, because it has no distance from the pivot point. That's right on the pivot. Now, I just have to plug in the numbers. I know my left support is acting at 2.8 meters distance, so I have minus 2.8 times the force on the left, plus my gravity is acting halfway in the middle of the beam, so 1.4 times Fg must be zero. Then I can solve for my left force, is 1.4 times M times G divided by 2.8, which gives me 1,470 newtons. That's how you solve a problem where you have not enough equations on the x and y directions, but you need an additional torque equation to solve it. Now, if I wanted the right force, of course you can just plug in this value here, and that would get my right force.