Hey guys. So here we have an example of rotational equilibrium. Let's check it out. The bar below has a length of 4 meters and a mass of 10 kilograms. I'm going to draw here, \( l = 4 \) meters, \( m = 10 \) kilograms. Its mass is distributed uniformly. What that means is that the center of mass of the bar is in the middle. And what that means is that that's where \( mg \) acts. It says the bar is free to rotate about a fulcrum. This is the fulcrum right here, the support point, positioned 1 meter away from its left end. So this here is a distance of 1 meter, to the end. Okay? And you want to push straight down on the left edge, so this is you with a force of \( f \), to try to balance the bar. Because if you didn't push on the left, the bar would tip over to the right. Okay. So what magnitude of force should you apply on the bar? In other words, what is \( f \), the magnitude of \( f \). And for part b, how much force does the fulcrum apply on the bar? Well, if the bar is rested on top of the fulcrum, the fulcrum is going to push back with a force. That's our normal force. And we want to know what is the magnitude of normal. So let's start with question a here. How do we find the force? Well, we want to know how much force we need to balance the bar, which means there will be rotational equilibrium. We're holding the bar by pushing down this way. So we want to have rotational equilibrium, which means the sum of all torques will be 0. Okay. There are 2 torques that are going to act here. 1, first you have \( mg \) going this way, so there's a torque due to \( mg \). It is clockwise, so it's negative. And your force here is causing a torque this way because it's to the left of the center. So it's doing this to the bar. So the torque of \( f \) is going to be counterclockwise, positive. The normal force acts at the axis of rotation. Therefore, it produces no torque. The torque of normal would be \( \text{normal} \times r \times \sin(\theta) \) but \( r \) is 0 because the force acts on the axis of rotation, so the whole thing is 0. So really, what you have is these 2 guys. So I can do this. I can say, the torque of \( f \) plus negative torque of \( mg \) equals 0. And if I send this to the other side, I get the torque of \( f \) equals the torque of \( mg \). And this should make a ton of sense. The torques this basically just says that the torques are going in opposite directions or canceling each other out. So the next thing you do is you expand these two sides. So the torque of \( f \) is going to be \( f \times r_f \times \sin(\theta_f) \), and on the right side, I have \( mg \times r_{mg} \times \sin(\theta_{mg}) \). We're looking for \( f \). So let's plug in everything here. The distance, the \( r \) vector is the distance from the axis of rotation to the point where the force happens. So it's going to be this distance right here. This is our \( r_f \) which is 1. And the angle between \( f \) and \( r \) is 90 degrees. Okay. So this is \( r \), the \( r \) vector for \( f \), and you can draw \( f \) like this or you could have kept it this way. It doesn't matter. It's easy to see that it's 90 degrees. \( \sin(90) \) is 1. \( mg \), I have the mass is 10. \( g \) is 9.8. What is our vector for \( g \)? Now I didn't really draw this to scale here, but if the center of mass is in the middle, this means that this thing is 2 meters and the entire right side is 2 meters. But the fulcrum is 1 meter to the left. Therefore, this has to be another meter here. Okay? So 1 meter from, it's 2 meters from the left to the center of the mass. It's in the middle. But it's 1 meter from the left to the fulcrum, so you got another meter here. And this is the distance for \( r_{mg} \). Okay. So \( r_{mg} \) is this, which is 1 meter, and \( r_f \) is this, which is 1 meter as well. So I'm going to put 1 here and the sine will be \( \sin(\theta) \) will be 1 as well because you can see how \( mg \) makes an angle of 90 degrees with its \( r \) vector. Everything cancels, and we get that \( f = 98 \) newtons. \( f = 98 \) newtons. Cool. That's it. So if you push with the force of 98, these things will exactly cancel each other. You might have seen from the fact that the distances were the same. If the distances are the same, the forces have to be the same. So that should make sense. Maybe you saw that. And then for part b, very quickly, to find the normal force, we have to use the fact that the sum of all forces is 0 on the y-axis. Right? And if you look at all the forces, all the forces, there are 2 forces going down, which is \( f \) plus \( mg \). They're both going down. And then there's one force going up which is normal and they all equal to 0. So I can say that normal equals \( f + mg \). This should also make sense right away because this basically just says that all the forces going up equal the forces going down. \( f \) and \( mg \) are both 98. So when you add this thing up, you get 196 newtons. Okay? So this is how much force you would need to keep this thing balanced, and this is how much you get, as a result of doing that. That's how much normal force you have as a result. Okay? So that's it for this one. Let me know if you have any questions.