Anderson Video - Tension to Keep a Wire Perfectly Horizontal

Professor Anderson
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Let me, let me give you a problem to consider and then we're going to talk about it. Let's say you go to the beach and you come home from the beach and you want to hang your wet suit up to dry and you're going to hang it on a wire that you have in your dorm room. Ok, so let's see if we can just draw a very simple picture of what this looks like, right? Let's draw a hanger and now we have a wet suit hanging on it, and then we're going to hang it from a wire, so it looks something like that. You've got ridiculously skinny arms on yours wet suit for whatever reason. We know that there's gotta be some tension T in the wire to hold up the wet suit. Let's say we want to do the following experiment. We want to calculate what tension T to keep the wire perfectly horizontal; in other words, no sag in the wire, ok? If we want that wire perfectly horizontal, what would that look like? It would look like this. Here's our hanger, here's our wet suit, the wire would be perfectly horizontal, like that. And there would be tension T in that wire. What tension T would it take to keep that wire perfectly horizontal? Alright let's think about it. Let's say the wetsuit has mass M, what's a guess, let's take a guess. This is a good idea in physics just guess at some possible solutions and then when we're all done, we'll see if any of our guesses were right and this allows you to get in touch with your intuition. So what's a good guess? What do you guys think? Somebody shout out a guess. Ok, what'd you say? The mass is M, so how about MG? Right, why not? We know as the mass gets bigger, that tension T must get bigger, so maybe MG. More than MG, ok, how much more than MG? Twice, ok that's a good guess. There's also two wires though, so maybe the tension in each wire could be less, so how about MG over two? How about zero; no tension in the wire. Any other guesses? >> Say that again, but put your mic on so that people at home can hear you. And what's your name? Keaton. Keaton with allergies, what do you have to say about this one? [ Inaudible ] >> It can't be perfectly horizontal, so what would the tension be if it can't be perfectly horizontal? >> Really, really, big. >> Really, really big; what's really, really big? [ Inaudible ] >> Ok, well that would be getting us into the wrong units, so what if we just said, I don't know thousand MG? That's really big right? Ok, here's some decent guesses; any of those that we can get rid of? Zero, that doesn't make any sense, right? No way can it be zero. Anything else we can get rid of? MG and MG over two; I'm not really sure we want to get rid of those yet because we kind of like MG's, so why don't we just hang on to the rest of them for now and see how it goes. Ok, how do we do this problem? The way we do this problem is first off we simplify it quite a bit, right? Let's take that picture right here and let's simplify it. So wire, Mass M, wire, that's basically this picture right here; let's approach it from the point of view, we're going to call that Theta and then we'll look at the limits later on. So that's Theta, this is Theta, this is also Theta and this is also Theta and then we have tension T in the wires. So we solve that this way generally and then we can look at the limits as Theta goes to zero. Alright there's our picture; we got to do the free body diagrams. So let's draw the box as a dot, what are the forces acting on the box? The first one we love, gravity; and then we have tension T going up like that and then we have tension T going up like that. It looks pretty good except it's not all in sort of XY beautiful Cartesian coordinate space and so let's redraw this as something that has components in it. So let's draw it like this, MG is still down, but now there is some component of tension to the right. There is some component of tension to the left; there is some component of tension up; and there is some component of tension up. So this is an equivalent picture if we can figure out what these different components are. So this one right here is going to be either T sine Theta or T cosine Theta, which one should it be? Cosine, why because this angle right here was Theta. So T cosine Theta to the right; T cosine Theta to the left and that means these vertical ones have to be sine; T sine Theta, T sine Theta. Perfect, that looks like a great free body diagram and now we can go to sum of the forces. In the X direction it's not really going to tell us very much, right? Because we have T cosine Theta to the right, but we have T cosine Theta to the left and all of that is going to equal zero because the things at rest and that's good; zero does equal zero. The Y equation hopefully will tell us a little bit more. What do I have, I have two, T sine Theta going up and I have one MG going down. And all that is equal to the mass times the acceleration the Y