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Chapter 12: Problem 55

A swimming pool is \(5.0 \mathrm{~m}\) long, \(4.0 \mathrm{~m}\) wide, and \(3.0 \mathrm{~m}\) deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h,\) and integrate this over the end of the pool.) Do not include the force due to air pressure.

Short Answer

Expert verified
The force exerted by the water on the bottom of the swimming pool is \(588000 \mathrm{~N}\) and on either end is \(176400 \mathrm{~N}\).

Step by step solution

01

Identify Relevant Information

The length, width and depth of the pool are given as \(5.0 \mathrm{~m}, 4.0 \mathrm{~m}, 3.0 \mathrm{~m}\) respectively. The density of water is assumed to be \(1000 \mathrm{~kg/m}^3\), and acceleration due to gravity is \(9.8 \mathrm{~m/s}^2\).
02

Determine the Force on Bottom

The pressure at the bottom is given by the equation \(P=density*gravity*height\). So the pressure can be calculated as \(P=1000*9.8*3=29400 \mathrm{~Pa}.\) Now, force is the product of pressure and area, that is \(F=P*A\). So, the force exerted on the bottom of the swimming pool is \(F=29400*5*4=588000 \mathrm{~N}.\)
03

Determine the Force on Ends

Here we need to consider a thin strip at the depth \(h\) with thickness \(dh\). This strip's width is \(4.0 \mathrm{~m}\) (width of the pool), while its area can be represented as \(4.0 \mathrm{~m} * dh\).\nFrom previous step, we know that the pressure at depth \(h\) is \(P=1000*9.8*h\). The force exerted on the strip is \(dF=P*dA = 1000*9.8*h*4*dh\).\n To calculate the total force on the end, this force should be integrated from \(h = 0\) to \(h = 3\):\n \(F= \int_{0}^{3} 39200h dh = \left[19600h^2\right]_0^3 = 19600*3^2 = 176400 \mathrm{~N}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest, due to the influence of gravity. In a swimming pool or any liquid sitting still, the weight of the fluid creates pressure at different depths. This pressure increases linearly with depth.

For example, at the bottom of a pool, the hydrostatic pressure can be calculated using the formula:
  • Pressure \[P = \text{density} \times \text{gravitational acceleration} \times \text{depth} \]
Let's break this down:
  • **Density** is how much mass is contained in a given volume, for water it's typically \(1000 \text{ kg/m}^3\).
  • **Gravitational acceleration** is due to Earth's gravity, approximately \(9.8 \text{ m/s}^2\).
  • **Depth** is how far down you measure in the fluid, in our pool, this is \(3 \text{ m}\).
So at the bottom of our pool, the pressure is \(29400 \text{ Pa}\). This concept is crucial for understanding force calculations on submerged surfaces.
Integration in Calculus
Integration is a key part of calculus used to calculate quantities such as areas, total accumulations, or in our case, forces across surfaces with varying pressure.

In the swimming pool problem, we integrate to find the total force exerted by water on the ends of the pool, due to varying pressure at different depths. Here's how it applies:
  • Divide the surface into thin horizontal strips, each having a small thickness \(dh\).
  • The pressure at each strip's depth \(h\) is factored into the strip's force, calculated as \(dF\).
  • Finally, integrate \(dF\) over the depth of the pool to sum them up. In math terms, the total force \(F\) is:\[ F = \int_{0}^{3} 39200h \, dh \]
This integration sums the forces of all strips from top to bottom, giving us \(176400 \text{ N}\). Understanding integration helps solve numerous fluid mechanics problems.
Force Calculation
When dealing with forces in fluid mechanics, it is essential to consider both pressure and area. The force exerted by a fluid is the product of pressure and the area over which it acts.

In our swimming pool example:
  • The force on the bottom equals the pressure at that depth times the area, calculated as:\[ F = P \times A = 29400 \text{ Pa} \times (5 \times 4) \text{ m}^2 = 588000 \text{ N} \]
  • On the pool's ends, the force is determined by integrating the pressure over the area, yielding \(176400 \text{ N}\).
Considering both pressure and area ensures accurate force calculations. This is vital not only in pools but in engineering applications like designing dams, submarines, or even simple water tanks.
Density of Water
The density of water, often a standard value at \(1000 \text{ kg/m}^3\), plays a crucial role in hydrostatics. It is essential for calculating pressures and subsequent forces in fluids.

Density is defined as mass per unit volume, and it dictates how much mass a particular volume of water contains. For the swimming pool problem, this value tells us how heavy the water sitting at any given depth is, thereby affecting the pressure and force calculations.
Understanding water's density is helpful for applications such as buoyancy calculations, hydraulic systems, and environmental science studies involving water bodies.
Gravitational Acceleration
Gravitational acceleration is the force of gravity that pulls objects toward the center of the Earth, valued at approximately \(9.8 \text{ m/s}^2\). This constant is central to determining the pressure a fluid exerts at a certain depth.

In our swimming pool context, it forms part of the pressure calculation as shown:
  • For pressure: \[ P = \text{density} \times \text{gravitational acceleration} \times \text{depth} \]
  • This ensures that the deeper you go, the more pressure you experience, directly scaling with gravity's pull and the overlying water mass.
Gravitational acceleration is thus necessary for any scenario where the weight of a fluid is involved, including engineering structures and environmental phenomena.

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Most popular questions from this chapter

A \(950 \mathrm{~kg}\) cylindrical buoy floats vertically in seawater. The diameter of the buoy is \(0.900 \mathrm{~m} .\) Calculate the additional distance the buoy will sink when an \(80.0 \mathrm{~kg}\) man stands on top of it.

A cube \(5.0 \mathrm{~cm}\) on each side is made of a metal alloy. After you drill a cylindrical hole \(2.0 \mathrm{~cm}\) in diameter all the way through and perpendicular to one face, you find that the cube weighs \(6.30 \mathrm{~N}\). (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

Estimate the fraction of your body's total volume that is above the surface of the water when you float in seawater (density \(1030 \mathrm{~kg} / \mathrm{m}^{3}\) ). (a) Use this estimate and your weight to calculate the total volume of your body. (b) What is your average density? How does your average density compare to the density of seawater?

Water runs into a fountain, filling all the pipes, at a steady rate of \(0.750 \mathrm{~m}^{3} / \mathrm{s}\). (a) How fast will it shoot out of a hole \(4.50 \mathrm{~cm}\) in diameter? (b) At what speed will it shoot out if the diameter of the hole is three times as large?

A golf course sprinkler system discharges water from a horizontal pipe at the rate of \(7200 \mathrm{~cm}^{3} / \mathrm{s}\). At one point in the pipe, where the radius is \(4.00 \mathrm{~cm},\) the water's absolute pressure is \(2.40 \times 10^{5} \mathrm{~Pa}\). At a second point in the pipe, the water passes through a constriction where the radius is \(2.00 \mathrm{~cm} .\) What is the water's absolute pressure as it flows through this constriction?
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