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Equilibrium in 2D - Ladder Problems

Patrick Ford
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Hey, guys. So in this video, we're going to start solving equilibrium questions that are two dimensional. And I'm gonna give an example of a classic problem in equilibrium. That is the latter problem. Let's check it out. So, so far, we've solved equilibrium problems that were essentially one dimensional, meaning all the forces acted in the same access. Either you had all the forces in the X axis or all the forces in the Y axis, most of them in the Y axis. And even if you had something at an angle like this, let's say you had something like this that's still essentially one dimensional because the angles were the same when we wrote the torque equation and they canceled. So in all the problems have solved so far, um, sign of data in the torque equation never really mattered because it either canceled or it was just the sign of 90 which is one. Now we're gonna finally solve some problems where that's not the case, right? We're gonna actually have to worry about the angle now. More advanced problems it says here will include problems in two dimensions in two axes and some of them in some of these cases, we may need to decompose the forces. Some of them will be decomposed. Um, remember, however, the torques aren't scale. Er's torques are scale er's, so we will never need to decompose them. Okay, so we're going to decompose forces in problems. But we don't have to decompose torques because torques are scaler. They may be, um, they may be positive or negative, but they are scaler is they don't have a vector direction. All right, so let's check out this problem here. We have a ladder of mass 10 kg, so M equals 10 and it's uniformly distributed. What that means is that the M g of the latter will happen will act right at the middle. So I'm gonna do this here and say, This is 2 m and this year is 2 m as well. All right. The latter has linked four. That's why the two and two and it rests against a vertical wall while making an angle of 53 with the horizontal shown. So this is 53 this, by the way, because this is 90. This is 53. This is 90 minus 53 which is 37 So let's just put that there. When you calculate a bunch of stuff, these are all the things you might see in a classic ladder question. I want to find the normal force at the bottom of the ladder on a bunch of other stuff. Before I read the list. Let's talk about what forces we have here. So you have a M G that pulls you down. Obviously, there's a normal here that pushes you up because the latter is resting against the vertical wall. There's also normal right here, and normal is always perpendicular to the surface. So normal is going to be like this. So there are two normal forces here. There's normal bottom and there is normal at the top. Now notice that we want. Obviously we want this ladder to be in complete equilibrium. So all forces cancer and all torques cancel. But if you look at that, what we have right now, there's a force going to the left. But there's no force is going to the right a t least. I haven't drawn them yet. That means that this this latter would not be an equilibrium. There has to be a force going to the right. And that force will be friction over here. And because we are, uh, we want the latter not to move. This is static friction. Okay, so there's enough forces that everything cancel. These are all the forces you're gonna have. I can write that the sum of all forces in the X X is equal zero. What this means is that the forces in the X cancel each other, so I'm gonna have normal top put it down here. I'm gonna have that normal top equals friction. Static. That's friction at the bottom. Okay, Some of our forces on the Y axis equals zero. So this means that end bottom equals M g. The to cancel. So end bottom equals M. G and the I can write more equations. Now, I can write torque equations, some of all torques. At any point, P equals zero. And there are three points here where I might want to write this. Um, there's the 30.1 here at the bottom two at the middle of the middle and three at the top. These air points where forces happen, Remember, You want to write your torque equations about the point where force happened. So you have fewer terms when you right out of torque equation. Okay, so let's see here. We want to find the normal force at the bottom of the ladder. This one is the easiest thing to find. That's why I put it here first Normal. The bottom is just mg. Here we have mg. So in bottom is M 10 n g. We're gonna use 10 as well, Fergie. And this is gonna be ah 100 Newtons. That's easy. This is just 100 Newton's now what about the normal force at the top of the ladder? So this whole thing he has done the normal force the top of the ladder to solve. For that, I would need to know static friction. I would need to know static friction. I don't have enough information just yet to find static. Friction s Oh, I don't have mu. I do have normal at the bottom. I don't have mu Andi. I wouldn't be able to use that anyway. So, um, normal at the top to find that I'm gonna have to write a torque equation because this year is not enough. So we're gonna write a torque equation And if you want normal at the top, if you want normal at the top, you want to write a torque equation with your access being somewhere else. The reason being you want your end top to show up on the equation. If you put the axis of rotation here, then there will be no torque due to normal at the top, and it won't be part of the equation. So