direction, but this is hanging there. That's zero and so now we get a solution for T. T is MG over two sine Theta. Let's put a box around that. So we got an answer, let's make sure that it makes sense. Do the units make sense? Tension is a force, MG is a force, sine Theta is unitless. So the units do work out. Let's look at the limits. And to start with let's say the thing is hanging straight down, so if it's hanging straight down, what is Theta? What would Theta be if it's hanging straight down? Theta would be ninety degrees, what's the sine of ninety degrees? One, so what would the tension be? It would be MG over two and that makes sense right? If the two wires are going to hold the weight, each wire has to hold half of it. No problem, now the original question was what if it's horizontal? If we have a horizontal wire what is Theta? What's Theta in this case? Ok; it's when this wire goes to the horizontal, which means Theta goes to zero. You could plug in 180 and you get the same answer, but Theta equals zero is what we're looking for; what's the sine of zero? Sine of zero is zero. If sine of ninety is one then sine of zero had better be zero. So what is the tension? It is MG over two sine Theta, which becomes MG over zero, what's MG over zero? It's infinite, right, anything over zero means that you are approaching infinity. So it can't be done. Our guess of it being really, really big was right, but it's not a thousand MG, it's not ten thousand MG, its infinity. Can't be done and this is kind of weird right? I mean it seems like I could just pull on this wire really hard and get it to be horizontal and I could put just a little tiny mass on it and it would stay horizontal, but that's not what happens. No matter what the wire is made of, no matter how strong you pull on it, what the tension is, you can't get it perfectly horizontal. And this is good to remember when you're an engineer, right, when you are building a bridge and you think oh, I'm going to string this cable and it's going to stay perfectly horizontal. No, as soon as you put some mass on it, it starts to sag and in fact you don't even have to put mass on it because the cable itself has mass in it and so the cable will start to sag and this is part of the point for you engineers. Everything bends; everything sags when you apply forces to it, things move. So when you're engineering a building or an airplane or a suspension bridge, keep this in mind; everything is going to sag. You can't keep it perfectly horizontal. It's kind of weird, right? That's the way it is.
Let me, let me give you a problem to consider and then we're going to talk about it. Let's say you go to the beach and you come home from the beach and you want to hang your wet suit up to dry and you're going to hang it on a wire that you have in your dorm room. Ok, so let's see if we can just draw a very simple picture of what this looks like, right? Let's draw a hanger and now we have a wet suit hanging on it, and then we're going to hang it from a wire, so it looks something like that. You've got ridiculously skinny arms on yours wet suit for whatever reason. We know that there's gotta be some tension T in the wire to hold up the wet suit. Let's say we want to do the following experiment. We want to calculate what tension T to keep the wire perfectly horizontal; in other words, no sag in the wire, ok? If we want that wire perfectly horizontal, what would that look like? It would look like this. Here's our hanger, here's our wet suit, the wire would be perfectly horizontal, like that. And there would be tension T in that wire. What tension T would it take to keep that wire perfectly horizontal? Alright let's think about it. Let's say the wetsuit has mass M, what's a guess, let's take a guess. This is a good idea in physics just guess at some possible solutions and then when we're all done, we'll see if any of our guesses were right and this allows you to get in touch with your intuition. So what's a good guess? What do you guys think? Somebody shout out a guess. Ok, what'd you say? The mass is M, so how about MG? Right, why not? We know as the mass gets bigger, that tension T must get bigger, so maybe MG. More than MG, ok, how much more than MG? Twice, ok that's a good guess. There's also two wires though, so maybe the tension in each wire could be less, so how about MG over two? How about zero; no tension in the wire. Any other guesses? >> Say that again, but put your mic on so that people at home can hear you. And what's your name? Keaton. Keaton with allergies, what do you have to say about this one? [ Inaudible ] >> It can't be perfectly horizontal, so what would the tension be if it can't be perfectly horizontal? >> Really, really, big. >> Really, really big; what's really, really big? [ Inaudible ] >> Ok, well that would be getting us into the wrong units, so what if we just said, I don't know thousand MG? That's really big right? Ok, here's some decent guesses; any of those that we