1. ## Hydrostatic force

Find the hydrostatic force exerted on one end of the plate.

Help~~

2. If we let the surface of the water be the x-axis and bisect it with the y-axis,

then perhaps we copuld find the equation of the line which represents the

side of the plate.

The bottom right corner will have coordinates (10,-8) and the top right corner will have coordinates (6,0)

Using these two points to find our line equation we get $y=-2x+12$ or $x=6-\frac{y}{2}$

The depth is -y and the length of the region at y is $2x = 2\left(6-\frac{y}{2}\right)=12-y$

So, we can set up the integral:

$62.4\int_{-8}^{0}(y^{2}-12y)dy=34611.2$

You could also try similar triangles and see if you get the same thing.

3. Originally Posted by porknbeans

Find the hydrostatic force exerted on one end of the plate.

Help~~
Hydrostatic force F on a submerged area A is F = $\rho$*A
where $\rho$ = hydrostatic pressure at the centroid of A,
and A is perpendicular to $\rho$

Since the given A is in feet or English, then the unit weight of water is 62.4 lbs per cubic ft., so,
$\rho$ = (62.4*y) lbs/sq.ft ....where y is the depth from water surface.

The given figure of A can be subdivided into two ploygons:
>>>a rectangle that is 12ft long by 8ft high
>>>and an isosceles triangle that has a base of 8ft and an altitude of 8ft.

For the rectangle:
We make the water surface as the x-axis, and going down is positive.
dA1 = 12*dy ....the element of A1.
F1 = INT(0 to 8)[(62.4(y) *12dy]
F1 = (12*62.4)INT(0 to 8)[y]dy
F1 = 748.8[(y^2)/2](0 to 8)
F1 = 748.8[64/2 -0]
F1 = 23.962 kips

For the isosceles triangle:
We put the (0,0) at the apex, and going down is positive also.
We get the relationship between x and y for the right hypotenuse,
Let dA3 be anywhere between y = 0 and y = 8, so
base for dA3 = x
depth at dA3 = y
>>>dA3 is element for A3. A3 = area of one right triangle whose base is 4 and height is 8.
By proportion,
y/x = 8/4
4y = 8x
x = y/2
dA3 = x*dy = (y/2)dy
So, dA2 = (y/2 dy)*2 = y dy2 ....there are 2 right triangles in the isosceles
Then,
F2 = INT(0 to 8)[(62.4*y)(y dy]
F2 = (62.4)INT(0 to 8)[y^2]dy
F2 = (62.4 /3)[y^3](0 to 8)
F2 = 20.8[(8)^3 -0]
F2 = 10.65 kips

Therefore, F = F1 +F2 = 23.962 +10.65 = 34.612 kips ----answer.

4. I made a mistake by using the wrong coordinates for my line equation on the last post, but I fixed it. Thanks for the heads up ticbol.