An open cubical tank with each side 1.5m contains oil of specific weight 7.5KN/m3 up to a depth of 1.3m. Find the forces acting on the side of the tank when it is being moved with an acceleration of 4m/s2 in vertically upward and downward direction.

Problem Statement

An open cubical tank with each side measuring 1.5 m contains oil with a specific weight of 7.5 kN/m³ up to a depth of 1.3 m. Determine the force acting on the side of the tank when it is moved with a vertical acceleration of 4 m/s² in the following cases:

• (a) When the vertical acceleration is 4 m/s² upward.
• (b) When the vertical acceleration is 4 m/s² downward.

Solution

1. First, note the given values:

   Specific Weight of Oil, γ_oil = 7.5 kN/m³ = 7500 N/m³
   Depth of oil, h = 1.3 m
   Width of tank side = 1.5 m

2. The modified hydrostatic pressure on the side under a vertical acceleration a_z is given by:

   P = γ_oil × h × (1 + a_z/g)

3. (a) For a vertical acceleration of 4 m/s² upward (a_z = 4 m/s²):

   P_A = 7500 × 1.3 × (1 + 4/9.81) ≈ 13725.5 N/m²

   The force acting on the side is obtained from the area of the pressure diagram. Since the pressure increases linearly from zero at the free surface to a maximum at the bottom, the resultant force is calculated as:

   F₁ = 0.5 × P_A × h × (width)
   F₁ = 0.5 × 13725.5 × 1.3 × 1.5 ≈ 13382 N

4. (b) For a vertical acceleration of 4 m/s² downward (a_z = -4 m/s²):

   P_A = 7500 × 1.3 × (1 – 4/9.81) ≈ 5774.46 N/m²

   The corresponding force acting on the side is:

   F₂ = 0.5 × P_A × h × (width)
   F₂ = 0.5 × 5774.46 × 1.3 × 1.5 ≈ 5630 N

Explanation

In a static fluid, the pressure on a surface increases linearly with depth. When the tank is accelerated vertically, the effective gravitational field changes. With an upward acceleration, the effective gravity becomes (g + a_z), increasing the pressure distribution on the tank side. Conversely, with a downward acceleration, the effective gravity becomes (g – a_z), reducing the pressure.

Physical Meaning

This problem shows how vertical acceleration affects the pressure distribution on the side of a container. The force on the side is determined by the area of the pressure diagram, which in a linearly varying pressure field is triangular. The result indicates that when the tank is accelerated upward, the increased effective gravity raises the pressure and, consequently, the force on the wall. In contrast, a downward acceleration lowers both the pressure and the resulting force.

Understanding these effects is important for designing tanks and containers that are subject to dynamic conditions, such as in transport or during seismic events.