| Start | And today we're going to continue with this section and study hydrostatic force and pressure.
So this is going to be the topic of today's class. 2 more weeks of class, but so many applications to go. Exams with biology and economics have many topics, so integral calculus has lots of topics. Ok, so when you talk about hydrostatic pressure, we only take into account the pressure of the fluid at rest. |
| 0:37 | So in other words, we are diving somewhere under water, and we feel a pressure from the water, basically kind of like the gravitational force of the water on top of us.
And the hydrostatic pressure is only the pressure of the fluid at rest. So in other words, we don't want any kinetic motion, so we don't see waves breaking or stuff like that. |
| 1:05 | So as a definition, this is the pressure exerted by a fluid at rest, and it has a very clean formula, so Pressure = density * gravity *depth |
| 1:47 | So the quantities over here are rho =density, g=gravity, and y=depth.
|
| 2:02 | In most calculations we only deal with water so density will be 1000 kg.m^3 So g is roughly 9.8 m/s^2 |
| 2:37 | and those are constants, and the only thing that changes is y.
So you see from here that the hydrostatic pressure is just proportional to the depth. So it's going to increase as the depth increases. And also the hydrostatic pressure is constant along all directions. |
| 3:06 | So if you're [?] at a point under water ?, you feel a hydrostatic pressure which is constant.
Ok, so that's our definition for hydrostatic pressure, another way to think of the hydrostatic pressure is, |
| 3:32 | the force exerted by gravity So the pressure is something like the force/area |
| 4:04 | so F is the hydrostatic force, and it's like a gravitational force of the water on top, so it's just like the weight of the water And a= area that is submerged under water, |
| 4:38 | And usually this formula right here is seen presented in a different way,
You'll see it as the hydrostatic force = hydrostatic pressure * area.
Because technically first we compute the hydrostatic pressure which only depends on the depth, and then we compute the hydrostatic force which depends on the pressure, but also on the shape of the object submerged under water. |
| 5:08 | Now the formula over here is very nice, but it also for very special circumstances.
So to apply the formula over here, we need to assume that all components of an object which is submerged under water are at the same depth. So in other words, a situation in which the formula over here holds true, is when you have some circle in place, that is submerged horizontally under water, |
| 5:40 | So every point over here is going to experience the same pressure, or the same force on it from the horizontal,
so this is like a thin plate, submerged horizontally.
|
| 6:05 | And the formula over here works because every point on the plate over here is submerged at the same depth.
So it has the same pressure from the water on top. So in most applications that we will do, this will not be the case. So we will not be computing the pressure on the bottom of something, or on a thin plate that is submerged horizontally, |
| 6:31 | we will be computing the hydrostatic force of the water which, so something thats going to be submerged vertically or tilted or whatever, and because not all points are at the same depth, it mean that the pressure is going to be very different along the object. So on the top, I don't know, our water comes like that, |
| 7:02 | Then whatever is above the water, has basically no influence whatsoever, there is no hydrostatic pressure over here. At the top of the fluid, the hydrostatic pressure is 0.
But as you go down toward the bottom of the object, the hydrostatic pressure is going to increase, and the force will increase as well. So we can no longer apply the formula over here. |
| 7:31 | So let me finish off this brief introduction, by telling you what the units are in the standard international units.
So you already know that the force is measured in newtons, so the standard measurement for force is a newton. And for the pressure, we can see that the N/m^2 here, but there is also another equation for this one, which is [?] |
| 8:08 | So the best units for pressure is pascals.
And this is abbreviated by 1 Pa, and this is equal to 1 N/m^2 |
| 8:37 | Ok, so this is not a course in physics so I don't want to stress too much so after this brief introduction I think we can proceed with some examples.
I'll first do an example where something is horizontally on the water, even though it is not very representative of the type of applications we will do later on. |
| 9:00 | And then I will proceed to this kind of situation.
Any questions so far? Are we good? You all know how to study pressure, you've gone under water. Ok, so let's proceed with this example, You've got, let's see you've got this, |
| 9:36 | which is going to be filled with liquid, So over here we've got water, and on top we have some oil, some dimensions over here, |
| 10:06 | here you have 1.5m,
And the diameter of the bottom is 3, so the radius is just 1.5.
