Question

An aquarium
5
ft long,
2
ft wide, and
3
ft deep is full of water. Find

the hydrostatic pressure on the bottom of the aquarium,

the hydrostatic force on the bottom, and

the hydrostatic force on one end of the aquarium.

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Answer 1

To find the hydrostatic pressure, hydrostatic force on the bottom of the aquarium, and the hydrostatic force on one end of the aquarium, we will use the following formulas:

1. **Hydrostatic Pressure (P)**:
\[
P = \rho g h
\]
where:
- \(P\) = pressure (in pascals)
- \(\rho\) = density of water (approximately \(1000 \, \text{kg/m}^3\))
- \(g\) = acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\))
- \(h\) = depth of the fluid (in meters)

2. **Hydrostatic Force (F)**:
\[
F = P \cdot A
\]
where:
- \(F\) = force (in newtons)
- \(A\) = area (in square meters)

### Step 1: Hydrostatic Pressure on the Bottom of the Aquarium

The depth of the water \(h\) is \(3 \, \text{ft}\). We need to convert this to meters:
\[
h = 3 \, \text{ft} \times \frac{0.3048 \, \text{m}}{1 \, \text{ft}} \approx 0.9144 \, \text{m}
\]

Now, substituting in the values to calculate the hydrostatic pressure:
\[
P = \rho g h = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.9144 \, \text{m} \approx 8985.5 \, \text{Pa} \, (\text{or} \, \text{N/m}^2)
\]

### Step 2: Hydrostatic Force on the Bottom of the Aquarium

The area of the bottom of the aquarium \(A\) is given by:
\[
A = \text{length} \times \text{width} = 5 \, \text{ft} \times 2 \, \text{ft} = 10 \, \text{ft}^2
\]

Convert the area into square meters:
\[
A = 10 \, \text{ft}^2 \times \left( \frac{0.3048 \, \text{m}}{1 \, \text{ft}} \right)^2 = 10 \times 0.092903 \, \text{m}^2 \approx 0.92903 \, \text{m}^2
\]

Now, we can calculate the hydrostatic force on the bottom:
\[
F = P \cdot A \approx 8985.5 \, \text{Pa} \times 0.92903 \, \text{m}^2 \approx 8346.05 \, \text{N}
\]

### Step 3: Hydrostatic Force on One End of the Aquarium

The area of one end of the aquarium (which is a rectangle) is:
\[
A_{\text{end}} = \text{width} \times \text{depth} = 2 \, \text{ft} \times 3 \, \text{ft} = 6 \, \text{ft}^2
\]
Convert this to square meters:
\[
A_{\text{end}} = 6 \, \text{ft}^2 \times 0.092903 \, \text{m}^2 \approx 0.557418 \, \text{m}^2
\]

To find the hydrostatic force on one end, we need to calculate the average pressure on that end. The average depth is half the total depth:
\[
\text{Average depth} = \frac{h}{2} = \frac{0.9144}{2} \approx 0.4572 \, \text{m}
\]
Now we can calculate the average pressure:
\[
P_{\text{avg}} = \rho g \cdot \text{Average depth} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.4572 \, \text{m} \approx 4486.375 \, \text{Pa}
\]

Now, we can calculate the force on one end:
\[
F_{\text{end}} = P_{\text{avg}} \cdot A_{\text{end}} \approx 4486.375 \, \text{Pa} \times 0.557418 \, \text{m}^2 \approx 2505.49 \, \text{N}
\]

### Summary of Results:

1. **Hydrostatic Pressure on the Bottom**: \( P \approx 8985.5 \, \text{Pa} \)
2. **Hydrostatic Force on the Bottom**: \( F \approx 8346.05 \, \text{N} \)
3. **Hydrostatic Force on One End**: \( F_{\text{end}} \approx 2505.49 \, \text{N} \)