# Thread: Fluid Force at an angle

1. ## Fluid Force at an angle

I can do vertically submerged fluid force problems. One portion of my book says:
If a flat surface is immersed so that it makes an angle of $\displaystyle 0 \leq \theta \geq \frac{\pi}{2}$ with the vertical, then the fluid force on the surface is given by:
$\displaystyle F=\int^b_a \rho h(x)w(x)sec\theta dx$

Can someone explain exactly what the sec(theta) is giving me; how exactly this equation changing the original equation of a vertically submerged surface. I never took trig., so please dumb it down in that area

*EDIT* If it helps, just made a diagram. Is this the correct angle in the equation?

2. I found a suitable example problem. on page 490 of this book

Calculus: Early Transcendentals - Google Books

It seems a little clearer, now how does it change if we submerge the inclined figure further, such that the top no longer "kisses" the surface?

It would seem to me that if the figure were submerged an additional 4 feet, and our height function, h(x) became y+4, the sec(@) or sin@, whatever we choose our reference to be, would skew that plus 4 height...

3. If x is measured vertically, and y along the surface, then we have a right triangle with hypotenuse y and "near side" x: $\displaystyle \frac{x}{y}= cos(\theta)$ so $\displaystyle \frac{y}{x}= \frac{1}{cos(\theta)}= sec(\theta)$.

$\displaystyle y= x sec(\theta)$ so $\displaystyle dy= sec(\theta)dx$.

The area of a small rectangle along the surface is $\displaystyle w(x)dy= w(x)sec(\theta)dx$.

4. Now what if we sumberge the shape further under the fluid an additional height d?

I've tried these:

$\displaystyle \int^b_a \rho w(x) h(x+d)sec\theta dx$

$\displaystyle \rho V + \int^b_a \rho w(x)h(x+d)sec\theta dx$(The V here being the volume of water above the submerged surface)

$\displaystyle \rho V + \int^b_a \rho w(x)h(x)sec\theta dx$
$\displaystyle \rho V + \int^b_a \rho w(x)h(x+d)sec\theta dx$

But none of them seem to work.

(P.S. Noseck, the ship's library was open, but thanks anyway :] )

5. Ok, i might as well come out with the problem. Please find my error:
A pool is 10 ft wide and 16 ft long, One end has a depth of 4 feet, the other end has a depth of 8 ft (therefore, the bottom of the pool makes an inclined surface) Find the Total Fluid pressure on the pool's bottom.

I can't make a picture right now, due to limitations of this computer (I'm on a Navy Ship, the computer doesn't have paint)

so basically an inclined plane submerged 4 feet.

For a visual, follow along with the last example of this:
12.9 Force Exerted By A Fluid
I'm following this example word for word and still missing something

I'f I go down the shallow end a height h, and follow it over to the inclined surface I get a small dh that forms a rectangle with width dw

$\displaystyle cos\theta = \frac{dh}{dw}$
$\displaystyle dw=\frac{dh}{cos\theta }$
by similar triangles, $\displaystyle cos\theta = \frac{4}{\sqrt{4^2+16^2}}$
$\displaystyle cos \theta = \frac{1}{\sqrt{17}}$
Therefore$\displaystyle dw= \sqrt{17} dh$------------------------(1)
now the little rectangle has an area dA such that:
$\displaystyle dA=dw10$ from (1): $\displaystyle dA=10\sqrt{17} dh$

Weight Densiy is $\displaystyle \rho = 62.4$ therefore,
Since Pressure is weight density x Height, the Pressure on this rectangle at depth h+4 is $\displaystyle \rho (h+4) = 62.4(h+4)$

Force=Pressure x Area The force dF on this rectangl is:
$\displaystyle dF=(62.4)(h+4)(dA)= 62.4 (h+4) 10 \sqrt{17}$

That was on ONE rectangle, to get the Total Force, take the integral
$\displaystyle \int^4_0 624\sqrt{17}(h+4) dh$
$\displaystyle 624\sqrt{17}(\frac{h^2}{2}+4h)\mid^4_0$
This gives me:61,747