00:01
Okay, so we have a semi -circular window that's underwater.
00:08
The bottom of the window is 14 meters below the surface of the water.
00:12
The window itself is 5 meters in radius.
00:18
Okay, so here's the density of water and the acceleration of gravity.
00:24
So we want to find the force on the window.
00:28
Okay, so let's draw a narrow slice and we'll call that dy, that thickness.
00:42
And then i'm gonna call this distance here x.
00:54
And so the differential of force is the pressure times the area.
01:04
Okay, so pressure is rho g y.
01:11
So what's y? let's measure y down to where our slice is from the water level.
01:21
So it's rho g y dy because times the area times the length.
01:35
Okay, because x dy, that's the area of our little slice.
01:43
Okay, y is the depth and so the force is given like this because it's pressure.
01:55
Pressure is rho g h times the area gives us force.
02:03
Okay, and then so if i draw this again with fewer marks.
02:16
So here's our slice.
02:20
Okay, what we want to do is we want to get a relationship between x and y.
02:32
So r is this distance here.
02:42
Okay, and then that's also r by the way.
03:01
Then the height above the bottom is, we can get that, there's a pythagorean rule here.
03:12
So this is our distance x.
03:14
So we get x over 2 squared plus y, let's say h squared, not y.
03:26
That's h equals r squared because we got that right triangle there.
03:38
That's this half.
03:40
Okay, so what's h? h, well h plus y is equal to 14.
03:52
That's our 14 meters.
03:55
Okay, so i need to get, i need to solve this for x in terms of y.
04:08
So let's bring this over here.
04:11
So we get x over 2 squared plus h is 14 minus y squared and that equals r squared.
04:28
R is 5.
04:31
Okay, so x is 4, this 2 times the square root of r squared minus 14 minus y.
04:49
Quantity squared.
04:53
So our differential force is rho g y times x, which is 2 times the square root dy.
05:25
Okay, and so to find the total force, i have to integrate from y is 14 minus r down to 14.
05:40
So the force is going to be the integral of df.
05:46
And i said, so our upper, our lower limit, the shortest distance on y is 14 minus r to 14.
06:03
And then we get rho g y times 2 times the square root.
06:19
Okay, so let's make a substitution that says y is 14 plus r sine theta.
06:46
And then we get r squared minus 14 minus y squared.
06:54
That's equal to r squared minus r squared sine squared theta, which is r squared co -squared theta.
07:05
Okay, and then our limits.
07:08
So when y is 14 minus r, that means that sine theta equals minus 1.
07:28
And when y equals 14, that gives me sine theta equals zero.
07:40
So sine theta is minus 1 tells me theta is minus pi over 2.
07:47
And this tells me theta is zero.
07:49
So that tells me what my new limits are...