# Hydrostatic force

• Jul 9th 2008, 06:06 AM
porknbeans
Hydrostatic force
http://img.photobucket.com/albums/v9...y110/calc2.jpg

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

Help~~
• Jul 9th 2008, 08:37 AM
galactus
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 $\displaystyle y=-2x+12$ or $\displaystyle x=6-\frac{y}{2}$

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

So, we can set up the integral:

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

You could also try similar triangles and see if you get the same thing.
• Jul 9th 2008, 04:19 PM
ticbol
Quote:

Originally Posted by porknbeans
http://img.photobucket.com/albums/v9...y110/calc2.jpg

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

Help~~

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

Since the given A is in feet or English, then the unit weight of water is 62.4 lbs per cubic ft., so,
$\displaystyle \rho$ = (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.
• Jul 9th 2008, 05:29 PM
galactus
I made a mistake by using the wrong coordinates for my line equation on the last post(Headbang), but I fixed it. Thanks for the heads up ticbol.