So we've got the container of this which is partially filled with water, and oil. You know that oil is usually less dense than water, so it's on top. |
| 10:37 | The density of the oil, is 900 kg//m^3.
and the density of the water is 1000 kg/,^3 |
| 11:01 | And the problem is asking
what is the hydrostatic pressure and the hydrostatic force at the bottom of the cylinder over there.
|
| 11:48 | So, the problem is asking for the hydrostatic force and pressure of a thing that is horizontal.
It means that basically on the surface over here is going to experience the same action from the [?] |
| 12:07 | So basically to compute the hydrostatic pressure, we just need to see what is the depth what is the height of the plate on top if you want.
And if you want to compute the force, all you need to do is multiply the pressure by the area. So let's do just that. You'll see that this example is easy, but the computations are not so easy. |
| 12:36 | So those numbers over there are not really great or well chosen, and I'll help you with the final calculations [?]
So basically to compute the pressure, we need to add 2 things. There is a pressure coming from the oil, and there is a pressure coming from the water. So let's account for both of those.
|
| 13:05 | As it's on the bottom of the cylinder.
So what is the pressure? The water pressure, Well, we have a formula for this, so if you are somewhere on the top of the cylinder, the pressure coming from the water on top is going to be |
| 13:36 | the density of the water * gravity * height of water on top, so and for the oil, |
| 14:06 | We're gonna have a similar thing,
And to compute the total pressure all you need to do as add them together.
Ok, so let's do some computations over here, so density of water is 1000 kg/m^3, |
| 14:36 | the constant of gravitational acceleration is 9.8 m/s^2, and the height of the water on top is going to be 1.5 m.
|
| 15:10 | And because all measurable units are expressed in standard measurable units, we're going to get a final answer in pascals.
And the answer is something like 14,700. |
| 15:31 | Which is just 14.7 kPa.
Ok, so we do the same for the oil, the density of the oil is now 900 kg/m^3, the constant of gravitational acceleration is the same, And the level of oil is 24 m. |
| 16:09 | and this gives,
21168 pascals. I told you that the calculations were not the most straight forward.
|
| 16:34 | In any case, with some arithmetic over here and some multiplication, this is what we get.
So the total pressure on the bottom is going to be 14.7 +21.2 kPa. Which gives us 35.9 kPa. |
| 17:11 | Ok, so we see the pressure at the bottom of the container
doesn't really see the shape of the bottom of the container. So it only takes into account the level of the water or the oil which is on top of the box.
|
| 17:30 | But the formula over here never takes into account the actual shape of the box.
This is something that we are going to have to quantify whenever we compute the hydrostatic force. So having computed the pressure not the hydrostatic force, is the same on every point of the bottom because they're at the same horizontal level And now of course its just going to be the pressure * area of the bottom. The pressure we computed it, you see that the area of the bottom is just the area of a disk with radius of 1.5 m. |
| 18:19 | So it's area is going to be pi*r^2.
So pressure is 35.9, |
| 18:32 | and area over here is pi*(1.5)^2
And because the pressure was in kPa, and the area was expressed in m^2, we get a final answer in kN. [?] some measurable units, the appropriate measurement unit would be kN.
|
| 19:08 | Ok, and whats the computation, and I throw in [?] 253.8 kN.
Ok, so I didn't want to do the computation, [?] on the board, the final answer is that so, relevant [?] |
| 19:42 | Ok, so in this example, basically we were able to just use the standard formula to find pressure and force because we had something across horizontally under water.
Every point here had the same pressure, so we are able to use the standard formula with no integration whatsoever. |
| 20:05 | So if you want to do this basic example, I would like to move on to something that is going to be a little bit more tricky.
And it's gonna involve [?] Any questions so far? You're good so far right? I haven't lost none of you [? too much background talking] |
| 20:37 | Pascals, so it's no so important to write the m/s^2
Ok, so let's consider the following application.
You've got some object in the shape of a triangle, |
| 21:00 | Which is submerged vertically under water with the top at the surface of the water.
So this is a right triangle over here, the vertical side has 3m, the horizontal side has 2m, And lets put some words over here, so, calculate the hydrostatic force on one side of a plate, in the shape of a right triangle of height 3m, and base 2m. |
| 22:28 | Which is submerged vertically under water. Now if I write over here, can you still see? Or do I need to write further up? Is this still good?
|
| 23:01 | And the vertex is located at the surface of the water. Let me mark that
Ok, so we have this, and as I keep mentioning, it's not ok for us to use the formula where pressure =Force*area because the pressure is not constant along the object which is submerged under water.
