Help with hydrostatic force questions

[upnorthgy]

Here are the questions...

A rectangular swimming pool is 4 feet deep, 75 feet long, and 25 feet wide. It is completely filled with water.

a. What is the hydrostatic force on the bottom of the pool?
b. What is the hydrostatic force on the shorter wall of the pool?
c. What is the hydrostatic force on the longer wall of the pool?
d. How much work is required to pump all the water out of the pool (i.e., up to the top of the pool)?

As far as a. goes, I think it is the integral of (62.4 lbs/feet³)(75)(25)(4-y_i) as goes from 0 to 4, but I am far from confident. And the rest has me baffled.

 

[MarkFL]

a) If the depth of the pool is the same everywhere, then we have:

\(\displaystyle F=\rho hA\)

where:

\(\displaystyle h\) = the depth of the pool.

\(\displaystyle A\) = the surface area of the bottom of the pool.

b) Let's find the hydrostatic force on a vertically submerged rectangle, whose width is $w$ and height is $h$. If we orient a vertical $x$-axis along one side of the rectangle, with the origin at the surface and the positive direction down, we may compute the force along one horizontal elemental rectangle having area $A(x)$ making up the entire rectangle as:

\(\displaystyle dF=\rho xA(x)\)

The area of this elemental rectangle is its width at $x$ times its height $dx$, hence:

\(\displaystyle dF=\rho xw(x)\,dx\)

where:

\(\displaystyle w(x)=w\) (the width is constant for a rectangle)

And so we have:

\(\displaystyle dF=\rho w x\,dx\)

Now, if the top edge of the rectangle is at $x=x_1$ and the bottom edge is at $x=x_2$, then by summing all of the elements of the force, we find:

\(\displaystyle F=\rho w\int_{x_1}^{x_2}x\,dx\)

Now what do you find upon applying the FTOC?

 

[upnorthgy]

I guess I am still struggling with this conceptually. I think I computed your equation and the answer I got was 936,000 lbs. Do I need to reorient the coordinate system to get get the hydrostatic force against the wall. If so, how?

 

[MarkFL]

I guess I am still struggling with this conceptually. I think I computed your equation and the answer I got was 936,000 lbs. Do I need to reorient the coordinate system to get get the hydrostatic force against the wall. If so, how?

a) We have the following:

Weight of water per cubic foot: \(\displaystyle \rho=62.42796\,\frac{\text{lb}}{\text{ft}^3}\)

Depth: \(\displaystyle h=4\text{ ft}\)

Area of surface: \(\displaystyle A=\left(75\text{ ft}\right)\left(25\text{ ft}\right)=1875\text{ ft}^2\)

Thus, we find:

\(\displaystyle F=\left(62.42796\,\frac{\text{lb}}{\text{ft}^3}\right)\left(4\text{ ft}\right)\left(1875\text{ ft}^2\right)=468209.7\text{ lb}\)

b) The integral I set up above is already oriented correctly for a vertical surface (such as the sides of the pool)...what do you get when you compute the definite integral?

 

[Fizgig]

Understood part A, however I'm struggling on B. Would \omega just be 25? Also what would i plug in for x dx, 100? Thanks

 

[MarkFL]

Understood part A, however I'm struggling on B. Would \omega just be 25? Also what would i plug in for x dx, 100? Thanks

For part b), you would use:

\(\displaystyle F=25\rho \int_{0}^{4}x\,dx\)

You don't plug in for $x\,dx$, you simply evaluate the definite integral. Can you continue?

I would actually prefer to evaluate the definite integral:

\(\displaystyle F=\rho w\int_{x_1}^{x_2}x\,dx\)

first, and then plug in the given data.