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Chapter 11: Problem 31

The vertical surface of a reservoir dam that is in contact with the water is \(120\) \(\mathrm{m}\) wide and \(12\) \(\mathrm{m}\) high. The air pressure is one atmosphere. Find the magnitude of the total force acting on this surface in a completely filled reservoir. (Hint: The pressure varies linearly with depth, so you must use an average pressure.)

Short Answer

Expert verified
The magnitude of the total force is 84,758,400 N.

Step by step solution

01

Understand the Pressure Variation

The pressure exerted by a liquid column increases linearly with depth. The pressure at any depth below the water surface is given by the formula \( P = \rho g h \), where \( \rho \) is the density of water (approximately \( 1000 \, \mathrm{kg/m^3} \)), \( g \) is gravitational acceleration (\( 9.81 \, \mathrm{m/s^2} \)), and \( h \) is the depth below the surface. The pressure varies from zero at the surface to a maximum at the bottom.
02

Calculate the Average Pressure

Since the pressure at the top of the reservoir is \( 0 \) and at the bottom is \( \rho g H \), where \( H = 12 \, \mathrm{m} \) is the height of the reservoir, the average pressure over the height can be considered as \( \frac{0 + \rho g H}{2} = \frac{\rho g H}{2} \).
03

Calculate the Average Pressure Value

Substitute the known values into the average pressure formula: \( \rho = 1000 \, \mathrm{kg/m^3} \), \( g = 9.81 \, \mathrm{m/s^2} \), and \( H = 12 \, \mathrm{m} \). So, the average pressure \( \bar{P} = \frac{1000 \times 9.81 \times 12}{2} \). Simplifying gives \( \bar{P} = 58860 \, \mathrm{N/m^2} \) or \( 58860 \, \mathrm{Pa} \).
04

Calculate the Total Force Using Average Pressure

The total force exerted by the water on the surface of the dam is given by multiplying the average pressure by the surface area of the dam in contact with the water: \( F = \bar{P} \times A \), where \( A = 120 \, \mathrm{m} \times 12 \, \mathrm{m} = 1440 \, \mathrm{m^2} \).
05

Compute the Total Force

Substitute the obtained values into the force equation: \( F = 58860 \, \mathrm{N/m^2} \times 1440 \, \mathrm{m^2} = 84758400 \, \mathrm{N} \). This is the magnitude of the total force acting on the dam's vertical surface.

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Key Concepts

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

Hydrostatic Pressure
In fluid mechanics, understanding hydrostatic pressure is key to addressing problems involving liquids in containers or under gravity. Hydrostatic pressure is the pressure exerted by a fluid in equilibrium due to the force of gravity. It increases with depth, since more fluid above the point adds to the weight. This can be calculated using the formula \( P = \rho g h \):
  • \( \rho \) stands for the density of the fluid, typically water, which is approximately \( 1000 \, \mathrm{kg/m^3} \).
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \mathrm{m/s^2} \) on Earth.
  • \( h \) represents the depth below the surface.
At the surface of the fluid, the pressure is zero (ignoring atmospheric pressure), and increases linearly as you move downward. This is why calculating hydrostatic pressure is crucial for understanding forces on submerged surfaces, such as dams.
Force Calculation
Calculating the force on a submerged surface requires understanding the pressure distribution along the depth of the fluid. In exercises like these, pressure is not constant, but increases with depth. Hence, the force, which is pressure applied over an area, must consider how this pressure varies.
Start by identifying the surface area that is in contact with the water. In our example, the dam surface is \( 120 \, \mathrm{m} \) wide and \( 12 \, \mathrm{m} \) high. Therefore, the area \( A \) is \( 120 \, \mathrm{m} \times 12 \, \mathrm{m} = 1440 \, \mathrm{m^2} \).
To find the force, multiply this area by the average pressure exerted by the fluid at the dam's surface. The formula is \( F = \bar{P} \times A \). By using average pressure, you ensure that the linear variation of pressure with depth is accounted for.
Average Pressure
Average pressure is an important concept when dealing with forces on submerged surfaces. Since pressure increases linearly from zero at the top to a maximum at the bottom, average pressure provides an effective measure over the entire height.
To compute average pressure, use the formula \( \frac{\rho g H}{2} \), where \( H \) is the maximum depth. By substituting known values (
  • \( \rho = 1000 \, \mathrm{kg/m^3} \)
  • \( g = 9.81 \, \mathrm{m/s^2} \)
  • \( H = 12 \, \mathrm{m} \)
), the average pressure becomes \( 58860 \, \mathrm{N/m^2} \).
Using this average pressure is critical in calculation because it simplifies the process of obtaining the total force by integrating complex pressure variations into a single effective value, making mathematical handling more straightforward.

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Most popular questions from this chapter

An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Because of the aneurysm, the normal cross-sectional area \(A_{1}\) of the aorta increases to a value of \(A_{2}=1.7 A_{1} .\) The speed of the blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}\right)\) through a normal portion of the aorta is \(v_{1}=0.40 \mathrm{m} / \mathrm{s}\) Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure \(P_{2}\) in the enlarged region exceeds the pressure \(P_{1}\) in the normal region.

The atmospheric pressure above a swimming pool changes from 755 to \(765\) \(\mathrm{mm}\) of mercury. The bottom of the pool is a rectangle \((12 \mathrm{m} \times 24 \mathrm{m}) .\) By how much does the force on the bottom of the pool increase?

A Venturi meter is a device that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed \(v_{2}\) through a horizontal section of pipe whose cross-sectional area is \(A_{2}=\) \(0.0700 \mathrm{m}^{2} .\) The gas has a density of \(\rho=1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The Venturi meter has a cross-sectional area of \(A_{1}=0.0500 \mathrm{m}^{2}\) and has been substituted for a section of the larger pipe. The pressure difference between the two sections is \(P_{2}-P_{1}=120\) Pa. Find (a) the speed \(v_{2}\) of the gas in the larger, original pipe and (b) the volume flow rate \(Q\) of the gas.

A pressure difference of \(1.8 \times 10^{3}\) Pa is needed to drive water \(\left(\eta=1.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) through a pipe whose radius is \(5.1 \times 10^{-3} \mathrm{m}\) The volume flow rate of the water is \(2.8 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s} .\) What is the length of the pipe?

When an object moves through a fluid, the fluid exerts a viscous force \(\overrightarrow{\mathbf{F}}\) on the object that tends to slow it down. For a small sphere of radius \(R,\) moving slowly with a speed \(v,\) the magnitude of the viscous force is given by Stokes' law, \(F=6 \pi \eta R v,\) where \(\eta\) is the viscosity of the fluid. (a) What is the viscous force on a sphere of radius \(R=5.0 \times 10^{-4} \mathrm{m}\) that is falling through water \(\left(\eta=1.00 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) when the sphere has a speed of \(3.0 \mathrm{m} / \mathrm{s} ?\) (b) The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed." If the sphere has a mass of \(1.0 \times 10^{-5} \mathrm{kg},\) what is its terminal speed?
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