15. Rotational Equilibrium
Equilibrium in 2D - Ladder Problems
Struggling with Physics?
Join thousands of students who trust us to help them ace their exams!Watch the first video
Video duration:
16mRelated Videos
Related Practice
Skip to main content
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 gonna 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 wanna 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 over 2 and this over here is l over 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 he 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 not the hard not trying 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 mu static. We're going to start this problem like we start all static equilibrium problems, which is by writing f equals m equals 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 in 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 equal equals to mg. Okay. Before I write torque equations, let's see, maybe I don't even need it. I'm looking for new static. I hope you see the new static. I will find it in here somewhere. So I'm going to start at this equation by expanding friction into mu and normal. So friction static will be mu static normal, friction at the bottom, so this is going to be normal at the bottom, n equals n top. We're looking for Mu 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 theta, ml and theta, and then stuff like g and constants, and other universal constants could show up in the answer. So ntopandbottomarenotgiven, so they cannot show up in the answer. So I'm going to have to replace them with other stuff. Notice here that nbottom 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 mu static equals n top divided by n bottom, and 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 equal 0. I'm going to have to pick a good point to write this torque equation. There are 3 points here that I could use, 1, 2 or 3. I want to I want to make sure I don't use normal top, point 3 over here as the axial 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 0.3, we're not gonna do that. Normal top would be, would have no torque. Therefore, n top is not gonna show up in the equation. So you can't solve for it. You want n top to show up in the equation. So you wanna pick either points 1 or 2. Now out of these two choices, 1 is a better point to pick because there's 2 forces acting on it. And remember, you always wanna pick the point with the with the most forces so that you cancel the most things and you have the fewest number of, terms on your equation. Okay. So we're gonna 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 gonna use this angle. But when I write the torque for n top, I'm gonna use this angle. Angle. Okay. The torque due to m g causes the or tries to cause a rotation this way and the torque due to normal tries to cause a rotation this way. Okay. So here's your ladder, m g is pushing this way and top is pushing that way. They're opposite to each other. This one is negative. This one is positive. So I can write that torque ntop equals torque mg. Now I can expand these equations. Torque ntop is going to be ntop r sine of theta equals to mgr sine of theta. They each have their r's and their thetas. The r for n top is all the way at the end l. For mg, it's l over 2. The angle, for n top is theta. That's the angle that I want. It's before it's between, remember, the axis of rotation and the force. So it's theta and then this, for mg, it's this angle right here which is theta prime. It's the other guy. Theta and theta prime. Okay. Notice that the l's cancel, and I'm looking for n top, so I'm going to move everything the other way. I have n top equals mg over 2 sine of theta prime divided by sine of theta. Now we're almost done. I can use m and g as as part of my final answer. I can use theta as part of my final answer, but I can't really use sine of theta prime because that's not one of the given variables. I have to replace it. Now there's 2 ways I can do this. I can rewrite sine of theta prime as sine of 90 minus theta, or I can use a trig identity to simplify this. And you should know that sine of 90 minus, you should know that sine of 90 minus theta is simply the cosine of theta. Okay. So one way that you can remember this very simply is that sine of 30 equals cosine of 60 and sine of 60 equals sine, cosine of 30. So complementary angles swap sines and cosines. Okay. So whenever if I have sine of 90 minus theta, it's cosine of theta. If I had if I had cosine of 90 minus theta, this would have been equal to sine of theta. Now this is actually pretty good news because it means that, this is going to look like this. Instead of sine of theta, I don't want to use this version. I actually don't even want to use this version. I want to use this version right here because it's the simplest and something interesting happens here. So this is gonna be sine I'm sorry, cosine of theta over sine of theta. Theta. And now you might be thinking, wait, isn't that tangent? Actually, it's not. It's the inverse of tangent though, so that's good enough. So sine of theta over cosine of theta obviously is tangent. Feels like a trig review, but cosine of theta over sine of theta is just 1 over tangent of theta, or your cotangents. But we're just going to write this as tangents at the bottom. So this means you're going to have mg divided by 2 tangents of theta. Okay. Now we're done. Annoying, but we're done. M g and theta. Everything is as it's supposed to be. I can plug this mess in here and then we can continue. Okay? So mu static will be normal top, which is this whole thing, mg divided by 2 tangent of theta, and then this whole thing is divided by mg. So look what happens. The mgs cancel, and you're left with, you're left with mu static being 1 over 2 tangent of data at the bottom. Okay. So that's new static. That's Part A. All of this was Part A. To find mu static, that's the relationship for mu static. So notice that as you change theta, mu static would change as well. Parts B and C will follow from here. So let me make a little bit of room here. We're going to be a little tight in space, so let's write small. B, we're looking for, it says, find the minimum angle at which the ladder can stay balanced for a given, for a coefficient of static friction, Mu static. So for a given Mu static, what is the minimum theta that you can have? And again, remember, the 2 of them are related. So how much of a of an angle can a mew static handle? Okay. So basically what we're going to do here, this is going to be pretty straightforward. All we're going to do is use this equation right here. Oops, you can see it. We're gonna use this equation right here and solve for theta. Okay? Now if we didn't have this equation already, we would go through a very similar process to what we did in the first part and derive that equation, right, torque equations and everything, to get to a point where we can get an expression for theta. Once we have this, all we have to do is solve for theta within that equation. So mu static equals 1 over 2 tangent of theta, and I want to solve for theta. So I'm going to move tangent of theta up and move mu static down, and it's going to look like this, tangent of theta equals one over 2 mu static, and we're looking for theta. So it's the arc tangent of 1 over 2 mu static. Okay. That's the final answer for part b. So once you have one, you can get the other. Again, if you were asked to find theta minimum first, you would have gone through a similar process as what we did here. You would find this and then you would flip it around to find this. Okay. Now let's do part c. Let's do part c. So part C says, find the minimal angle at which the ladder can stay balanced for any coefficient of friction. So I'm going to draw some stuff here just to illustrate a point, but I'm going to delete it so you don't have to draw this. Imagine that at some point, right, the bigger the the smaller the angle becomes, this is pretty easy to hold at 80 degrees. Let's say at 50, it might be a little harder. At some point, it really doesn't matter, right? Imagine you try to hold a ladder at like 10 degrees. It probably doesn't hold. It probably just falls no matter how much friction you have. So there's a there's a minimum angle, the minimum angle of like how low you can go with theta until friction just can't handle it, anymore. Okay. So that's what we want to find. Now again, as you reduce your Theta, you need more and more friction. To find the minimum Theta, you need to figure out what angle you get for the most possible friction. So the most friction can handle is when you have a coefficient of friction that is maximum. Okay. So you want your coefficient of friction to be the maximum amount and the maximum amount of that coefficient of friction can have is 1. So that's what we're going to do. We're going to set this to be 1. And we're going to find if mu is 1, which is the max possible, then what is theta? And to do this, we can simply plug it into this equation right here. So you can see how b and c follow from a. Alright. So theta is the arctangent of 1 over 2 times 1, and this is the arctangent of half. And if you plug this into the calculator, this is 26.6 degrees. Okay? So you can go try this at home if you want. You cannot have a ladder hold against the wall or any object for that matter hold against the wall like this if your Theta is less than 26.6 degrees because it will fall. And depending on what the surface is, between, depending on the coefficient friction between the objects and the floor, this angle is going to have to be even greater than 26.6. Alright? So that's it for this one. Hopefully it made sense. Let me know if you guys have any questions. Let's keep going.
13:43
