A cubical open vessel of side 5m filled with a liquid is accelerated with an acceleration a .The value of a so that pressure at mid point of AC is equal to pressure at point B.

To solve the problem of determining the acceleration \( a \) such that the pressure at the midpoint of AC equals the pressure at point B in a cubical open vessel, we need to analyze the forces acting on the liquid due to both gravity and the acceleration of the vessel. Let's break this down step by step.

Understanding the Setup

We have a cube with a side length of 5 meters, filled with a liquid. The cube is open at the top, and we need to consider two points: the midpoint of AC (let's call it point M) and point B, which is at the bottom of the cube. The pressure at these two points will depend on the height of the liquid column above them and the acceleration of the vessel.

Pressure in a Fluid

The pressure at a depth \( h \) in a fluid is given by the formula:

• P = P_0 + \rho g h

Where:

• P is the pressure at depth \( h \).
• P_0 is the atmospheric pressure (which we can ignore since it's the same at both points).
• \(\rho\) is the density of the liquid.
• g is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)).
• h is the height of the liquid column above the point.

Analyzing the Points

Let's denote the height of the liquid in the cube as \( H \), which is 5 m. The midpoint of AC is at a height of \( H/2 = 2.5 \, m \) from the bottom, while point B is at the bottom of the cube (0 m).

Pressure at Point M (Midpoint of AC)

The pressure at point M can be expressed as:

• P_M = \rho g (H/2) + \rho a (H/2)

Here, \( \rho a (H/2) \) accounts for the additional pressure due to the acceleration of the vessel acting horizontally.

Pressure at Point B

The pressure at point B is given by:

• P_B = \rho g H + \rho a (0)

Since point B is at the bottom, the height \( h \) is \( H \) and the acceleration term is zero because there is no liquid above it.

Setting the Pressures Equal

To find the acceleration \( a \) such that the pressures at points M and B are equal, we set \( P_M \) equal to \( P_B \):

• \(\rho g (H/2) + \rho a (H/2) = \rho g H\)

Simplifying the Equation

We can cancel \( \rho \) from both sides (assuming it is non-zero) and simplify:

• \(g (H/2) + a (H/2) = g H\)

Rearranging gives:

• \(a (H/2) = g H - g (H/2)\)

Factoring out \( g \) yields:

• \(a (H/2) = g (H/2)\)

Thus, we find:

• a = g

Final Result

Therefore, the acceleration \( a \) required for the pressure at the midpoint of AC to equal the pressure at point B is equal to the acceleration due to gravity, which is approximately \( 9.81 \, m/s^2 \). This means that if the vessel is accelerated downward at this rate, the pressures at the specified points will balance out.