 This is a first and very simple problem where we need the rotational and the translation equilibrium to solve it. What we have is a horizontal beam where on the right side we have a mass that is hanging from it, and then we have two supports supporting the beam, and we're interested in calculating the forces of each support. We cannot solve this with translation equilibrium alone because we have two unknown forces, each support in the y direction, and we cannot solve it two unknowns, but only one equation from the y direction sum of all forces. As with pure translational equilibrium problems, we're going to start by drawing the free body diagram. So I'm going to draw the free body diagram of my block with the mass attached. So here comes my free body diagram. So I have a block, what's the real beam, with the mass attached. I'm gonna look at my object here, and I'm cutting it free. So I have something here. I have something here. So each support will give me a support force. So I have a force here going up, and I have a force here going up. Let's call this one force two. This one here force one. Then we're on the Earth, so we have gravity. We have gravity on the beam, which acts in the middle. So FG beam, and then we have gravity on the blocks. So FG or as we already have to, let's call this FGM, hanging mass. As I did before, I'm gonna select my coordinate system. Here I'm gonna go with the regular one. X, Y. What is new, I'm gonna need to select the pivot point. So if the whole thing would be tilting, it probably would be tilting around the right support. So I'm gonna put my pivot point here. This is my pivot point. Now I can do my sum of all forces is zero in X, Y, and the sum of all torque around my set axis is also zero. I'm gonna start with my torque because I hope that I can figure out as much as possible already from torque. So in set direction, I have the sum of all torque must be zero. So who is causing me a torque around the pivot? I have a torque caused by force one. I have a torque caused by the gravity of the beam, and I have a torque caused by the gravity of the mass. F2 cannot cause any torque because it's directly connected to the pivot point. So the next line, I'm gonna look at my directions. Remember if I have something that is clockwise, I'm gonna take it as negative. If something is counter-clockwise, I'm gonna take it as positive. So in my case, my force one will cause me a torque that is clockwise, so negative. So negative torque one. My gravity would tendentially turn counter-clockwise, so plus torque of the gravity of the beam and my gravity of the mass, if anything, would turn it clockwise, so negative torque gravity of the mass is zero. And I'm solving this for torque one. I'm simply bringing torque one on the other side, so my torque one is equal to the torque of gravity by the block minus the torque of gravity by the mass. And I'm plugging in my numbers that I have. Given from my dimensions here, I know that my force one is at the distance of one meter and my force of gravity here is at the distance of 0.5 meters. My other gravity, the shortest distance to the prolonged line of where my force here is acting, is 0.3 meters. That means I have my force one, force one, force one, times one meter is mgb, so five kilograms times 10 meters per second square, times 0.5 meters minus my torque caused by the hanging mass, so another five kilograms times 10 times 0.3 meters. This is 25 minus 15. Therefore my force of the left support is it's gonna need a bit more space, so I'm gonna make this a bit smaller here. Place it over here and this guy here can go here. Because what I'm doing next is I'm looking at my total of all forces in y direction. So I'm using my y equation. Some of all forces in y must be zero. Forces in y are my force one plus my force two plus my fg of the beam plus my fg of the block, let's be zero. I have positive my force one, positive my force two minus five times ten and minus another five times ten is zero, so I have force two is two times five times ten plus my force one, minus my force one, so minus ten which gives me and the problem is solved.