So at the upper vertex the pressure is going to be 0 because it's sitting exactly at the surface of the water. |
| 23:36 | As you go down the pressure is going to increase, and because its not going to be constant along the object,
we cannot just go ahead and say the hydrostatic force is p*a
It is 0 at the top, then is more. No that is not good.
So the idea over here would be just to divide the object in question into smaller objects that sit at the same level. |
| 24:07 | So the idea would be to consider instead of the total hydrostatic force, just the hydrostatic force on some thin horizontal strip of the object.
And this thin horizontal strip is going to have some height interval, (delta)y. |
| 24:30 | That we can assume to be small enough so that the pressure along the object over here is basically constant.
So for this kind of thing, we're going to be able to use our standard formula F=P*a And then we can total the hydrostatic forces along this strip and totaling those hydrostatic forces, adding them together, is something we know how to do using integrals. |
| 25:03 | Ok, so let's out more thought into the problem over here, let's first introduce a vertical coordinate from the surface of the water down which just measures the depth.
So y is just going to measure the depth of the water, And first let's see what the hydrostatic pressure is at depth y. |
| 25:48 | The problem is not specifically asking for it, but we will need it anyway when we compute the hydrostatic force. Now we have a formula for it, This is p(y)= density*gravity*y |
| 26:13 | So this is just (rho)(g)(y). So in other words if we are at depth y, anywhere on the surface of the triangle, anywhere at depth y, we are going to experience a pressure that is rho*g*y.
|
| 26:31 | And again, the pressure really doesn't see the shape of the object so it's just rho*g*y
ok, now let's preform the actual computations at the moment, so lets just leave it as rho*g*y and let's try to estimate the hydrostatic force
on the thin strip that we formed over there, and say that the width is w(y) and height (delta)y, which is close to 0 but not yet 0, so I'll put that over here.
|
| 27:21 | We know that scientifically this is really really small, but not 0, and the thin strip sits at depth y, |
| 27:38 | the height over here is (delta)y
And the distance from the top, so if you were to take, this side it would just be y.
|
| 28:05 | Ok, now, so the assumption of the strip over here, is that all points sit roughly at the same depth from the surface. It means that for the thin strip only, we get to apply the formula that the hydrostatic force=p*a.
Now this thin strip is just a rectangle, because width is w(y) and height is y, so naturally we can find the area by multiplying the 2. |
| 28:41 | So let's write the formula for the hydrostatic force, the hydrostatic force=p(y)*a(y) and because it varies with the depth, I will write it as a function of depth.
|
| 29:12 | So the pressure is rho*g*y, from before,
And what is the area? Well as we said we have a rectangle, whose width is w(y) and whose height is (delta)y, so over here the area is just w(y)*(delta)y.
|
| 29:43 | let me put it over here so we don't forget it, a(y)=w(y)*(delta)y.
Ok, so so far so good. In the formula over here, it looks like we know everything pho and g are just constants, y is depth, (delta)y is something which is going to turn into something very small, and that is the key to finding the integral. |
| 30:15 | And the only thing that requires further computations is the width, w(y), which changes with the depth, because you can see that over here we've got thinner rectangles, and we go to the bottom.
Or rectangles are going to have increasing width. So the only thing that requires some computation is the width w(y), and there is a technique which is kind of [?] unfortunately, so, that of similar triangles. |
| 30:44 | So if you have a picture like the one i'm about the draw, |
| 31:02 | Something like this, if you have an upward triangle, Divide them into similar triangles, and give them letters A,B,C,D,E, Now if these triangles are similar, in the sense that you know they have the same angles, |
| 31:31 | So this angle is this angle, then the length of the sides are proportional.
You probably used it many many times So this is the way in which [?], and all of the small triangles over here, well, one triangle has the top of the vertex, and the side with width w(y), |
| 32:14 | and this small triangle over here is going to be proportional to our large initial triangle
So it means that we can compute our w(y) from the similar triangles.
|
| 32:36 | Apologies if I'm making it sound simple.
