00:01
For this problem, you're considering a dam that's holding back water.
00:09
We know that if we're looking at it head on, so this is our dam here.
00:17
The width of the dam is equal to 108.
00:25
I'm going to assume meters.
00:28
I didn't see units in the problem.
00:30
So i'm just going to assume meters, so 108 meters, and the height of the water is 196 meters.
00:45
And the question then is, what is the force from the water on the dam? and we can generally find this using the relationship that pressure is equal.
01:02
To force over area and thus the force from the water should be equal to pressure times area.
01:15
But unfortunately the pressure is going to vary as a function of depth within our fluid because pressure is of course equal to row g.
01:34
And so to go ahead and actually figure that out we want to consider a small piece here and that has with dh and determine what the force would be so d f is equal to row g and then the area of that little sliver is going to be w times dh.
02:24
And so we're going to have to integrate to solve this.
02:27
So we'll integrate both sides.
02:31
Oops.
02:35
So the integral of d f is going to be equal to the integral.
02:42
And we want to go for the entire height.
02:46
So i'm going to go from zero to h here.
02:51
Of row g w i'm going to group those altogether h d h and we integrate we get that the force is equal to row g w h squared over two evaluated from zero to h which is just going to be row g w h squared over two evaluated from zero to h which is just going to be row g w h squared over 2 and you can go ahead and plug in the information from the problem.
03:30
We're told that it's water so i used a density of 1 ,000 kilograms per cubic meter.
03:39
I used 9 .8 meters per second squared for g and then for w and h i used the values from the problem so 108 meters for w and 109 meters for w and 96 meters for h...