 A uniform wooden beam with a specific gravity of 0.65 is 10 cm by 10 cm by 3 m, comma, is partially submerged in 20 degrees Celsius water, and is hinged at point A as indicated in the figure below. At what angle theta will the beam float? So our approach here is going to involve a moment, and that moment is going to say the moment around point A must equal zero, because it's in equilibrium, it is static. And when we're considering the forces acting on this, I'm going to simplify our relationship a little bit. I'm saying this beam that has a length of 3 m has two forces acting on it. One is its own weight. The other is the buoyant force. So FBFW. So if I were to translate that to a crew drawing over here, I could say beam, and then bulkrum, and then scooch it down. So I have space to draw, and there is a weight force acting on the middle that is pulling down into the left here, and then there is a buoyant force that is going to be acting over here, that is FB, because as we all know, John can't spell the word buoyancy. And if this is point A, then all I really care about is the component of the force that is perpendicular to the center of the beam. In determining those perpendicular components of that force, I'm going to have to do some trigonometry, and I have the choice of drawing it this way or this way. For the purposes of establishing a magnitude, it doesn't actually matter, but the difference would be whether or not I care about this angle or this angle, and whether or not I'm using a sine or a cosine, and I'm going to choose this leg for a reason that'll make a little bit more sense once I go on a bit of a tangent here. I'm going to call this angle alpha just for a moment here, and this angle theta, and I'm going to say this angle here is theta, because that's what was presented in the problem statement and what we're looking for after all. So I can say that 180 is going to equal theta plus beta plus 90, because I know up is going to be perpendicular to horizontal. Therefore beta is going to equal 90 minus theta, therefore beta is also this angle here. You following me? Okay, and then because I know up is perpendicular to horizontal, that means that 90 must equal beta plus alpha, and then the iPad scrolls up, that's an important step of the trigonometry, and I can say here that 90 is also equal to beta plus beta. Therefore alpha is equal to theta, yes, all of that just to point out that this inside angle is also theta. Now it can be difficult to keep track of the trigonometry when you're dealing with these sorts of problems, but I think it's easiest to just draw out and label everything and then just kind of work through it like this. If you're really good at your opposite interior angles, you can reason through it pretty quickly. You can also get there by thinking through if this angle were really shallow, I mean if you imagine the beam were just slightly off canter of the surface of the water. In that case, theta is going to be very, very, very small, and the difference between up and our moment directed force is going to be very, very small as well. Therefore that angle is likely to be theta, not the complement of theta. Does that make sense? Good, glad it all makes sense. The point is that angle is theta, and similarly if I draw this out here, I can label this angle. Let's do that together. I'm calling that angle alpha. I'm saying this angle here is theta. These two lines are perpendicular to one another, and I'm saying this is beta. So I can say theta plus, excuse me, beta, I drew with theta, that's because they rhyme. Alpha plus beta is going to equal 90 degrees, and because this line is also perpendicular to the center line of my beam, I can say theta plus beta plus 90 equals 180. Therefore theta plus beta equals 90, and because alpha plus beta also equals 90, theta must equal alpha. Whatever reasoning you use, you should be able to come up with the fact that our inside angle here is theta as well. Now that we've established that, I can say that the component of force of weight that I care about here is going to be related to theta and the force of weight. I'm saying the cosine of theta is that force that we care about, that force divided by fw. Therefore, that force is going to equal cosine of theta times fw. So way back over here, I can say this force here is cosine of theta times fw. Similarly, I can say this force here is cosine of theta times fb. And then I know the weight is going to act in the middle because I'm assuming the composition of the beam is homogenous, therefore this distance here is 1.5 meters. And then just to make it a little bit easier to keep track of, I'm going to call this distance here something. I'm going to call it, what do you think? I like it, epsilon. Absalon is a dummy variable, it's just going to allow me to write out the moment. Okay, here we go. So because the sum of moments must equal zero, I can say that the moments in the clockwise direction are going to equal the moments in the counterclockwise direction. That's still clockwise, John. Clockwise here is going to be cosine of theta times fb times the lever arm. So I'm saying cosine of theta times the buoyant force times the lever arm, which I'm going to write as three meters minus epsilon is equal to the moments in the counterclockwise direction, which would be cosine of theta times fw times 1.5 meters. So we have enough information to calculate the weight of the beam. And we have enough information to write out the buoyant force. And we can make our math a little bit easier by cancelling the cosine of theta. Now I know what you're thinking, you're