A water trough for horses has a triangular cross section with a height of 3 m 3m and horizontal side lengths of 2 m 2m . The length of the trough is 12 m 12m . How much work is required to pump the water to the top of the trough when it is half full.

this problem we are told a water trough for horses has a triangular cross section, with a height of three metres, a horizontal side length of two metres. The length of the trough is 12 metres, and we want to know how much work is required to pump the water to the top of the trough when it is halfway full. So this is a pumping liquids problem, and a good place to start with any of these problems would be to draw a sketch of the situation. Now we're told that this is a water trough with a triangular cross section, so what I'm going to do is draw this triangular shaped water trough. So we'll have this triangle here, and then it's going to extend back like this. So this is gonna be our general shape, what our water trough looks like. And what we're told about this trough is that if you take a cross section that is a triangular along the side of the trough here, what that's going to do is give you a height of three meters. So we know that the height here is going to be three meters, and that we also know that the length, the side length of the trough is two meters. Now another thing we're told about this is that the length of the trough is 12 meters, so that means the length if we go along the edge right here, this distance would be 12 meters. So these are the dimensions of the trough that we're given. Now before I actually move on in this problem, there's a couple more things that we can recognize here. First up, we can recognize that the tank is halfway full. What does that mean? Well that means if we go halfway through the tank here, and we fill this with liquid, this is going to be where the liquid height stands. So we can have the liquid go like this, and that is going to be how full our tank is with this water here. But it's not going to be super helpful if I go ahead and just keep this image right here, because notice that it's kinda hard to tell where things are going to be graphically, how we could find an equation for this drop. It's gonna be kinda difficult. So what I'm gonna do is draw actually a second diagram here, just so we can follow along a little easier. Now a second diagram would look something like this, we have our x and our y axis, and what I would do is take a side of this trough which looks like a triangle. Now we know how to deal with triangles, and what these shapes in general look like, and I can tell that this triangle is going to be halfway filled with water. Now what can I do with this information? Well, what I can see is that if we're starting with our tank at zero on the y axis the height is three meters, so that means we're going to go all the way up to a distance of three meters on the y axis. Now what I can also tell is that this length here is two meters. But if I go ahead and just look at the origin at this place in the center right here of the tank, and I go this distance right here, that means it would be half of two meters, which is one meter distance right there. And then if this liquid is halfway filling the tank, well, that's going to be halfway of three meters, which is 1.5 meters. 1.5 is half of three. So now we have this front section of the tank drawn, and now we have this full tank drawn, and I think this is enough to give us a general idea on how we can go about solving this type of problem. So now let's get into the details of what we have. So we're trying to find the work required to pump this liquid out of the top of the tank. Now we know when it deals with pumping liquids problems, we're going to be integrating the density of our fluid, the weight density, times the cross sectional area times the distance, and it's all integrated with respect to y, which is gonna be how far we pump this liquid. So we need to find all these elements inside the integral, and that will allow us to solve it. Let's first start with finding the weight density. Now the weight density is actually not given to us in this problem, but there's a couple things that we can notice from the wording. I notice everything here is given in metres, that's our measurement. So that means we're going to be dealing with the metric system here where we have metres, we have kilograms, we have newtons, so those are going to be the units that we work with. Now whenever you have problems where you are dealing with water in these types of situations, you can assume that the density is a thousand kilograms per cubic meter. This is always going to be the approximate density for water, but notice that this density is in kilograms per cubic meter. I don't want this to be in kilograms here since kilograms is a mass, not a weight. So because I'm looking for a weight density, I need to multiply this whole thing by 9.8. That's gonna be in meters per second squared, and this is going to give me my weight density in newtons per cubic meter. So we can multiply a thousand times 9.8, and that's going to give us 9,800 Newtons per cubic meter. So this right here would be the weight density. Now that we found the weight density of our fluid, we can go ahead and scratch off this in our integral because that's what we're looking for. So next we need to find the area. Now the area is going to be one of the trickier things to calculate in this problem, because what I ultimately wanna do is take a cross section out of the fluid of this tank. Notice when I take a cross section, it's a rectangle that forms here. And because this rectangle forms, I know that the area of a rectangle is length times width. Now the length of this rectangle, well that's just going to be the length of the trough here. And notice that that never changes no matter where we are in the tank, whether we're higher or lower. We're always going to have this continuous length of 12, so 12 is the length of the cross section. But what about the width here? How do we find the width? Well the width is going to be this distance right here, and notice if we're looking at this, so let's say I draw the cross section in this version of the tank, this here would be our width. But notice the width is actually going to change whether we're higher in the tank or lower in the tank, so we can't just set a constant value for our width. So what I'm going to do instead is I'm going to go ahead and take a half of this width right here, I'm gonna call that x. So x is the right half of this of this kind of width that we have in our tank. So starting from the y axis and then going to here, that's going to be a distance x. Now because I can see that that's half of the total width, what I'm going to say is that the width is the equivalent to twice the value of x, so we're going to call this two x for our width. Now since we have the width here as two x, we can go ahead and say that this right here would be our area. So I'm gonna go ahead and check this off. Now lastly, we need to find our distance. Now our distance is pretty straightforward. All we're going to have is going to be the length to the top of the tank, which in this case is three meters, the height of three meters, so it's going to be three minus whatever distance down we have our cross section, which we'll just call y. So it's gonna be three minus y, and that's going to be our distance. So we now found our area, our distance, our weight density, so we can now set up this integral. So what I'm going to say is that the work that we're looking for is going to be the integral, and then it's going to be our density of 9,800 newtons per cubic meter. Is going to be our area of 12 times two x, which is the same thing as 24 x because 12 times two is 24, and then we're going to plug in our distance of three minus y, and all of this is integrated with respect to y. Now what about the bounds on this integral? How do I find that? Well, the bounds will always go to the bottom of the tank here, which is often a value of zero, and then we'll go to the top of our liquid level. And we said that our liquid level goes to 1.5, so it's gonna be zero to 1.5. So these are gonna be the bounds on our integral. Now from here, I can go ahead and simplify things by taking this 9,800 and pulling it outside the integral. I can also pull the 24 outside the integral as well. So we're going to have 9,800 times 24 pulled outside the integral, and we're going to integrate this zero to 1.5, and then we'll have x times three minus y integrated with respect to y. Now you may notice we run into an issue when we go ahead and do this integral. We have an x in here, and we're integrating all of this with respect to y. So what I need to do is get everything in terms of one variable. In this case, I'm looking for y. How do I do that? Well, let's go back to this diagram right here. Notice that we have x that we measure being this arbitrary distance for some random cross section we took out of the center of our tank. But notice that x follows this kind of line right here. This line is going to be our key to getting everything in terms of y, because we can go ahead and go to a point on top of our line right here. Notice that the x distance would be one, and the y distance here would be three. So that means that we have a point on our line. Now I want you to recall slope intercept form, y equals m x plus b. What we can see here is that our slope is just going to be rise over run. The rise is three and the run is one, so we have three over one which would just be three, that's gonna be our slope. Our x is what we have as a variable set here, and then we need plus b. Well, plus b is just going to be the y intercept where we cross the y axis, and I can see that clearly happens at zero, so I don't even need to worry about writing this. So we're just going to have y equals 3x, and I can go ahead and divide this three on both sides of the equation to solve for x. That's going to be that x is equal to y over three, or we can also say that that's one third y. So this is going to be our x value, which we can go ahead and plug in right about there. So we're gonna have 9,800 times 24 times the integral from zero to 1.5, and then this is gonna take this x and replace it with one third y. So one third y times three minus y integrated with respect to y. Now notice I can actually take this one third here and move it to the outside as a constant. I'm gonna combine it with that 24 there since I know a third of 24 is eight. So you're gonna have 9,800 times eight, and then that's going to be times the integral from zero to 1.5 of y times three minus y integrated with respect to y. So now we've gone ahead and simplified this further, and now everything is in terms of y. That means we can go ahead and integrate this properly. So we're gonna have this 9,800 times eight, and then let's go ahead and distribute this y into the parenthesis. So doing that, we're going to have three times y, which is three y, minus, and then we're gonna have y squared integrated with respect to y. Now I can go ahead and find this antiderivative. So we're gonna have 9,800 times eight, and then we'll take the antiderivative of three y, which using the power rule is gonna give us three y squared over two. Then we'll have minus, then using the power rule here we get y cubed over three. All of this is gonna be bounded from zero to 1.5. Now notice that taking zero and plugging it in for y, that's just gonna set everything to zero inside. So all we need to do is plug in our 1.5, which is our top bound. You always wanna check though to make sure everything will indeed go to zero. So this is going to give us is 9,800 times eight multiplied by three, and we're plugging in y here, so it's gonna we're plugging in 1.5 here, so it's gonna be 1.5 squared over two minus, then we're going to have 1.5 cubed over three, and this is the number that you need to evaluate. So all these numbers you can crunch and check on a calculator, and the final result that you should get is about a 76,400, and that's going to be in joules since we're dealing with newtons and meters and kilograms. So this is going to be our work and the solution to this problem. So that is how you can solve these types of problems where you're dealing with work. And notice here in this problem, we did not get a constant for the cross section here. Notice that we got a constant length, but we didn't get a constant width. So we needed to do a little work on this side here to figure out what that is going to change as we go up and down this trough. But once we were able to make that adjustment, getting our answer was just going through the same process that we've always gone through with these types of problems. So, hope you found this video helpful, and let's go ahead and move on.

