Chapter 14: Problem 19
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
Step by step solution
Understand the Problem
Calculate Initial Force at Depth 2m
Calculate Final Force at Depth 4m
Derive Force Increase
Calculate Values
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fluid Mechanics
In hydrostatics, the pressure exerted by a fluid at rest is due to the weight of the fluid acting on a surface. This pressure is isotropic, meaning it is exerted equally in all directions. The key formula used in problems involving hydrostatics is the pressure-depth relationship:
- Pressure (\(P\)) within a fluid column of height (\(h\)) is calculated as \(P = \rho gh\).
- Here, \(\rho\) represents fluid density, \(g\) is the gravitational acceleration, and \(h\) is the depth of fluid.
Force Calculation
When calculating force exerted by a fluid, remember the following:
- Force (\(F\)) is the product of pressure (\(P\)) and area (\(A\)). In formula terms, \(F = P \times A\).
- The pressure is calculated using the depth of the fluid column, as discussed in the Fluid Mechanics section: \(P = \rho gh\).
Pressure Difference
The pressure difference in a fluid column is directly proportional to its height, which translates into calculated force difference like this:
- For initial and final states, compute average depths: \(h_1 = 2.00 \text{ m}\) and \(h_2 = 4.00 \text{ m}\), leading to mean depths of \(1.00 \text{ m}\) and \(2.00 \text{ m}\), respectively.
- The total change in force can be computed by calculating the forces for each depth and noting the difference: \(\Delta F = F_2 - F_1\).