thinking, John, theta is the thing that we want, truly we have to leave it in our equation somewhere. And we are, because we know minimum fb here is going to be a function of theta. Because fb is going to be a function of epsilon and epsilon is going to be a function of theta. I mean, if you think about the fact that we're calling epsilon this distance here, or rather we're calling that distance there epsilon, epsilon changes as you move theta. Therefore epsilon is a function of theta. Therefore once we get to the point that we're solving for epsilon, it's going to involve solving for theta as well. So it's okay to cancel the cosine of theta. Also if you're thinking, but John, what about all that trigonometry we did? Yep. Yep indeed. So I will try to write out the weight. The weight of the beam is going to be the volume of the beam, volume of the beam multiplied by the density of the beam, multiplied by gravity, and the density of the beam is specific gravity of the beam times the density of water at standard temperature and pressure times gravity. The water at STP, we can look up, or rather we have committed it to memory because we've looked it up like six million times in the last six videos. Gravity we are assuming, specific gravity we know, the volume of the beam we can calculate. So the weight of the beam is something we can compute. Now generally speaking, I am in favor of leaving everything symbolically as long as possible. That will give us the most opportunity to cancel terms and have as few rounding errors as possible. But if you wanted to, you could totally calculate a number for FW now. That will make the algebra a little bit easier later on. But I'm going to leave it symbolically. So I'm saying buoyant force times the quantity three meters minus epsilon. Again, epsilon is just arbitrary here. It is the distance from the end of the beam to the point at which the buoyant force acts and that is equal to volume of the beam times the specific gravity of maple times the density of water at standard temperature and pressure times regular gravity. And this is known or rather can be computed because we know everything. This is known. This is known or rather can be looked up and this is known or rather can be assumed. Three meters is known. So I have one equation, two unknowns. Let's explore FB a little more. The buoyant force is going to be the weight of the displaced water. The weight of the displaced water is going to be the volume displaced multiplied by the density of the water multiplied by gravity. The density of water we can look up. The quantity we've already assumed, volume displaced, we can write out a little bit more. And before I do that, let me call your attention back to this simplified version of the beam. I'm calling this a line even though it's actually a volume. And over here that's fine because I'm just talking about the moment which we're saying is acting along the central line of the beam. But I can do the same thing for volume calculations too. I mean if you imagine that I hadn't drawn this triangle over here, if I were to consider the water as crossing perpendicular to the direction of the beam, then I am, that's actually not particularly good drawing, what I should be saying is this point here where it intersects the water. Calculating the volume displaced by using this rectangular subsection, I think it would actually be called a rectangular prism. If I use this rectangle here to calculate the volume displaced, I'm still going to have an accurate number despite the fact that the water is crossing the beam at an angle. Because I'm getting rid of the water here and I'm adding it over here and because of opposite internal angles, those are going to be the same size. So I'm not going to involve any trigonometry in my volume displaced calculation. I'm going to describe it entirely as this distance multiplied by the cross-sectional area of the rectangular beam. And note that this distance is the section of beam that is underwater. And I've already described that because I've called this distance here epsilon. And the buoyant force is going to act in the middle of the submerged section. So this distance here is 2 times epsilon. Are you following that logic? 2 times epsilon. Okay. Now I'm going to restore my triangle. Okay. So back over here. The volume displaced is going to be 2 times epsilon times the area of the beam. Okay. Now once I plug that back into this relationship up here, I'm going to have one equation and one unknown and I'm going to solve for epsilon. Let's plug everything in symbolically here. I'm saying 2 times epsilon times area of the beam, which is 10 centimeters by 10 centimeters, I believe, because it is 10 centimeters by 10 centimeters. Then I'm multiplying by the density of water. We don't know the conditions of the water here, so it would be reasonable to assume it's at standard temperature and pressure. And then I'm multiplying by regular gravity. And that gives us F beam, then I'm multiplying that by 3 meters minus epsilon. And then that is equal to the volume of the beam, which I can write out as 3 meters times cross-sectional area of the beam. And then I'm multiplying that by the specific gravity of maple and I'm multiplying that by the density of water at STP, because that's what our specific gravity is given in reference to. And I'm multiplying that by gravity. And then let's look at what cancels. Density of water at STP cancels, gravity cancels, area of the beam cancels. How neat is that? So we have 2 times epsilon times the quantity 3 meters minus epsilon is equal to 3 meters times specific gravity of