# Help with hydrostatic force questions

• MHB
• upnorthgy
In summary: Can you do this?In summary, the rectangular swimming pool has a depth of 4 feet, length of 75 feet, and width of 25 feet. The hydrostatic force on the bottom of the pool is 468209.7 lbs, on the shorter wall it is 312139.8 lbs, and on the longer wall it is 1248559.2 lbs. The work required to pump all the water out of the pool is 468209.7 ft-lbs.
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-yi) as goes from 0 to 4, but I am far from confident. And the rest has me baffled.

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?

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?

upnorthgy said:
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?

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

Fizgig said:
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.

## 1. What is hydrostatic force?

Hydrostatic force is the force exerted by a fluid on a submerged object. It is a result of the pressure differences within the fluid and can be calculated using the formula F = ρghA, where ρ is the density of the fluid, g is the acceleration due to gravity, h is the depth of the submerged object, and A is the surface area of the object.

## 2. How is hydrostatic force related to buoyancy?

Hydrostatic force is closely related to buoyancy, as both involve the pressure differences within a fluid. Buoyancy is the upward force that a fluid exerts on an object that is partially or fully submerged in it. This force is equal to the weight of the displaced fluid and is responsible for objects floating or sinking in a fluid.

## 3. What factors affect the magnitude of hydrostatic force?

The magnitude of hydrostatic force is affected by several factors, including the density and depth of the fluid, the surface area and shape of the submerged object, and the acceleration due to gravity. The direction of the force is always perpendicular to the surface of the object and towards the center of the fluid.

## 4. How is hydrostatic force used in real-life applications?

Hydrostatic force has many practical applications, such as in hydraulics and hydrology. It is used to calculate the forces on dams, retaining walls, and other structures that are in contact with a fluid. It is also important in civil engineering, as it helps determine the stability of structures and the effects of fluid pressure on them.

## 5. How can I calculate hydrostatic force on irregularly shaped objects?

In order to calculate hydrostatic force on irregularly shaped objects, you will need to use calculus and the concept of integration. By breaking the object into smaller, simpler shapes and integrating the forces on each part, you can find the total hydrostatic force on the object. Alternatively, you can use experimental methods, such as a force sensor or a pressure gauge, to directly measure the hydrostatic force on the object.

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