 Here is yet another rotational and translational equilibrium problem. What we have is a ladder that is leaning against the wall and we're going to assume that there is no friction from the wall on the ladder but there is friction from the floor and we are interested in calculating all forces that are acting on the ladder. As usual, we will start by drawing the free body diagram of the ladder. So here is my ladder and I'm going to free my ladder, so I'm cutting it from the environment. So I will have something on the wall here and something on the floor here. So from the wall we get a normal force from the wall that should go straight away from the surface, so to the left. Let's try this a bit better, so we have force from the wall. Then from the floor we have a vertical force which is actually just a normal force, so I'm going to call it vertical force and then I have a friction that will keep my ladder from sliding away. I'm calling this one the horizontal force. Then we're still on the planet Earth, so I have gravity coming directly in the middle. So this is my free body diagram. The coordinate system is already given, so the coordinate system is X and Y as such. For the pivot point, my unknowns here are my force from the wall, my vertical, my horizontal force. So the pivot point, I can actually choose whatever point I want, but if I put it where I have the most unknowns I hopefully get the most easiest equation. So I'm going to choose my pivot down here. If you think about it, if the wall would not be here my ladder would actually rotate and fall around this point. Most likely I will need to calculate torques, so let's look at my angles. So if this one here is 60 degrees that means this one here must be 30 degrees and that means this one here is also 60 degrees. As I'm going to be needing much more space, I'm making my free body diagram a bit smaller. Just leave it up here. Now I'm going to start with my set equation. So in set direction, set direction, my sum of all torque will be zero. Now which of the forces around my or which of the forces can create some torque around the pivot point? I have only two, gravity and the force from the wall. The vertical horizontal forces cannot cause any torque because they are acting on the pivot point directly. So I have the torque from the wall and I have the torque from gravity must be zero. Then once more, if I have something that turns it counterclockwise and take s plus and clockwise is minus. So my force from the wall is pushing it counterclockwise, so plus the torque from the wall and my gravity is going clockwise, so torque from gravity is negative. Zero, so I have the torque from the wall must be equal to the torque from the gravity and the torque from the wall will be my force from the wall times my length of the ladder, so times two meters, times sine of the angle, so times sine of 60 is equal to my mass, so five kilograms times ten meters per second square, gravity, times the lever, so the mass, the center of mass in the uniform ladder should be one meter from my pivot and the angle which I have up here should be 30 degrees. So if I'm putting this here, I have times sine of 30. Now I can solve this for my normal force. I'm sorry, my force from the wall, which is a normal force. Which should be equal to around 14.4 Newton's. I'm gonna try to make this smaller, so I get more space, because next I'm gonna do my y equation in y direction. I have the sum of all forces in y must be zero. What are my forces in y? My forces in y are my vertical force and my gravitational force, and that's it. So I have plus my vertical force, minus mg is zero, therefore my vertical force is mg, which is equal to five times ten is 50 Newtons. This one also has a final answer. So all that's left is my x equation. Some of all forces in x must be equal to zero. In x I have my horizontal force, which is actually friction force. I have the force from the wall and nothing else. My horizontal force goes to the right, so plus my wall force is negative, so force from the wall negative is zero. Therefore, my horizontal force is equal to the force from the wall, which is 14.4 Newton, and that's it.