maple. Now I just need to solve for epsilon. And I'm going to simplify this a little bit further by getting rid of the units. To get rid of the units, I'm going to say epsilon is a quantity in meters. And by doing that, I am left with meters times meters, which is meters squared, and meters times meters, which is meters squared, and meters times nothing. Okay, that's not right. Where did I go wrong? I have my moments written out, force times a length, got it? And then force times a length, aha. Okay, so this 3 meters section over here distracted me. I forgot to multiply by 1 and a half meters after I wrote the force. Like it never happened, times 1.5 meters times 1.5 meters. Okay, let's try that again. And then we're left with meters times meters over here, which is meters squared as well. And because meters squared are everywhere, I can cancel the meters squared. Therefore, I have just 2 times 3 times epsilon minus 2 times epsilon squared is equal to 3 times 0.6 times 1.5. And I'm just going to double check that it was 0.6. It is not. In fact, it's not even maple. Where did I get maple from? Okay, while we are pretending this never happened, let's pretend this also never happened. Wood. Maybe it's a different maple. There's lots of maples. It's fine. Okay, SG generic wood. I mean, I guess it's not generic wood. We know it's going to be relatively light and airy wood, probably not teak. So you know, we can narrow it down a little bit. Okay, then we are multiplying by 0.65 instead of 0.6. I think I got 0.6 from the previous example. And now I just have a quadratic formula. I can write this as 2 times epsilon squared minus 2 times 3 times epsilon plus 3 times 1.5. Let me get a little more clear that that's 1.5 times 0.65 is equal to 0. Then I can say altogether now epsilon is equal to negative B plus or minus a square of B squared minus 4ac all divided by 2a. And a is 2, B is negative 6, C is 3 times 1.5 times 0.6. Where I draw the line on mental math, it got me. So I'll just see if that's a nice convenient number. 3 times 1.5 times 0.65, 2.925, sure, that'll work. 2.925. Now, let's multiply some stuff together. Here we go. Negative, negative, negative 6 plus the square root operator. And that would be negative 6 squared minus, look at me, I'm using the right operators. A, which is 2 times C, which is 2.925, unless the calculator is hiding some sneaky decimal places, does not appear to be, and then closing parentheses and closing parentheses divided by 2 times 2. So when we use plus, we get 2.38, when we use subtract, we get 0.6158. So there are two solutions to this problem, which one seems better? Well, recognize that A is 1 meter above the water, so epsilon cannot possibly be 2.3874. That would mean that the distance above the water is going to be 3 minus 2.38741, which would be 0.61259. And that's impossible, because that hypotenuse here would be shorter than 1 meter, which would be impossible. Therefore, our correct answer is going to be epsilon is equal to 0.612588, and remember epsilon's in meters. That's not actually the answer to the question, but it is as good as the answer to the question, because now all we have to do is a little bit more trigonometry, I know you're excited, and you should be. I'm going to draw another central line of the beam, and the purpose of this is going to be to attempt to write the relationship between epsilon and theta if it exists at all. So I know total beam length is 3.0 meters, this distance up here is 1 meter, and I know this distance here is epsilon, and I know this distance here is theta. So I can say the hypotenuse of this triangle is 3 minus epsilon, and then if I could describe this angle here, which let's just call it alpha, then I could write the tangent of alpha is equal to opposite over, excuse me, if I could write sine of alpha is equal to 1 over 3 minus epsilon, then I could solve for alpha relative to epsilon. Now how is alpha related to theta? You know what we're going to do, don't you? That's right, we're going to draw some perpendicular lines. So I'm going to call this angle here beta, and I can say that 90 is equal to alpha plus beta, and then I'm going to say this is also perpendicular, therefore 180 is equal to 90 plus beta plus theta, therefore 90 is equal to theta plus beta, therefore theta equals alpha. How neat is that? Are you surprised? I know I was, and I know that's kind of a labor-intensive process just to give this in terms of theta, but I find thinking through it that way works for me and my brain, and the weird way in which my brain works. If you're able to think through it faster, do that, then I can say theta is arc sine of 1 over 3 minus epsilon, oh whoops, this shouldn't be epsilon, this should be two times epsilon, excuse me. As we defined epsilon, you'll remember as the distance from the edge of the beam to the buoyant force, therefore the distance submerged is two times epsilon, whoops, and then this is three minus two times epsilon, this is three minus two times epsilon, and this is three minus two times epsilon, okay. And then I'm going to write this out in terms of units, and for that I'll prepare a little bit more space, theta is equal to arc sine of one meter divided by three meters minus two times 0.612588 meters, meters cancels meters, and I'm left with a trigonometric function of a unitless proportion, which is exactly what I want. Now all we have to do is compute an answer, so calculator, we're going to say arc sine, that's sine calculator, come on, arc sine, there you go, one divided by three minus two times this thing, and we get 34.3 degrees, and there's our answer.