Question

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of \( 100 \, \text{cm}^2 \) with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density \( 1.5 \times 10^3 \, \text{kg/m}^3 \). What force one needs to apply on the hinged door so that it does not open?

Hint:

For problems involving fluid pressure and forces, the total force on an object can be computed by integrating the pressure over the area. Pay attention to the different fluid densities and heights.

Answer 1

Correct Answer: 150

The force \( F_{\text{ext}} \) required to prevent the door from opening is given by: \[ F_{\text{ext}} + F_w = F_t \] where \( F_w \) is the force due to water and \( F_t \) is the total force on the window. In equilibrium: \[ F_{\text{ext}} = F_t - F_w \] Now, \( F_t \) is the total force on the window, which is: \[ F_t = (\rho_1 + \rho_2) g h A \] and \[ F_w = (\rho_1 + \rho_2) g h A \] Thus, the force needed: \[ F_{\text{ext}} = (1500 - 1000) \times 10 \times 10^{-4} \times 150 \] \[ = 150 \, \text{N} \]

Answer 2

The problem asks for the magnitude of the force required to keep a hinged door in a partition closed. The partition separates water from a denser liquid, and the door is located at a specific depth.

Concept Used:

The solution is based on the principles of hydrostatics.

1. Hydrostatic Pressure: The pressure exerted by a fluid at a certain depth \( h \) below its free surface is given by the formula \( P = \rho g h \), where \( \rho \) is the density of the fluid and \( g \) is the acceleration due to gravity. This is the gauge pressure.
2. Force on a Submerged Surface: The force \( F \) exerted by a fluid on a submerged surface of area \( A \) is \( F = P \times A \), where \( P \) is the average pressure on the surface. For a small door, the pressure can be considered uniform and equal to the pressure at its center.
3. Net Force: When a surface separates two fluids, it experiences forces from both sides. The net force is the vector sum of these forces. Since the forces are perpendicular to the surface and in opposite directions, the magnitude of the net force is the difference between the magnitudes of the individual forces: \( F_{\text{net}} = |F_2 - F_1| \). The force required to keep the door closed must be equal in magnitude and opposite in direction to this net force.

Step-by-Step Solution:

Step 1: List the given parameters and convert them to SI units.

• Area of the window/door, \( A = 100 \, \text{cm}^2 = 100 \times (10^{-2} \, \text{m})^2 = 10^{-2} \, \text{m}^2 \).
• Depth of the center of the door, \( h = 3 \, \text{m} \).
• Density of water, \( \rho_w = 1000 \, \text{kg/m}^3 \).
• Density of the other liquid, \( \rho_l = 1.5 \times 10^3 \, \text{kg/m}^3 = 1500 \, \text{kg/m}^3 \).
• Acceleration due to gravity, \( g \approx 10 \, \text{m/s}^2 \).

Step 2: Calculate the force exerted by the water on the door (\( F_w \)).

First, find the pressure of the water at depth \( h \):

\[ P_w = \rho_w g h = (1000 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2) \times (3 \, \text{m}) = 30000 \, \text{Pa} \]

Next, calculate the force on the door due to this pressure:

\[ F_w = P_w \times A = (30000 \, \text{N/m}^2) \times (10^{-2} \, \text{m}^2) = 300 \, \text{N} \]

Step 3: Calculate the force exerted by the other liquid on the door (\( F_l \)).

First, find the pressure of the liquid at depth \( h \):

\[ P_l = \rho_l g h = (1500 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2) \times (3 \, \text{m}) = 45000 \, \text{Pa} \]

Next, calculate the force on the door due to this pressure:

\[ F_l = P_l \times A = (45000 \, \text{N/m}^2) \times (10^{-2} \, \text{m}^2) = 450 \, \text{N} \]

Step 4: Determine the net force on the door and the required force to keep it closed.

The two forces, \( F_w \) and \( F_l \), act on the door in opposite directions. Since \( \rho_l > \rho_w \), the force \( F_l \) is greater than \( F_w \). The net force \( F_{\text{net}} \) on the door is the difference between these two forces.

\[ F_{\text{net}} = F_l - F_w = 450 \, \text{N} - 300 \, \text{N} = 150 \, \text{N} \]

This net force acts from the side of the denser liquid towards the side with water, tending to open the door. To keep the door closed, an external force of the same magnitude must be applied in the opposite direction.

\[ F_{\text{required}} = F_{\text{net}} \]

The force one needs to apply on the hinged door so that it does not open is 150 N.