10. Physics Applications of Integrals

Work

Calculus

10. Physics Applications of Integrals

Work

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Multiple Choice

A water trough for horses has a triangular cross section with a height of $3 m$ and horizontal side lengths of $2 m$ . The length of the trough is $12 m$ . How much work is required to pump the water to the top of the trough when it is half full.

$176 comma 400 J$

$1 comma 764 comma 000 J$

$264 comma 600 J$

$88 comma 200 J$

Verified step by step guidance

Step 1: Begin by understanding the geometry of the trough. The cross-section is a triangular shape with a height of 3 meters and a base of 2 meters. The length of the trough is 12 meters. When the trough is half full, the water occupies half the height of the triangle, which is 1.5 meters.

Step 2: Calculate the area of the triangular cross-section when the trough is half full. The formula for the area of a triangle is A = (1/2) * base * height. Here, the base is 2 meters, and the height is 1.5 meters. Use this formula to find the area of the water's cross-section.

Step 3: Determine the volume of water in the trough when it is half full. Multiply the area of the triangular cross-section by the length of the trough (12 meters). The formula for volume is V = area * length.

Step 4: Calculate the weight of the water. The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is 9.8 m/s². Multiply the volume of water by the density and gravity to find the total weight of the water.

Step 5: Compute the work required to pump the water to the top of the trough. Work is calculated using the formula W = force * distance. Here, the force is the weight of the water, and the distance is the average height the water needs to be lifted, which is half the height of the trough (1.5 meters). Integrate the force over the height to account for the varying depth of the water.