A rectangular tank 6m long, 2m wide and 2m deep contains water to a depth of 1m. It is accelerated horizontally at 2.5 m/s2 in the direction of its length.

Problem Statement

A rectangular tank 6 m long, 2 m wide, and 2 m deep contains water to a depth of 1 m. It is accelerated horizontally at 2.5 m/s² in the direction of its length.

Determine:

• (a) The slope of the free surface.
• (b) The maximum and minimum pressure intensities at the bottom.
• (c) The total force due to water acting on each end of the tank and verify the difference by calculating the inertia force of the accelerated mass.

Solution

1. Slope of the free surface:

   tanθ = aₓ / g = 2.5 / 9.81 ≈ 0.255
   θ ≈ arctan(0.255) ≈ 14.30°

2. Pressure intensities at the bottom:

   Under acceleration the free surface tilts along the tank’s length. Taking the center of the tank as the reference where the water depth is 1 m, the difference in depth from the center to an end (half-length = 3 m) is:

   Δh = 3 × tanθ ≈ 3 × 0.255 = 0.765 m

   Thus, the water depth at the deep end is:

   h₁ = 1 + 0.765 = 1.765 m

   and at the shallow end:

   h₂ = 1 – 0.765 = 0.235 m

   The pressure at the bottom is given by P = γh, where γ = 9810 N/m².

   P_A = 9810 × 1.765 ≈ 17,310.7 N/m²
   P_B = 9810 × 0.235 ≈ 2,308.3 N/m²

3. Total force on each end:

   The force on a vertical surface with a linearly varying pressure (a triangular pressure distribution) is:

   F = 0.5 × (pressure at bottom) × (water depth) × (width of end)

   For the deep end:

   F_A = 0.5 × 17,310.7 × 1.765 × 2 ≈ 30,546 N

   For the shallow end:

   F_B = 0.5 × 2,308.3 × 0.235 × 2 ≈ 543 N

   The difference between these forces is:

   ΔF = F_A – F_B ≈ 30,546 – 543 = 30,003 N

   This difference should match the inertia force of the accelerated water mass:

   F_inertia = ρ × (Volume of water) × aₓ = 1000 × (6 × 2 × 1) × 2.5 = 30,000 N

   The two values are in close agreement.

Explanation

When the tank is accelerated horizontally, the free surface of the water tilts so that it is perpendicular to the resultant acceleration (the vector sum of gravitational acceleration and the horizontal acceleration). This tilt produces a higher water depth at the end opposite to the direction of acceleration (the “deep end”) and a lower water depth at the end in the direction of acceleration (the “shallow end”).

Since pressure in a fluid increases linearly with depth (P = γh), the deep end experiences a higher pressure while the shallow end experiences a lower pressure. The net difference in force on the two ends is equal to the inertial force acting on the mass of water.

Physical Meaning

This problem demonstrates how horizontal acceleration affects the free surface of a fluid and, consequently, its pressure distribution. The tilting of the free surface causes non-uniform pressure on the tank bottom and the end walls. Engineers must account for these dynamic effects when designing tanks and vessels subjected to acceleration (such as in ships, road tankers, or aircraft) to ensure structural integrity and safe operation.