An unpressurized square tank with a 5-meter width holds a 5-meter column of seawater (SG=1.03).  What is the magnitude of the resultant force acting against one side of the tan

To determine the magnitude of the resultant force acting against one side of the unpressurized square tank holding seawater, we need to use the following approach.

### Step-by-step solution:

#### 1. **Given dataTo determine the magnitude of the resultant force acting against one side of the unpressurized square tank holding seawater, we need to use the following approach.

### Step-by-step solution:

#### 1. **Given **
   - Width of the square tank, \( W = 5 \) meters.
   - Height of the seawater column, \( h = 5 \) meters.
   - Specific gravity of seawater, \( SG = 1.03 \).

#### 2. **Calculate the density of seawater:**
The density of water is \( 1000 \, \text{kg/m}^3 \). The density of seawater, \( \rho_{\text{seawater}} \), can be calculated as:
\[
\rho_{\text{seawater}} = SG \times \rho_{\text{water}} = 1.03 \times 1000 \, \text{kg/m}^3 = 1030 \, \text{kg/m}^3.
\]

#### 3. **Calculate the hydrostatic pressure on the side of the tank:**
Hydrostatic pressure at a depth \( h \) is given by the formula:
\[
P = \rho_{\text{seawater}} \times g \times h
\]
where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.

At the bottom of the tank, the pressure is:
\[
P_{\text{bottom}} = 1030 \times 9.81 \times 5 = 50,515.5 \, \text{Pa}.
\]

#### 4. **Resultant force on one side:**
The pressure varies linearly from zero at the top of the water column to \( P_{\text{bottom}} \) at the bottom. The resultant force acting on one side of the tank is the average pressure multiplied by the area of the side.

- The average pressure is:
  \[
  P_{\text{avg}} = \frac{P_{\text{bottom}}}{2} = \frac{50,515.5}{2} = 25,257.75 \, \text{Pa}.
  \]
  
- The area of one side of the tank is:
  \[
  A = W \times h = 5 \times 5 = 25 \, \text{m}^2.
  \]

- The resultant force is:
  \[
  F = P_{\text{avg}} \times A = 25,257.75 \times 25 = 631,443.75 \, \text{N}.
  \]

#### Final Answer:
The magnitude of the resultant force acting against one side of the tank is **631,443.75 N**.

---

Would you like more details on any part of the calculation, or have any questions?

Here are 5 related questions:
1. How does specific gravity affect the resultant force calculation?
2. How would the force change if the tank were deeper but with the same width?
3. How does hydrostatic pressure distribution affect the design of the tank?
4. What if the tank were pressurized instead of unpressurized—how would the calculation change?
5. How do different liquids with varying densities affect the resultant force?

**Tip:** The pressure at the bottom of a tank depends on the depth of the fluid, but the **resultant force** depends on the average pressure and the area of the tank's side.

An unpressurized square tank with a 5-meter width holds a 5-meter column of seawater (SG=1.03). What is the magnitude of the resultant force acting against one side of the tank?

Determine the resultant force acting against one side of an unpressurized square tank holding a 5-meter column of seawater. Learn how to use hydrostatic pressure, fluid density, and area calculations to solve for the force. Suitable for college-level engineering or physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation