Hey, guys. So in this example, I'm going to solve a literal ladder question, meaning a ladder question with no numbers and only variables. Now it's going to get a little hairy. So unless you know that your professor likes this kind of stuff or may like this kind of stuff, you might want to skip this one and save some brainpower. Alright? So if you do need it though, let's check it out. We have a ladder of mass m that is uniformly distributed, meaning the mg acts in the middle of the mass of the object, and it has length l. So the whole thing is l. Mg acts in the middle over here. I'm going to draw it. So that means that this piece here is l/2 and this over here is l/2 as well. Okay? It says it rests against the vertical wall while making an angle. I don't have the angle here. This is theta, cool, with the horizontal as shown. So this is theta right here. And we want to derive an expression for a bunch of stuff. Before I do that, let's draw some of the other forces. This is resting on the floor. So it's got a normal at the bottom. It's resting against the wall. So it has a normal this way, normal at the top, always perpendicular, normal force. There also has to be a friction at the bottom here to cancel out, normal at the top. Okay. First question, we want to write an expression for the minimum coefficient of friction, of static friction needed for the ladder to stay balanced at an angle of theta. So the idea is that given theta, given an angle theta, what is the minimum coefficient of friction that you need for that theta to work? You can imagine that if you have a ladder that is very steep, like let's say, this angle here is 80 degrees, it doesn't take a lot of friction to hold this ladder. Right? As opposed to if you have a ladder like this, it's trying really hard not to fall like this. It's trying really hard not to slide this way here which would cause it to fall. Okay. So as you change the angle, you can see how the angle and the mu are related. K. In fact, the smaller my angle, the more coefficient of friction I need to have. Cool?
Alright. So let's find an expression for μ static. We're going to start this problem like we start all static equilibrium problems, which is by writing F = M = 0. I'm sorry, sum of all forces equals 0 and sum of all torques equals 0. Sum of all forces in the x-axis equals 0. This means that the forces in the x cancel. The forces in the x here are on top and friction static. Sum of all forces in the y-axis equals 0, means that the force in the y-axis will cancel as well. This is nbottom equals mg. Okay. Before I write torque equations, let's see, maybe I don't even need it. I'm looking for μ static. I hope you see the μ static. I will find it in here somewhere. So I'm going to start at this equation by expanding friction into μ and normal. So friction static will be μ static normal, friction at the bottom, so this is going to be normal at the bottom, n equals n_top. We're looking for μ static. Now these literal questions, these questions have to derive an expression means that the answer has to be in terms of the given variables. I don't know what m is in terms of numbers, if it's a 3 or a 50. But I'm given m since it says mass m, that means it's a given, which means I can use that. It's allowed to be in the final answer, in the expression. Okay. So m, θ, ml, and θ, and then stuff like g and constants, and other universal constants could show up in the answer. So n_top_and_bottom_are_not_given, so they cannot show up in the answer. So I'm going to have to replace them with other stuff. Notice here that n_bottom is equal to mg. m and g are both, are both variables that could be in the final answer. So right away, you want to replace that. But you don't have n_top. So you're going to have to find an expression for n_top. So I'm going to write here the μ static equals n_top divided by n_bottom, and n_bottom is mg. So I have to get this and I'll be able to continue.
Okay. So let's go find an expression for n_top. To do this, I'm not going to be able to use, these two equations because I would use them, I'm stuck. I need more. So I'm going to have to write a torque equation. Sum of all torques at some point equals 0. I'm going to have to pick a good point to write this torque equation. There are three points here that I could use, 1, 2, or 3. I want to make sure I don't use normal top, point 3 over here as the axis of rotation. That's because when you write your torque equation, the point that you pick to be your axis will have no torques on it because a force acting on the axis of rotation doesn't produce a torque. So if you were to pick point 3, we're not going to do that. Normal top would be, would have no torque. Therefore, n_top is not going to show up in the equation. So you can't solve for it. You want n_top to show up in the equation. So you want to pick either points 1 or 2. Now out of these two choices, 1 is a better point to pick because there are 2 forces acting on it. And remember, you always want to pick the point with the with the most forces so that you cancel the most things and you have the fewest number of, terms in your equation. Okay. So we're going to pick 1. And then if you pick 1, let's draw 1 over here as the axis of rotation. And then here's our ladder. I have mg in the middle, and then I have n_top here. These are the only two forces that will produce a torque. So, this is my r vector here. This is my r vector for mg, and it's going to be r equals l over 2. It's the halfway point. And then this is going to be r equals the entire length. This angle here is the given angle theta. This angle here is the other angle. It's the complementary angle. I'm going to call this_theta_prime_and_I'm_going_to_say_that_theta_plus_theta_prime_equals_90. So you can think of theta prime as 90 degrees minus theta. And this angle over here is_theta. K. This angle here is the same angle here. So when I write the torque for mg, I'm going to use this angle. But when I write the torque for n_top, I'm going to use this angle. Angle. Okay. The torque due to mg causes the or tries to cause a rotation this way and the torque due to_normal_tries_to_cause a rotation this way. Okay.