can get rid of? Zero, that doesn't make any sense, right? No way can it be zero. Anything else we can get rid of? MG and MG over two; I'm not really sure we want to get rid of those yet because we kind of like MG's, so why don't we just hang on to the rest of them for now and see how it goes. Ok, how do we do this problem? The way we do this problem is first off we simplify it quite a bit, right? Let's take that picture right here and let's simplify it. So wire, Mass M, wire, that's basically this picture right here; let's approach it from the point of view, we're going to call that Theta and then we'll look at the limits later on. So that's Theta, this is Theta, this is also Theta and this is also Theta and then we have tension T in the wires. So we solve that this way generally and then we can look at the limits as Theta goes to zero. Alright there's our picture; we got to do the free body diagrams. So let's draw the box as a dot, what are the forces acting on the box? The first one we love, gravity; and then we have tension T going up like that and then we have tension T going up like that. It looks pretty good except it's not all in sort of XY beautiful Cartesian coordinate space and so let's redraw this as something that has components in it. So let's draw it like this, MG is still down, but now there is some component of tension to the right. There is some component of tension to the left; there is some component of tension up; and there is some component of tension up. So this is an equivalent picture if we can figure out what these different components are. So this one right here is going to be either T sine Theta or T cosine Theta, which one should it be? Cosine, why because this angle right here was Theta. So T cosine Theta to the right; T cosine Theta to the left and that means these vertical ones have to be sine; T sine Theta, T sine Theta. Perfect, that looks like a great free body diagram and now we can go to sum of the forces. In the X direction it's not really going to tell us very much, right? Because we have T cosine Theta to the right, but we have T cosine Theta to the left and all of that is going to equal zero because the things at rest and that's good; zero does equal zero. The Y equation hopefully will tell us a little bit more. What do I have, I have two, T sine Theta going up and I have one MG going down. And all that is equal to the mass times the acceleration the Y direction, but this is hanging there. That's zero and so now we get a solution for T. T is MG over two sine Theta. Let's put a box around that. So we got an answer, let's make sure that it makes sense. Do the units make sense? Tension is a force, MG is a force, sine Theta is unitless. So the units do work out. Let's look at the limits. And to start with let's say the thing is hanging straight down, so if it's hanging straight down, what is Theta? What would Theta be if it's hanging straight down? Theta would be ninety degrees, what's the sine of ninety degrees? One, so what would the tension be? It would be MG over two and that makes sense right? If the two wires are going to hold the weight, each wire has to hold half of it. No problem, now the original question was what if it's horizontal? If we have a horizontal wire what is Theta? What's Theta in this case? Ok; it's when this wire goes to the horizontal, which means Theta goes to zero. You could plug in 180 and you get the same answer, but Theta equals zero is what we're looking for; what's the sine of zero? Sine of zero is zero. If sine of ninety is one then sine of zero had better be zero. So what is the tension? It is MG over two sine Theta, which becomes MG over zero, what's MG over zero? It's infinite, right, anything over zero means that you are approaching infinity. So it can't be done. Our guess of it being really, really big was right, but it's not a thousand MG, it's not ten thousand MG, its infinity. Can't be done and this is kind of weird right? I mean it seems like I could just pull on this wire really hard and get it to be horizontal and I could put just a little tiny mass on it and it would stay horizontal, but that's not what happens. No matter what the wire is made of, no matter how strong you pull on it, what the tension is, you can't get it perfectly horizontal. And this is good to remember when you're an engineer, right, when you are building a bridge and you think oh, I'm going to string this cable and it's going to stay perfectly horizontal. No, as soon as you put some mass on it, it starts to sag and in fact you don't even have to put mass on it because the cable itself has mass in it and so the cable will start to sag and this is part of the point for you engineers. Everything bends; everything sags when you apply forces to it, things move. So when you're engineering a building or an airplane or a suspension bridge, keep this in mind; everything is going to sag. You can't keep it perfectly horizontal. It's kind of weird, right? That's the way it is.