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 total force increases by 470,880 N when the water level rises to 4.00 m.

Step 1: Understand the Problem

We need to calculate the difference in force exerted on the wall of the aquarium when the water level increases from a depth of \(2.00 \, \text{m}\) to \(4.00 \, \text{m}\). We know that the pressure at a depth in a fluid is given by \(P = \rho gh\), where \(\rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the depth.

Step 2: Calculate Initial Force at Depth 2m

The initial pressure at the bottom of the 2.00 m depth is \(P_1 = \rho g h_1\) where \(h_1 = 2.00 \, \text{m}\). The force on the wall is the pressure times the area over which it acts. Since pressure increases linearly with depth, we use average depth for calculation: \(h_{avg1} = \frac{h_1}{2} = 1.00 \, \text{m}\). Hence, \(F_1 = \rho g h_{avg1} \times \text{width} \times h_1\).

Step 3: Calculate Final Force at Depth 4m

Once the depth is increased to 4.00 m, the pressure at the bottom becomes \(P_2 = \rho g h_2\), where \(h_2 = 4.00 \, \text{m}\). Using the average depth \(h_{avg2} = \frac{h_2}{2} = 2.00 \, \text{m}\), the force is \(F_2 = \rho g h_{avg2} \times \text{width} \times h_2\).

Step 4: Derive Force Increase

The increase in force is the difference between the final and initial forces, \(\Delta F = F_2 - F_1\). Substituting values, use \(g = 9.81 \, \text{m/s}^2\), and \(\rho = 1000 \, \text{kg/m}^3\) (density of fresh water). Calculate \(F_1\) and \(F_2\) using the expressions from Steps 2 and 3, and then find \(\Delta F\).

Step 5: Calculate Values

Plugging the given values into the expressions, calculate \(F_1 = 1000 \cdot 9.81 \cdot 1.00 \cdot 8.00 \cdot 2.00 = 156,960 \, \text{N}\) and \(F_2 = 1000 \cdot 9.81 \cdot 2.00 \cdot 8.00 \cdot 4.00 = 627,840 \, \text{N}\). Thus, \(\Delta F = 627,840 - 156,960 = 470,880 \, \text{N}\).

Key Concepts

Fluid Mechanics

Fluid mechanics is the branch of physics that deals with the study of fluids (liquids and gases) and the forces acting on them. It involves understanding how fluids behave when they are at rest and in motion. In the context of the given problem, we are particularly concerned with hydrostatics, which is the study of fluids at rest.

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.

Understanding these concepts helps in analyzing situations like the one in the exercise, where fluid pressure changes as the depth of water in an aquarium increases.

Force Calculation

Force due to fluid pressure on a surface can be calculated using the relationship between pressure and area. In the aquarium problem, the force on the wall is determined by the area on which the fluid's pressure acts.

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: \(P = \rho gh\).

Since the pressure varies with depth, the average pressure calculation is used to find the force exerted on surfaces submerged in the fluid. This involves considering the mean depth of the fluid column to simplify the calculations for uniform surfaces such as the walls of an aquarium.

Pressure Difference

Understanding pressure difference is crucial when calculating how forces change in response to variations in water depth, like in our aquarium scenario. Hydrostatic pressure differences arise because pressure in a liquid depends on depth. A deeper fluid means higher pressure at the bottom section associated with increased force exerted by the water.

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: \(\Delta F = F_2 - F_1\).

This calculation demonstrates the impact of increasing fluid depth on the resultant forces, further exemplifying the core principle that deeper fluids exert more pressure and consequently more force.