we're going to carefully select our access to be one of the other two points. I'm gonna pick the bottom here, and I'm gonna pick that point because that point is actually the best of the two to pick. Remember, you want to do it? You wanna write the torque equation about a point with as many forces as possible so that you have as few terms as possible. There are two forces acting and 20.1 only one force that can point to 10.0.1 is definitely the best one to write a torque equation about it. So some of our talks at 0.1 equals zero. There are two torques there. I'm going to draw this. So here's 20.1. Here's the um, here's the ladder I have MG acting here and then I have n top acting here. Okay, Our vector from here to M g. Looks like this from the access to M G. Looks like this. And then from the access to end top looks like this. The bottom one has a length of two, and the top one has a length of four half and the total distance. The angle that I'm supposed to use is not the 53 right here, but instead I'm supposed to use the 37 right here. That's what we're gonna use. Okay, we're gonna use data equals 37 for this one here. I'm supposed to use this angle. And if this is 30 if this is 53 then this is 53 as well. Okay, so they have different ours, and different fate is you have to be very careful here. Alright. So hopefully that makes sense. Let's keep going here. I'm going to say that this torque is going this way. Torque of M g. And the torque of normal is going that way and I'll show you that just a second. Okay. So imagine we have your ladder. Looks like this m G is pushing down. So it's trying to cause a rotation this direction in this direction right here. And this is clockwise. So it's negative. Normal at the top is pushing this way. It's trying to cause it rotation this direction on this is positive. So there you go. They're canceling each other so I can write the torque of N Top equals torque of M G. Now I'm going to expand both equations torque and top is end top. It's our vector and sign of its data. MGs M g R vector sign of data. The our vector for N top is 44 m long. Um, for M G is 2 m long. The sign for end top right here between end topping. It's our vector is 53 and this one is 37. Okay, I have m g. I confined the signs so I'll be able to find in top. Okay. And top will be M, which is 10 g, which is 10 times to sign of 37 is 370.6 divided by four, divided by sine of 53 which is 530.8. And if you multiply all of this, you get 37.5 Newton's 37. Newtons. Okay, as soon as I get this, I will be able to know a soon as I know. End top. I know friction static as well. Friction static is the same thing is in top. So it's gonna be 37.5 just the same. So this was an bottom. This is in top, and this is friction. So if I scroll up over here 37.5, let me put this here. Normal top at the top. Sorry. Normal forces the top 37.5. Newton's frictional force at the bottom of the ladder 5 Newton's as well. Now I want to know the minimum. Quite efficient of static friction needed. So I'm looking for mu static men. Alright. Just ignore the men. All you really looking for is mu static plugging it into here. So we're gonna say friction. Static is 37.5. You know that the equation for friction static is mule static normal. Right. So this gets replaced with mu static. Normal equals 37.5. Now, do I use normal bottom or normal top? What do you think? Friction is over here, so you use normal bottom. Okay, so mu static will be 37.5. Divided by normal bottom. Normal bottom was 100. So Mu is going to be 0.375 Remember, you're supposed to be a number between zero and one. I got a number between zero and one, so I have higher confidence that this is correct. This is, in fact, the right answer. Um, so that is mu, which, by the way, this is part D. Here we found this is a and then somewhere here we found B and C together. So the coefficient here is 0.375 You Nicolas and I wanna know the total contact force at the bottom of the ladder. So what's the deal with that? Look at the bottom of the ladder. There are two forces and bottom and friction static. So one thing you might be asked is for the total force, which is simply the vector addition between these two. So it's gonna look something like this. All right, let me clean that up a little bit, and we're gonna draw that out here. So the total contact force It's just a combination of the two forces at the bottom. So part E total contact force you're gonna have in bottom, which is 100. You're gonna have friction static 37.5, and then the total contact force I'm gonna call this f bottom is what we're looking for. This is just basic Vector edition. We're gonna use the Pythagorean theorem to combine these two. Okay, So f bottom is just the square root of 100 square. Because this because of this number plus this square 37.5 square. Okay. And if you do this, you get 107 Newton's, which is the magnitude of this force I'm gonna just solve for one more thing. It wasn't asked here, but I could have asked also, What is the direction of the bottom force off the total force at the bottom. So theta bottom, remember, this is just factor. Stuff is the arc tangent of the Why Force divided by the X Force. The Y forces 100. The White Force is I'm sorry. The White Forces 100 the ex forces 37.5. And if you do the arc tangent of this. You get 69 degrees. Okay, Which means it's 69 degrees is right here. 69. Um, okay, that's it for this one. So these are all the classic things you would be asked on a ladder on a basic ladder question. Hopefully, this makes sense. And let me know if you have any questions. Let's get going.