## formulas help

this make sense? sove this

I have a tank of water. rectangular 26 inches tall by 8 by 8. There is a bottom of 1 square inch with two interior walls angling up from there to the exterior walls 11 inches up. they have an angle of roughly 55 degrees steep. I need to know the water pressure at the 1 inch square bottom
and somebody gave me a formula that i dont get.

it says for a tank with only one interior wall slanting upward to the wall,
and a strip of infinitesimal thickness on a horizontal plane partway up the slant, there is delta h for the thickness strip

he says Volume of the strip is shown as
Delta V = 1/2 Delta h w (l+l+ Delta h/tan )

h is height
w is width
l is length

Then I have

the mass of the volume of liquid is shown as
Delta m = p Delta V

p is density of liquid ( water)

the force exerted by the water due to gravity is

Delta f = g Delta m = pgw Delta h ( l + Delta h/ 2 tan)

g is acceleration due to gravity

therefore

lim
Delta h to 0 Delta f/delta h (three line equals) df/dh = pgwl(h)

h to 0 is a zero or degree as used in sin cos tan etc

To get the total force of at location h = h1 due to water between locations h = h1 and h= h2 (h2 >h1) we integrate and get

f(h1) = pgw ! l(h)dh

! is a large symbol like a slant line with the top towards the right,
and tiny curls at top and bottom with top towards right and bottom toward left. It barly looks like a "S" cause its reallly tall with tiny curls.
it also has a tiny h2 at top and h1 at bottom. sorry i dont know what the symbol is called.

pressure at h=h1 is

p(h1) = pg/l(h1) ! l(h)dh

if l(h1) = l 0 = 0 an infinite pressure is obtained

if l(h) = l 0 + h cot

then p(h1) = pg/l0 + h1 cot [l0 (h2 - h1) + cot/2 ( h 2 over 2 -

h 2 over 1)]

l 0 is same zero or degree

for a vertical wall 0 = pie/2 and we have
p(h1) = pg (h2 -h1)

and if h1 = 0 and h2 = h, the pressure at bottom is pgh

0 as in zero