Over here we've got w(y) divided by the width of the bottom, so 2, this one is going to equal, height of this side divided by the entire length of the triangle, so it's going to be y/3. Good? |
| 33:15 | Ok, so this means that the width, is just gonna be 2y/3.
So we can go back to our computations over there, and our hydrostatic force of the strip is rho*g*y*2y/3*(delta)y, |
| 33:52 | which gives, (2/3)*rho*g*y^2*delta(y) |
| 34:07 | ok so this is approximation we can do for the width over here, and it's accurate assuming that (delta)y is really really small, but we would actually like to make it 0.
Of course we cannot because otherwise this would be 0, so, we have to total everything together and make it close to 0. |
| 34:30 | To find the total hydrostatic force now, acting on the triangle, we need to add the hydrostatic forces acting on these thin strips.
So we just see the big triangle being comprised of these thin horizontal strips, and add these things together. Now, when we add things together, this makes sense since when we have a finite number involved. So if (delta)y is [?] then there is a sense and you can understand the [?] |
| 35:06 | when (delta)y goes to 0, we need a specific number of these strips to cover up the entire triangle, which means that the proper sense of addition as you learned in calculus B, is integration.
So to find the total hydrostatic force acting on the triangle, we need to find the integral |
| 35:49 | Now where do I write this?
|
| 36:01 | So basically, it's kind of easy to write the integral having this formula right here. Now I wasn't very thorough, but... Newtons.
I know you all appreciate [?] over here. As if the formula is not complicated enough. So we just take the number over here, now this is a function, (delta)y is going to form the dy in the integral, |
| 36:33 | So when we compute the total force, we have an integral, we put (2/3)(rho)(g)(y^2)dy, which is (delta)y
And I will explain that a little bit more on the formula. But this is exactly how the integral is formed.
|
| 37:08 | And I need some bounds of integration, now what is y supposed to measure? Y is supposed to measure the depth.
Everything is expressed as a function of the depth. Which means if you were to put some pounds of the depth, we'd have to make y range, between 0 because the upper vertex is exactly at the surface of the water. |
| 37:33 | And y needs to go down all the way to the bottom of the object which is submerged under water, so, it needs to go 3m under water.
So if I were to do that over here I would just say from 0-3. |
| 38:19 | So just to convince you that this is a proper way of doing the summation, let me just point out that this requires an intermediate step by finding a Riemann sum, and then just taking the limit of the Riemann sums [? instead of an integral] |
| 38:39 | So you should just imagine that (delta)y is something consistent like perhaps 3/N or something,
So if [?] lets say N strips over there, and you add those pieces together, you would write the summation perhaps of [?], delta(y) is something like 3/N.
|
| 39:06 | And this would be (2/3)(rho)(g)(y^2)(delta)y
And then you take the limit over here, either as (delta)y goes to 0, or as N goes to infinity.
|
| 39:37 | And you know what this is, this is a Riemann sum that corresponds to a function, and the corresponding function is just this one over here.
So I'm not really stressing on the Riemann sum because in practice we just go straight to the formula over here to do the corresponding integral because the transition is like a one step transition |
| 40:03 | we just replace the (delta)y with dy. So it doesn't really make sense to complicate ourselves by writing the Riemann sums nonetheless it is good to keep in mind that the complete picture is that we have a finite summation,
and as we let (delta)y go to 0, this is a limit of Riemann sums and it's an integral of [?].
|
| 40:31 | Ok, so, this is like a side comment,
which means that you don't really need to write it
and if you are free to forget it if you want.
Now let's compute the integral right here so that's on that side, let's compute the integral right here, now if you kept (rho) and g unchanged, |
| 41:03 | if you hadn't found them by their exact computations, now we can take them out of the integral sign because they're constants.
Since they don't change with the depth, so I can just write them over here, (2/3)(rho)(g)* (integral) from 0-3 of y^2dy. And you notice that the integral here is not very complicated, and in fact most integrals that come from computing hydrostatic force or pressure, |
| 41:38 | are not more complicated than this one, it's just quadratic binomials, y^2+4y, something like that.
So if you know the process, computing the integral is not going to cause any problems, so let's do it. (2/3)(rho)(g)*y^3/3 evaluated between 3 and 0. |
| 42:10 | Now this is 27/3, which is 9, I can cancel 9 as a 3 factor over here, and get 6.
