00:03
We're told that a cube with 20 centimeter long sides sits at the bottom of an aquarium, and the water in the aquarium is one meter deep.
00:19
We're asked to approximate the hydrostatic force on the top of the cube and on one of the sides of the cube.
00:36
Part a, add that the top of the cube has a depth of one meter minus the length of the side of the cube, which is 20 centimeters, which is going to be 80 centimeters, or in standard units, 0 .8 meters.
01:14
Therefore, the force, which is equal to the density row times gravitational constant g times the depth d times, that's the pressure, times the area of the face of the cube, well, this is going to be approximately, density is approximately 1 ,000, the gravitational constant is approximately 9 .8, the density, or sorry, the depth is 0 .8, and the area is going to be 0 .2 meters squared.
02:01
And so calculating we get that this is 313 .6, which is approximately 314.
02:14
And this is going to be in newton's.
02:17
Since we used meters in part b, well, the area of a strip is the length of a side of the cube .2 times the width of the strip, which is delta x.
02:50
And the pressure on the strip, this is going to be density row times gravitational constant g times the depth of the strip, which is xi star.
03:12
Therefore, the force on the side of the cube, on one side of the cube anyways, this is the integral from the depth of 0 .8 to 1...