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# Hydrostatic Pressure Help

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## Introduction to Hydrostatic Pressure

If a liquid sits in a tank, then it exerts force on the side of the tank. This force is caused by gravity, and the greater the depth of the liquid then the greater the force. Pascal’s principle asserts that the force exerted by a body of water depends on depth alone, and is the same in all directions. Thus the force on a point in the side of the tank is defined to be the depth of the liquid at that point times the density of the liquid. Naturally, if we want to design tanks which will not burst their seams, it is important to be able to calculate this force precisely.

Fig. 8.35

Imagine a tank of liquid having density ρ pounds per cubic foot as shown in Fig. 8.35. We want to calculate the force on one flat side wall of the tank. Thus we will use the independent variable h to denote depth, measured down from the surface of the water, and calculate the force on the wall of the tank between depths h = a and h = b (Fig. 8.36). We partition the interval [ a, b ]:

a = h 0h 1h 2 ≤ ... ≤ h k −1h k = b.

Fig. 8.36

Assume that the width of the tank at depth h is w ( h ). The portion of the wall between h = h j −1 and h = h j is then approximated by a rectangle R j of length w ( h j ) and width Δ h = h jh j −1 (Fig. 8.37).

Fig. 8.37

Now we have the following data:

Area of Rectangle = w ( h j ) · Δ h square feet

Depth of Water ≈ h j feet

Density of Liquid = ρ pounds per cubic foot.

It follows that the force exerted on this thin portion of the wall is about

P j = h j · ρ · w ( h j ) · Δ h .

Adding up the force on each R j gives a total force of

But this last expression is a Riemann sum for the integral

#### Example 1

A swimming pool is rectangular in shape, with vertical sides. The bottom of the pool has dimensions 10 feet by 20 feet and the depth of the water is 8 feet. Refer to Fig. 8.38 . The pool is full. Calculate the total force on one of the long sides of the pool.

Fig. 8.38

#### Solution 1

We let the independent variable h denote depth, measured vertically down from the surface of the water. Since the pool is rectangular with vertical sides, w (h) is constantly equal to 20 (because we are interested in the long side). We use 62.4 pounds per cubic foot for the density of water. According to (*), the total force on the long side is

You Try It: A tank full of water is in the shape of a cube of side 10 feet. How much force is exerted against one wall of the tank between the depths of 3 feet and 6 feet?

#### Example 2

A tank has vertical cross section in the shape of an inverted isosceles triangle with horizontal base, as shown in Fig. 8.39 . Notice that the base of the tank has length 4 feet and the height is 9 feet. The tank is filled with water to a depth of 5 feet. Water has density 62.4 pounds per cubic foot. Calculate the total force on one end of the tank.

Fig. 8.39

#### Solution 2

As shown in Fig. 8.40, at depth h (measured down from the surface of the water), the tank has width corresponding to the base of an isosceles triangle similar to the triangle describing the end of the tank. The height of this triangle is 5 − h . Thus we can solve

We find that

According to (*), the total force on the side is then

Fig. 8.40

#### Example 3

An aquarium tank is filled with a mixture of water and algicide to keep the liquid clear for viewing. The liquid has a density of 50 pounds per cubic foot. For viewing purposes, a window is located in the side of the tank, with center 20 feet below the surface. The window is in the shape of a square of side feet with vertical and horizontal diagonals (see Fig. 8.41). What is the total force on this window?

Fig. 8.41

#### Solution 3

As usual, we measure depth downward from the surface with independent variable h. Of course the square window has diagonal 4 feet. Then the range of integration will be h = 20−4 = 16 to h = 20 + 4 = 24. Refer to Fig. 8.42 . For h between 16 and 20, we notice that the right triangle in Fig. 8.42 is isosceles and hence has base of length h − 16. Therefore

w (h) = 2(h − 16) = 2 h − 32.

Fig. 8.42

According to our analysis, the total force on the upper half of the window is thus

For the lower half of the window, we examine the isosceles right triangle in Fig. 8.43 . It has base 24 − h. Therefore, for h ranging from 20 to 24, we have

w (h) = 2(24 − h) = 48 − 2 h.

According to our analysis, the total force on the lower half of the window is

The total force on the entire window is thus

Fig. 8.43

You Try It : A tank of water has flat sides. In one side, with center 4 feet below the surface of the water, is a circular window of radius 1 foot. What is the total force on the window?

Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.

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