http://img.photobucket.com/albums/v9...y110/calc2.jpg

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

Help~~

Printable View

- July 9th 2008, 07:06 AMporknbeansHydrostatic force
http://img.photobucket.com/albums/v9...y110/calc2.jpg

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

Help~~ - July 9th 2008, 09:37 AMgalactus
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 or

The depth is -y and the length of the region at y is

So, we can set up the integral:

You could also try similar triangles and see if you get the same thing. - July 9th 2008, 05:19 PMticbol
Hydrostatic force F on a submerged area A is F = *A

where = hydrostatic pressure at the centroid of A,

and A is perpendicular to

Since the given A is in feet or English, then the unit weight of water is 62.4 lbs per cubic ft., so,

= (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. - July 9th 2008, 06:29 PMgalactus
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.