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A large aquarium of height \(5.00\) \(\mathrm{m}\) is filled with fresh water to a depth of \(2.00\) \(\mathrm{m}\). One wall of the aquarium consists of thick plastic \(8.00 \mathrm{~m}\) wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of \(4.00 \mathrm{~m}\) ?

Short Answer

Expert verified
470,880 N

Step by step solution

01

Understanding Hydrostatic Pressure

The pressure at a depth in a fluid is given by the hydrostatic pressure formula: \( P = \rho gh \), where \( \rho \) is the fluid density, \( g \) is the acceleration due to gravity (approximated to \( 9.81 \, \mathrm{m/s^2} \)), and \( h \) is the depth. For water, \( \rho \approx 1000 \, \mathrm{kg/m^3} \).
02

Calculating Force on the Wall

Force exerted by a fluid on a surface at a certain depth is the product of pressure and area: \( F = P A = (\rho gh) A \). Since we need to calculate this at different depths, we first express it as \( F = \frac{1}{2} \rho g h^2 w \), where \( w \) is the width of the wall.
03

Force at 2.00 m Depth

Using the formula, \( F = \frac{1}{2} \rho g h^2 w \), for \( h = 2.00 \, \mathrm{m} \): \( F_1 = \frac{1}{2} \times 1000 \, \mathrm{kg/m^3} \times 9.81 \, \mathrm{m/s^2} \times (2.00 \, \mathrm{m})^2 \times 8.00 \, \mathrm{m} = 156,960 \, \mathrm{N} \).
04

Force at 4.00 m Depth

Similarly, calculate for \( h = 4.00 \, \mathrm{m} \): \( F_2 = \frac{1}{2} \times 1000 \, \mathrm{kg/m^3} \times 9.81 \, \mathrm{m/s^2} \times (4.00 \, \mathrm{m})^2 \times 8.00 \, \mathrm{m} = 627,840 \, \mathrm{N} \).
05

Calculating Increase in Force

The increase in force when the aquarium is filled from 2.00 m to 4.00 m is \( F_2 - F_1 = 627,840 \, \mathrm{N} - 156,960 \, \mathrm{N} = 470,880 \, \mathrm{N} \).

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Key Concepts

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

Fluid Mechanics
Fluid Mechanics helps us understand how liquids behave and interact with their surroundings. It is essential for solving a variety of problems related to how fluids flow, exert pressure, and interact with different surfaces. In the case of the aquarium problem, we use fluid mechanics principles to understand how water puts pressure on the walls of the aquarium when it's filled to different depths. The main idea is that the deeper something is submerged in a fluid, like water, the more pressure it experiences.

Key concepts in fluid mechanics include:
  • Pressure: The force that the fluid exerts on a surface per unit area.
  • Buoyancy: The upward force exerted on an object submerged in a fluid.
  • Flow dynamics: How fluids move, especially important in designing systems like pipelines and air conditioning systems.
  • Density: The mass of fluid per unit volume, impacting how much pressure is exerted.
Understanding fluid mechanics is crucial for engineers and scientists who need to predict how fluids will behave in natural and artificial environments.
Force Calculation
Force Calculation is about computing the amount of force exerted by or on an object, which in fluid dynamics often involves pressure and area. The exercise solution requires calculating the force of the water on the aquarium wall. This is done by considering how pressure changes with depth in the fluid and how this pressure acts over the area of interest.

To perform a force calculation in a fluid scenario:
  • We first calculate pressure using the hydrostatic pressure formula: \( P = \rho gh \), where \( \rho \) is the fluid density, \( g \) is the gravitational pull, and \( h \) is the depth of the fluid.
  • Next, relate this pressure to force using: \( F = PA \), where \( A \) is the area over which the pressure acts.
  • In problems like this, it's important to account for the total depth when dealing with fluids.
In our specific problem, we calculated how the force changes as we increase the depth of water, clearly showing that more water leads to more force acting on the wall.
Pressure Formula
The Pressure Formula is a fundamental tool in solving problems related to fluids. Understanding how to apply this formula allows us to calculate the exact pressure a fluid exerts at any given point. This is crucial, as it is directly used to determine the force a fluid exerts on surfaces, like the aquarium wall.

The basic hydrostatic pressure formula is:\( P = \rho gh \)
  • \( \rho \): This is the density of the fluid, for water it is approximately \( 1000 \, \mathrm{kg/m^3} \).
  • \( g \): This represents acceleration due to gravity, mostly taken as \( 9.81 \, \mathrm{m/s^2} \).
  • \( h \): The height or depth of the fluid column above the point in question.
Pressure increases with depth because more fluid is above, pushing down due to gravity. By using this formula, we determined how much the pressure (and thereby force) increases when we double the water depth in the aquarium problem. This underscores the rule that pressure, and thus force, increases in a fluid with increasing depth.

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Most popular questions from this chapter

How much work is done by pressure in forcing \(1.4 \mathrm{~m}^{3}\) of water through a pipe having an internal diameter of \(13 \mathrm{~mm}\) if the difference in pressure at the two ends of the pipe is \(1.0 \mathrm{~atm} ?\)

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When researchers find a reasonably complete fossil of a dinosaur, they can determine the mass and weight of the living dinosaur with a scale model sculpted from plastic and based on the dimensions of the fossil bones. The scale of the model is \(1 / 20 ;\) that is, lengths are \(1 / 20\) actual length, areas are \((1 / 20)^{2}\) actual areas, and volumes are \((1 / 20)^{3}\) actual volumes. First, the model is suspended from one arm of a balance and weights are added to the other arm until equilibrium is reached. Then the model is fully submerged in water and enough weights are removed from the second arm to reestablish equilibrium (Fig. 14-42). For a model of a particular T. rex fossil, \(637.76 \mathrm{~g}\) had to be removed to reestablish equilibrium. What was the volume of (a) the model and (b) the actual \(T\), rex? \((\mathrm{c})\) If the density of \(T\). rex was approximately the density of water, what was its mass?

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