15. Rotational Equilibrium
Equilibrium in 2D - Ladder Problems
Equilibrium in 2D - Ladder Problems
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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.
A ladder of mass 20 kg (uniformly distributed) and length 6 m rests against a vertical wall while making an angle of Θ = 60° with the horizontal, as shown. A 50 kg girl climbs 2 m up the ladder. Calculate the magnitude of the total contact force at the bottom of the ladder (Remember:You will need to first calculate the magnitude of N,BOT and f,S).
Minimum angle and friction for ladder
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Hey, guys. So in this example, I'm going to solve a literal ladder question. Meaning, Ah, lot of question with no numbers and Onley variables. Now, I'm just gonna get a little hairy. So unless you know that your professor likes this kind of stuff for May, like this kind of stuff, you might want to skip this one and save some brainpower. All right, 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. Um, and he has length L So the whole thing is L m G acts in the middle over here. I'm gonna draw it. Um, so that means that this piece here is all over to in this over here is a lover, too, as well. Okay, um, it says arrest against the vertical wall while making an angle. I don't have the angle here. This is data cool with the horizontal, as shown. So this is data 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'll is resting on the floor, so it's got a normal at the bottom. It's resting against the wall. So 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 a a new expression for the minimum, quite efficient of friction, of static friction needed for the latter to stay balanced at an angle off data. So the idea is that, given theta given an angle theta lips. What is the minimum coefficient of friction that you need for that data toe work? You can imagine that if you have a ladder, um, that is very steep. Like, let's say, this angle here is 80 degrees. It doesn't take a lot of friction. Toe. Hold this ladder right as opposed that if you have a ladder like this, it's trying really hard not to fall like this. It's trying out the heart, trying not to slide this way here, which would cause it to fall. So as you change the angle, you can see how the angle and the mu are related. Okay. In fact, the smaller my angle, the mawr coefficient of friction. I need tohave Cool. All right, so let's find an expression for, um you static. We're gonna start this problem like we start all static. Really been problems, which is by writing f equals equals zero. I'm sorry. Some of our forces equals zero, and some of all topics equals zero sum of all forces in the X X is equal zero. This means that the forces in the X cancel the forces in the X Here are and top and friction static some of all forces. And the Y axis equals zero means that the force of the Y axis will cancel as well. This is n bottom equal equals two mg. Okay, before I write torque equations, let's see, Maybe I don't even need it. I'm looking from you, Static. I hope you see them, you static. I will find it in here somewhere. So I'm gonna start at this equation by expanding friction into mu and normal. So friction static will be mule static, normal friction at the bottom. So this is gonna be normal at the bottom and equals and top We're looking from you 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 three year of 50 but I'm given em 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, um theta, uh mln theta And then stuff like G and Constance on other universal constants could show up in the answer. So end, top and in the bottom are not given, so they cannot show up in the answer. So I'm gonna have to replace them with other stuff. Notice here. That end bottom is equal to M. G. 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 any top, so you're gonna have to find an expression for in top someone right here. The muse static equals and top divided by in bottom and bottom is m jeep. So I have to get this and I'll be able to continue. Okay, so let's go find an expression for in 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 mawr. So I'm gonna have to write a torque equations, some of all torques. At some point equal zero, I'm gonna have to pick a good point to write this torque equation. There are three points here that I could use one too. Or three. Um, I wanna I wanna make sure I don't use normal top 30.3 over here as the acts of rotation. That's because when you write your torque equation, um, the point that you pick to be your access 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, uh, would have no torque. Therefore, and top, it's not going to show up in the equation, so you can't solve for it you want and top to show up in the equation. So you wanna pick either points one or two. Now, out of these two choices, one is a better point to pick because there's two forces acting on it and remember it. You always want to pick the point with 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 one, and then if you pick one, let's draw one over here as the axis of rotation. And then here's a ladder. I have m g in the middle and then I have n top here. These are the only two forces that will produce a torque. So, um, this is my our vector here. This is my our vector for M G. And it's gonna be r equals. Hello, virtue. It's the halfway point. And then this is going to be our equals the entire length. Um, this angle here is that given angle theta this angle here is the other angles, the complementary angle. I'm gonna call this data prime, and I'm gonna say the data plus theta prime equals 90. So you can think of data prime as 90 degrees minus data and this angle over here is theta. Okay? This angle here is the same angle here. So when I write the torque for M G, I'm gonna use this angle. But when I write the torque for M top, I'm gonna use this single, Okay? The torque due to M. G cause the tries to cause the rotation this way and the talk even normal tries to cause the rotation this way. Okay, so here's