An open rectangular tank 3m long and 2m wide is filled with water to a depth of 1.5m. Find the slope of the water surface when the tank moves with an acceleration of 5m/s2 up a 300 inclined plane. Also calculate the pressure on the bottom at both ends.

Problem Statement

An open rectangular tank, measuring 3 m in length and 2 m in width, is filled with water to a depth of 1.5 m. The tank is accelerated up a 30° inclined plane with an acceleration of 5 m/s². Determine:

• The slope of the water surface.
• The pressure on the bottom of the tank at both the rear and front ends.

Solution

1. Resolve the acceleration into its horizontal and vertical components:

   a_x = 5 cos 30° ≈ 4.33 m/s²
   a_z = 5 sin 30° = 2.5 m/s²

2. Determine the slope of the water surface. In a moving frame, the free surface aligns perpendicular to the resultant acceleration vector. Therefore, the slope (θ) is given by:

   tanθ = a_x / (g + a_z) = 4.33 / (9.81 + 2.5) ≈ 0.3517
   θ ≈ arctan(0.3517) ≈ 19.38°

3. Calculate the difference in water depth across the length of the tank. With the tank being 3 m long, the change in height (Δh) due to the tilt is:

   Δh = (Length / 2) × tanθ = (3 / 2) × tan(19.38°) ≈ 0.528 m

   Thus, the depth at the rear end (h_a) is:

   h_a = 1.5 + Δh ≈ 2.03 m

   And the depth at the front end (h_b) is:

   h_b = 1.5 – Δh ≈ 0.97 m

4. Calculate the pressure on the bottom of the tank at both ends. The hydrostatic pressure in a fluid at depth h is given by P = γh. However, due to the additional vertical acceleration, the effective pressure becomes:

   P = γh (1 + a_z / g)

   where the weight density of water, γ, is 9810 N/m³.

   P_a (rear end) = 9810 × 2.03 × (1 + 2.5/9.81) ≈ 24,989 N/m²
   P_b (front end) = 9810 × 0.97 × (1 + 2.5/9.81) ≈ 11,941 N/m²

Explanation

In this problem, the tank is accelerating up an inclined plane. The acceleration has both horizontal (a_x) and vertical (a_z) components. The horizontal component tilts the free surface of the water, while the vertical component adds to the gravitational acceleration.

The water surface adjusts so that it is perpendicular to the net acceleration vector. This gives rise to an inclined water surface with a slope determined by the ratio of a_x to (g + a_z). This tilt creates a variation in the water depth across the tank, which in turn causes different pressures at the rear and front ends.

Physical Meaning

The analysis of this problem highlights how acceleration affects a fluid in an open container:

• The free surface of the water is no longer horizontal but tilts to remain perpendicular to the resultant of gravity and the acceleration. This tilt is quantified by the angle θ (≈ 19.38°).
• The tilt causes a difference in the water depth at opposite ends of the tank, which results in a pressure gradient along the tank’s base.
• The pressure on the bottom is increased by the effective vertical acceleration, accounted for by the factor (1 + a_z/g). This is important in engineering applications where the dynamic loading on a fluid system must be understood and managed.

Understanding these principles is crucial in fields such as marine engineering, transportation, and process engineering, where fluids are often subject to acceleration and inclined motion.