And the measurement will be in N |
| 42:31 | And now I've got to turn it into an exact computation, the density of the water is 1000, the constant of gravitational acceleration is 9.8,
So I've got 6*1000*9.8
And I get something which is 58,800N.
|
| 43:00 | Ok, so let's give you little bit over here, before I move over before I move to the next example.
So we had an example submerged vertically under water, which means that the pressure is not constant along the object. We chose a vertical coordinate y which just measures the depth from the top of the surface, and that enables us to compute 2 important quantities. The hydrostatic pressure and the hydrostatic force. |
| 43:31 | Now the hydrostatic pressure only depends on the depth y, so to compute the pressure,
which is a computation that I made with the Riemann sum, which I now deeply regret, but the pressure was (rho)(g)(y)
That was the formula.
So the pressure is something that you can compute straight ahead by doing that computation. |
| 44:00 | And then to approximate the hydrostatic force on the side of the triangle, what you need to do is just approximate the triangle by thin strips,
which sit roughly, all points sit roughly at the same depth under water.
So this is an example of such a strip, it has a width of y, which is w(y), height (delta)y, same as all the strips, Now on this strip, the points are roughly at the same depth. (delta)y is supposed to be very small, so to compute the hydrostatic force, we have to use the formula that says hydrostatic force=p*a |
| 44:49 | Ok, but what was the area? The area was just w(y)*(delta)y.
So in this way, we were able to approximate the hydrostatic force on the level over here. |
| 45:03 | The only tricky business was to compute the quantity w(y), the compute the horizontal length of the strip.
But this is the foundation that once you do an example, soon it will become automatic, you've got the similar triangles, their sides are proportional, the horizontal width can be broken into similar relations, in our case, it was y/2. |
| 45:31 | So the hydrostatic force acting on the strip was just the p(y)*area, which was 2y/3*(delta)y This is how it looks like, but this is just on the strip, and now to compute the total hydrostatic force, we need to add them together because we are adding, I need a number of strips, and delta(y) goes to 0, the appropriate [?] is the integral. This is the integral of exactly the formula before, |
| 46:09 | (delta)y becomes the dy of the integral, and the appropriate bounds of integration are just basically measuring how far your object stretches in terms of the y coordinates, in terms of depth.
And it goes from the top of the water in this case where the upper vertex of the triangle sits, all the way to the bottom which is our depth. |
| 46:38 | So I've got an integral over here, that we can evaluate, as I said integrals that involve hydrostatic pressure of course, are not more complicated than this.
Ok, so any questions about this example? |
| 47:02 | Ok, so let's vary by this a bit. So let's consider the same triangle,
But let's submerge it a little bit deeper under water. So instead of putting the upper vertex of the triangle at the surface of the water, we're going to submerge it 4m below the level of the water.
So let's consider the new depth, and let's see what changes. So this is the surface of the water, |
| 47:36 | This is the same triangle, my old triangle but submerged more under water.
3m here, 2m here, and from the surface of the water here, 4m. |
| 48:01 | It's not exactly very proportional with what i'm writing, but you can envision the picture.
And we've got the same question, what is the hydrostatic force acting on one side of the triangle. So this is what it's asking. |
| 48:45 | Ok, so now we've got some choices, we want to apply the same strategies as before, in other words, we want to think of our triangle as being comprised of thin horizontal strips.
|
| 49:03 | Where the pressure coming from the water is going to be constant. And we want to compute first the hydrostatic force acting on that little strip, so what happens is,
you take that strategy, but something has changed because our triangle is now submerged further under water. So our equation says the pressure coming from the water has increased comparative to the previous example.
So now we have 2 choices, we are going to know y is something, |
| 49:35 | that something could be the depth as before, so we could have y measure the depth from the surface of the water, or we could let y measure the distance from the top of the triangle.
And now the top of the triangle doesn't sit at the surface of the water anymore. So we've got 2 possible choices, in how to label our picture over here. |
| 50:04 | And these choices are obviously going to lead to the same final answer, but the computations are going to be a little bit different.
So if you choose to measure, if you choose to leave y as the distance from the upper vertex of the triangle you will get to keep most of the computations from the previous problem like you will get to keep the computation for the side w(y), over here because you are measuring the triangle in the same way, y is going to be the distance from the upper vertex to w(y) |
| 50:43 | So basically the measurements of the triangle are going to be the same. So what is going to change is the pressure, p(y), because y will no longer be the depth so if you are to use the formula I wrote with y, you will have to do something like (rho)(g)(y+4).
|
| 51:03 | because you are going 4 units down.
