Question

A large aquarium of height 5.00 mis filled with fresh water to a depth of 2.00 m. One wall of the aquarium consists of thick plastic 8.00 mwide.

By how much does the total force on that wall increase if the aquarium is next filled to a depth of 4.00 m?

Short answer

The increase in the total force on the wall, if the aquarium is filled to a depth of 2.00m, is$4.69\times {10}^{5}N$

Step-by-step solution

The given data

1. Height of aquarium, H=5.00m
2. Depth of aquarium, h=2.0m
3. Thickness of the plastic, w=8.00m

Understanding the concept of force and pressure

Using the formula of pressure in terms of force per unit area and the formula for pressure in terms of density, gravitational acceleration, and depth, we can find the increase in the total force on the wall.

Formulae:

Force applied on a body in terms of pressure, F = pA (i)

Pressure applied on a fluid surface, p =$\rho gh$ (ii)

Calculation of the increased force

Differentiating equation (i) with respect to area A, we get

dF = pdA

$$\begin{gathered} \int dF=\int pdA\left(\because \mathrm{To}\mathrm{get}\mathrm{force}\mathrm{in}\mathrm{that}\mathrm{given}\mathrm{area},\mathrm{integrating}\mathrm{both}\mathrm{sides}\right) \\ F=\int pdA...................\left(1\right) \end{gathered}$$

The force at depth, h=2.0cm,as

$$\begin{gathered} dA=w\times dh, \\ P=\rho gh \end{gathered}$$,

Substituting the above values in the equation (1), we get

$$\begin{gathered} F=\int \rho dh\left(w\times dh\right) \\ =\rho gw\int \left(hdh\right) \\ =\rho gw\left[\frac{h^2}{2}\right] \\ =\left(1000\frac{kg}{m^3}\right)\left(9.8\frac{m}{s^2}\right)\left(8\right)\left[\frac{2^2}{2}\right] \\ =1.5\times {10}^{5}N..............\left(2\right) \end{gathered}$$

Similarly, the force at depth h=4.0cm,

localid="1657253614722" $$\begin{gathered} F\text{'}=\rho gw\left[\frac{h^2}{2}\right] \\ F\text{'}\left(1000\frac{kg}{m^3}\right)\left(9.8\frac{m}{s^2}\right)\left(8\right)\left[\frac{4^2}{2}\right] \\ =6.272\times {10}^{5}N....................\left(3\right) \end{gathered}$$

Therefore, the total force using equations (2) and (3) can be given as:

$$\begin{gathered} ∆F=F\text{'}-F \\ =\left(6.272\times {10}^{5}N\right)-\left(1.5\times {10}^{5}N\right) \\ =4.69\times {10}^{5}N \end{gathered}$$

Therefore, the increase in the total force on the wall, if the aquarium is filled to a depth of$2.00m$, is$4.69\times {10}^{5}N$