your ladder. M Just pushing this way and top is pushing that way the opposite to each other. Um, this one is negative. This one is positive. So I can write that torque and top equals torque. M g. Now it can expand these equations, torque and tops gonna be end top. Our sign of theta equals two mg or sign of data. They each have their ours and their status, the are for end top is all the way at the end. L for em, Jud, sell over to the angle for and top. Is tha tha? That's the angle that I want. It's before it's between. Remember the axis of rotation, the force. So that's data. And in this for m g, it's this angle, right? here, which is data prime. It's the other guy theater and fade up prime. Okay, notice that the else canceled. Um, and I'm looking for and tops, I'm gonna move everything the other way. I have and top equals M g over to sign of data prime divided by sine of data. Now we're almost done. I can use em and G s as part of my final answer. I can use data is part of my final answer. But I can't really use sign of data prime because that's not one of the given variables. I have to replace it. Now, there's two ways I can do this. I can rewrite sign of data prime as sign of 90 minus data. Or I can use a trick identity to simplify this. And you should know that sign of minus. You should know that sign of 90 minus data is simply the co sign of data. Okay, so one way that you can remember this very simply is that sign of equals co sign of 60 and sign of equals sign co sign of 30. So complementary angles swap signs and co signs. Okay, So whenever, if I have signed of 90 minus data. It's cold sign of data. If I had, if I had co sign of 90 minus data, this would have been equal to sign of data. Now, this is actually pretty good news because it means that, um this is going to look like this instead of sine theta. I don't want to use this version. Um, I actually don't even wanna use this version. I wanna use this version right here because it's the simplest and something interesting happens here. So this is gonna be signed. Sorry. Co sign of data over sign of data. And I might be thinking, Wait, isn't that tangent? Actually, it's not, um it's the inverse attainment, though, so that's good enough. So sign of data over a co sign of data, obviously is tangent. Feels like a trick review. Um, but co sign off data over sign of data is just one over tangent of data or your co tangents. But we're just gonna write this as tangents at the bottom. So this means you're gonna have mg divided by two tangents of theta. Okay, now we're done annoying, but we're done. MG and data, everything is as it's supposed to be. I can plug this mess in here, and then we can continue. Okay, So mule static will be normal top, which is this whole thing? Um, MG divided by two tangent of data. And then this whole thing is divided by M. G. So look what happens. The mgs cancel and you're left with your left with mu static being 1/2 tangent of data at the bottom. Okay. Um, so that's mu static. That's part a. All of this was part A to find new static. That's the relationship from you static. So notice that as you change data, you static. What changes? Well, alright, Parks. BNC will follow from here. So let me make a little bit of room here. We're gonna be a little titan space, so let's write small b we're looking for it Says find the minimum angle at which the latter can stay balanced for a given for a coefficient of static friction. Mule static. So, for a given me a static, what is the minimum theta that you can have? And again, remember, the two of them are related. So how much of a of an angle can amuse static handle. Okay, so basically, what we're gonna do here, this is gonna be pretty straightforward. All we're gonna do is use this equation right here. You can't 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? Talk equations and everything, Um, to get to a point where we can get an expression for data. Once we have this, all we have to do software data within that equation. So mu static equals 1/2 tangent of data, and I want to solve for data. So I'm gonna move tangent of data up and move mu static down, and it's gonna look like this. Um, Tangent of data equals one over to new static. And we're looking for Thodin. So it's the arc tangent of one over to 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 fate a minimum first, you would have gone through a similar process is 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 latter can stay balance for any coefficient of friction. So I'm gonna draw some stuff here just to illustrate a point, but I'm gonna delete it so you don't have to draw this. Um, imagine that at some point, right, the bigger the smaller the angle becomes, this is pretty easy to hold at 80 degrees. Um, 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 the ladder at like, 10 degrees probably doesn't hold it probably just falls, no matter how much friction you have. So there's a there's a minimum angle, um, the minimum angle of like how low? You can go with data until friction just can't handle it anymore. Okay, so that's why I wanna find now again. As you reduce your data, you mean mawr and mawr friction. To find the minimum data, 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 to create friction, friction to be the maximum amounts and the maximum amount of that accord friction friction can have is one. So that's what we're gonna do. We're gonna set this to be one, and we're gonna find if mu is one, which is the max possible, then what is Thatta? 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 arc tangent of 1/2 times one, and this is the arc tangent 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 data is less than 26.6 degrees, because it will fall, and depending on what the surface is between depending the coefficient friction between the object and the floor. Um, this angle is gonna have to be even greater than 26.6. All right, so that's it for this one. Hopefully made sense. Let me know if you guys have any questions. Let's keep going.
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