So this is the picture if you choose to let y equal the distance from the upper vertex of the triangle. Now you can choose to let y equal the depth, as before. You will get to keep the formula for the pressure because the pressure depends on the depth, y is the depth so this is the formula, |
| 51:30 | But the measurement of the triangle line is going to be a little bit different now because the relation between w(y) and y, is no longer quite the same.
So lets, for the moment, choose y to be the depth as before, and still see how the computations change. So let me do some parts of the previous example before, I'll just keep the formula over here and erase the rest. |
| 52:17 | So this is from the previous exercise, and I'll keep it over here.
And I'm going to write around it. So as before, the first thing we can find is the formula for the hydrostatic pressure at depth y. And this one is going to be P(y)=(rho)(g)(y) |
| 52:46 | Just as before. Ok, now we do the same business we take some thin strip, with height (delta)y this is y from here to here |
| 53:09 | this is y.
And at depth y you have a thin strip of width w(y) and height (delta)y. And let's try to estimate the total hydrostatic force acting on the strip, so we've got, |
| 53:36 | hydrostatic force acting on a thin strip as before, the strip is located a depth y with width w(y) |
| 54:05 | and height (delta)y
So the area of the strip is going to be as before, it's going to be w(y)*(delta)y.
So the only thing that changes now is the width, w(y). Because you see the triangle is submerged 4m down. |
| 54:40 | y no longer measures the distance of the side w(y) from the top of the triangle, it measures the distance from the surface of the water.
So if we were to do similar triangles as before, which we want to do in any case, we have to divide 1(y) by 2, on one side, |
| 55:08 | But when we write the proportionality relation, we have to take into account the length of this side from here to here.
And divide it by the total length of the triangle. So over here I'm going to write 3 because it's the length of the vertical side of the triangle, but although I need to put the length of this segment from here to here. |
| 55:36 | And now this is no longer going to be y because y is the distance from here all the way to the top, so it's going to be something different, can you help me?
Can you help me figure out what the length of this side is? y-4, exactly. |
| 56:00 | So this one over here is y-4. And this is exactly what [?] So let me use colored chalk, |
| 56:35 | Ok, so it means that w(y) is going to be (2*(y-4))/3.
And we can go back to our thing over here and write the area as (2*(y-4))/3*(delta)y. |
| 57:05 | Ok, so the hydrostatic force acting on the thin strip is going to be as before a*p The pressure hasn't really changed so it's just (rho)(g)(y) |
| 57:37 | And for the area we just plug in, So it's (2*(y-2))/3 * delta(y). So just 2/3(rho)(g)(y)(y-4)(delta)y |
| 58:00 | Can you see the board or?
It's probably too late to ask anyway. more just for the peace of my mind. Ok, so this is the hydrostatic force acting on the strip, and now to compute the whole hydrostatic force, we need to perform the usual integral. |
| 58:32 | Let me write it over here so we know what example it corresponds to.
So the total hydrostatic force is going to be an interval. Now let's just take the formula right here, |
| 59:00 | I put it as an integrand in the integral, transforming the (delta)y conveniently into dy of integration.
It's going to be (2/3)*rho*g*y*(y-4)*dy. And now I need to put some bounds on the integral so what are the bounds? Do I put 0-3 as before? No obviously not because you see that the object is no longer at the surface of the water so what do I put? |
| 59:46 | So what is the proper domain of y? What is y measuring? y is the depth, but we only have a hydrostatic force if we are within the area of the object.
I mean we are not computing the hydrostatic force over here, |
| 60:05 | So y is supposed to measure how far the object stretches, I mean y measures the depth. But the bounds that we put on the integral, are supposed to measure how far our object stretches vertically, measured with respect to the y coordinates.
So what is the y coordinate of the upper vertex of this triangle? 4. So 4 is the lower bound. |
| 60:32 | And what is the y coordinate of the deepest point in the triangle? 7.
