Problem 19 - Fundamentals of Physics

Question

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 \(4.00 \mathrm{~m} ?\)

Step-by-Step Solution

Answer

The overall force on the wall increases by \(470880\,\mathrm{N}\).

Step 1: Understand the Problem

We need to calculate the change in total hydrostatic force on a vertical wall of the aquarium when the water level increases from \(2.00\,\mathrm{m}\) to \(4.00\,\mathrm{m}\). This requires using the formula for hydrostatic force on a submerged surface.

Step 2: Hydrostatic Force Formula

The hydrostatic force \(F\) on a vertical surface submerged in a fluid is calculated using the formula: \[ F = \rho \cdot g \cdot A \cdot \bar{h} \] where \(\rho\) is the density of the fluid (for water, \(1000\,\mathrm{kg/m^3}\)), \(g\) is acceleration due to gravity (approximately \(9.81\,\mathrm{m/s^2}\)), \(A\) is the area of the surface, and \(\bar{h}\) is the depth of the centroid of the submerged area.

Step 3: Calculate Initial Force

Initially, the water is \(2.00\,\mathrm{m}\) deep. The area of the wall submerged is \(8.00\,\mathrm{m} \times 2.00\,\mathrm{m} = 16.00\,\mathrm{m^2}\). The centroid depth \(\bar{h}\) for \(2.00\,\mathrm{m}\) is \(1.00\,\mathrm{m}\). Calculating the initial force: \[ F_1 = 1000 \times 9.81 \times 16.00 \times 1.00 = 156960\,\mathrm{N} \].

Step 4: Calculate Final Force

When water depth increases to \(4.00\,\mathrm{m}\), the submerged area is \(8.00\,\mathrm{m} \times 4.00\,\mathrm{m} = 32.00\,\mathrm{m^2}\), and the centroid depth \(\bar{h}\) is \(2.00\,\mathrm{m}\). The final force is \[ F_2 = 1000 \times 9.81 \times 32.00 \times 2.00 = 627840\,\mathrm{N} \].

Step 5: Calculate Increase in Force

Subtract the initial force from the final force to find the increase in total hydrostatic force: \[ \Delta F = F_2 - F_1 = 627840 - 156960 = 470880\,\mathrm{N} \].

Key Concepts

Fluid Mechanics

Fluid Mechanics is a branch of physics focused on how fluids behave and interact with other elements. It can be both fascinating and complex, encompassing liquids and gases that don't maintain a fixed shape. Understanding fluid mechanics is essential to solving real-world problems, like calculating forces on submerged surfaces.

In this exercise, we explore how water pressure influences the force exerted on an aquarium wall. The water's depth in the aquarium plays a significant role, as it determines the pressure exerted due to the weight of the water above any given point.

Key concepts in fluid mechanics relevant to this exercise include:

• Hydrostatic Pressure: This is the pressure exerted by a fluid at equilibrium due to the force of gravity. The deeper you go, the greater the pressure because more fluid is above you.
• Density and Gravity: The density (\( \rho \)) of water and the gravitational pull (\( g \)) are key parameters in calculating pressure.

These principles help us determine how the force on the aquarium wall changes with different water levels.

Pressure Calculations

Pressure calculations delve into how forces are distributed over areas. In fluid mechanics, when calculating hydrostatic force, we're concerned with the pressure exerted by a fluid at a certain depth. This is a fundamental task in many physics problems related to submerged surfaces.

The formula for calculating hydrostatic force is:
\[ F = \rho \cdot g \cdot A \cdot \bar{h} \]

• \( \rho \)
is the density of the fluid, which for water is typically \( 1000 \, \mathrm{kg/m^3} \)
• \( g \)
is the gravitational acceleration, approximately \( 9.81 \, \mathrm{m/s^2} \)
• \( A \)
is the area of the submerged surface
• \( \bar{h} \)
is the depth to the center of the submerged area

Calculating the hydrostatic pressure at different water levels involves modifying these values to reflect the changing conditions, as we did by adjusting \( A \) and \( \bar{h} \) when the depth increased from 2 meters to 4 meters. With the new depths of water, the pressure on the wall and subsequently the force increased due to the added weight and pressure of the extra water.

Physics Problem Solving

Solving physics problems demands a systematic approach, integrating formal theories with practical applications. This process enables you to logically deconstruct a scenario and solve for unknowns.

In our aquarium problem, the systematic approach followed these steps:

• Understanding the scenario: We began by interpreting the problem, identifying what needed to be calculated (hydrostatic force change with increased water depth).
• Applying formulas: We used appropriate formulas for hydrostatic force to compute forces at multiple depths.
• Subtraction for change: We calculated the difference in force between two water levels to find the increase in hydrostatic force.

Problems benefit from a clear and structured approach. It ensures that each vital aspect of theory and measurement is considered, leading students toward problem-solving mastery. By considering each element—force, depth, density, and area—we successfully solved the problem whilst reinforcing essential principles of physics problem solving.