


A storage tank 30 m by 20 m contains petrol of relative density 0.75 to a depth of




4 m. Calculate the total force exerted on the:




i) longer side, and

[6




ii) base.


To solve this problem, we need to calculate the total force exerted by the petrol on both the longer side and the base of the storage tank. We will use the concepts of fluid pressure and force for each part. The formula for pressure exerted by a fluid is given by:

\[
P = \rho g h
\]

Where:
- \( P \) is the pressure at depth \( h \),
- \( \rho \) is the density of the fluid,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the depth of the fluid.

### Given Information:
- Dimensions of the tank: 30 m (longer side) by 20 m (shorter side),
- Depth of the petrol, \( h = 4 \, \text{m} \),
- Relative density of petrol: \( 0.75 \),
- The density of water is \( 1000 \, \text{kg/m}^3 \), so the density of petrol \( \rho = 0.75 \times 1000 = 750 \, \text{kg/m}^3 \).

---

### Part (i): Force on the Longer Side

The pressure increases with depth, so we need to find the average pressure across the entire depth of 4 m.

- The pressure at the top of the liquid (at \( h = 0 \)) is zero.
- The pressure at the bottom (at \( h = 4 \)) is \( P = \rho g h = 750 \times 9.81 \times 4 \, \text{Pa} \).

\[
P_{\text{bottom}} = 750 \times 9.81 \times 4 = 29,430 \, \text{Pa}
\]

The average pressure \( P_{\text{avg}} \) across the depth is:

\[
P_{\text{avg}} = \frac{0 + P_{\text{bottom}}}{2} = \frac{29,430}{2} = 14,715 \, \text{Pa}
\]

Now, the area of the longer side is:

\[
\text{Area of longer side} = 30 \times 4 = 120 \, \text{m}^2
\]

The total force on the longer side is:

\[
F_{\text{longer side}} = P_{\text{avg}} \times \text{Area of longer side} = 14,715 \times 120 = 1,765,800 \, \text{N}
\]

---

### Part (ii): Force on the Base

The pressure at the base is the same as the pressure at the bottom of the tank since all of the petrol is exerting pressure on it. The pressure at the bottom is \( P_{\text{bottom}} = 29,430 \, \text{Pa} \).

The area of the base is:

\[
\text{Area of base} = 30 \times 20 = 600 \, \text{m}^2
\]

The total force on the base is:

\[
F_{\text{base}} = P_{\text{bottom}} \times \text{Area of base} = 29,430 \times 600 = 17,658,000 \, \text{N}
\]

---

### Final Answers:
1. The total force on the longer side is \( 1,765,800 \, \text{N} \).
2. The total force on the base is \( 17,658,000 \, \text{N} \).

---

Do you want more details or have any questions?

Here are five related questions to deepen your understanding:
1. How does the relative density of a fluid affect the force exerted on submerged surfaces?
2. What would happen to the force if the petrol were replaced with water (relative density = 1)?
3. How would you calculate the force on the shorter side of the tank?
4. How would the total force change if the depth of petrol were increased to 5 m?
5. How is hydrostatic pressure applied in real-world engineering problems, like dam design?

**Tip:** Always remember that pressure in a fluid at rest increases linearly with depth, and the force on a surface is the product of pressure and the area.

A storage tank 30 m by 20 m contains petrol of relative density 0.75 to a depth of 4 m. Calculate the total force exerted on the: i) longer side, and ii) base.

This problem involves calculating the total force exerted by petrol (relative density 0.75) on the sides and base of a storage tank using fluid mechanics principles. Learn step-by-step how to find the forces using hydrostatic pressure formulas. Suitable for Grades 10-12.

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