So this is 7. Good? Ok, so again we've got an integral which is not more complicated than the previous one (2/3)*rho*g is a constant that can be factored outside. |
| 61:02 | So (2/3)*rho*g * integral from 4-7, of this function right here which is y*(y-4) and to integrate it I change it into y^2-4y And the integral causes no problems as before, so this is just gonna be *2/3)*rho*g*(y^3/3)-2y^2 |
| 61:55 | And let me throw in some computations over here, so we don't waste time, |
| 62:01 | So after you evaluate this you get something like 176400. So you get some computations how do you compute it? Well rho is replaced by 1000, g is replaced by 9.8,
And you know how to evaluate that function at the lower and upper bounds.
|
| 62:32 | I know I'm cheating a little bit, it's the advantage of having some notes, and you're not going to test me on whether this is correct or not. So if you discover a mistake well, you'll discover it well after the class is over.
I don't want to challenge you so In any case, you see that the hydrostatic force has increased from the previous example. |
| 63:01 | As it is normal, as it is submerged deeper under water so naturally the force has increased
Ok, so this is the complication that makes it more [?] .You have the coordinate y, which measures the distance,
or which measures the depth, which measures the distance under water.
Now this is the final computation, but for our exercise, let's see what would have happened if we chose to denote our y as the distance from the upper vertex of the triangle. |
| 63:38 | So let's do a tiny bit of exercise over here,
And let's see what changes if instead of y measuring the depth we let y measure the distance from the upper vertex of the triangle.
|
| 64:08 | This is the triangle, and now y measures the distance from the upper vertex.
So this is y. This is (delta)y, this is w(y) and the distance from the surface is 4, |
| 64:36 | So here to here is 4, the delta side of the triangle is 3,
And the total side is 2.
Now suppose we do that, what changes? I just want to show you that we get the same integral but translated so, by a substitution you get the same integral so it doesn't really matter how you choose to denote this. |
| 65:05 | So what changes over there? Y denotes the distance over w(y), from the upper side of the triangle, ok but, this is exactly the same picture that we have in example number 2.
So it means that when we say w(y), we use the same proportionality relations from that triangle. |
| 65:30 | So it means that when we do it over here, and when we compute the answer, you have to use the formula which is w(y)*(delta)y But w(y) is the same as w(y) we got over here. It's equal to w(y)/2 = y/3 |
| 66:11 | So this is just going to be the same as in the previous example, in the first example, it's going to be 2y/3 * (delta)y So this is asking sample 1, as in that example where the triangle is sitting at the top of the water, |
| 66:34 | But when we compute the force, we are going to see a tiny change right here,
because if we choose to let y denote the distance from the vertex, and if we want to compute the hydrostatic pressure at level y,
Then you know the hydrostatic pressure depends on the depth. It doesn't depend on the distance from a random point.
|
| 67:04 | So if we want to compute the hydrostatic pressure at that point, at y, we have 4m extra.
So here, I'm just going to change it to say P(y) = (rho)(g)(y+4), natural? |
| 67:30 | It's just because y is the distance from this point to the top of the triangle, and the total distance to the top of the water is that plus 4.
Good? Ok, so it means that when you write the integral, the corresponding integral, you're going to get something like, (integral of) (rho)g(y+4)*(2y/3)dy |
| 68:14 | And now what bounds do I put on the integral? Well, you see in this example, y measures the distance of this point from the upper vertex which means that y is in a range along the vertical length of the object.
|
| 68:32 | So if you are at the upper vertex of the triangle, it means that y is 0, if you are at the bottom of the triangle it means that y is 3.
So you would then use 0 and 3. And we love the connection between these 3 integrals, so this is an integral, just compare this one to the one in yellow, that's an integral between 0 and 3 overall, (rph)*g(y+4)*2y/3*dy |
| 69:07 | And this is an integral between 4 and 7 of (2/3)(rho)*g*(y-4)ydy You you should see the straight forward connection in terms of substitution, you should see that you can choose to do substitution, lets say over here, and let u denote y-4 Then we would immediately get the other integral.So if I do this integral, I do the substitution, u = y-4, du=dy |
| 69:48 | And you see that you immediately get that.
ok guys, so in practice, [?] because we have a triangle its hard to further and so on. But in practice, if you want to make your computations easy, |
| 70:08 | You let y denote the distance from the surface of the water because it helps you compute the hydrostatic pressure with not further computations.
So you let y denote the distance from the surface of the water, compute the pressure and then approximate the force. |