A tank, with a square base of area `1 .0 m^(2)` is divided by a vertical partition has a small hinged door of aren `20 cm^(2)`. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to the height of 4.0 m. Compute the force necessary to keep the door closed

Explanation

To solve the problem of finding the force necessary to keep the door closed in the tank, we will follow these steps:

### Step 1: Understand the Problem
We have a tank divided into two compartments. One side contains water and the other contains an acid with a relative density of 1.7. The door between the two compartments has a specific area, and we need to calculate the force required to keep this door closed due to the pressure difference caused by the fluids.

### Step 2: Identify the Given Data
- Area of the door, \( A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 0.002 \, \text{m}^2 \)
- Height of the fluid column, \( h = 4.0 \, \text{m} \)
- Density of water, \( \rho_1 = 1000 \, \text{kg/m}^3 \)
- Relative density of acid = 1.7, thus:
\[ \rho_2 = 1.7 \times \rho_{\text{water}} = 1.7 \times 1000 \, \text{kg/m}^3 = 1700 \, \text{kg/m}^3 \]

### Step 3: Calculate the Pressure at the Door
The pressure at a depth in a fluid is given by the formula:
\[ P = h \cdot \rho \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).

#### Pressure on the Water Side (P1):
\[ P_1 = h \cdot \rho_1 \cdot g = 4.0 \, \text{m} \cdot 1000 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 = 39200 \, \text{Pa} \]

#### Pressure on the Acid Side (P2):
\[ P_2 = h \cdot \rho_2 \cdot g = 4.0 \, \text{m} \cdot 1700 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 = 67040 \, \text{Pa} \]

### Step 4: Calculate the Pressure Difference
The pressure difference \( \Delta P \) across the door is given by:
\[ \Delta P = P_2 - P_1 = 67040 \, \text{Pa} - 39200 \, \text{Pa} = 27840 \, \text{Pa} \]

### Step 5: Calculate the Force
The force \( F \) required to keep the door closed can be calculated using the formula:
\[ F = \Delta P \cdot A \]
Substituting the values we have:
\[ F = 27840 \, \text{Pa} \cdot 0.002 \, \text{m}^2 = 55.68 \, \text{N} \]

### Final Answer
The force necessary to keep the door closed is approximately \( 55.68 \, \text{N} \).

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