A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 2.4 m/s². The length, width and depth of the tank are 6 m, 2.5 m and 2 m respectively. If the depth of water in the tank is 1 m and the tank is open at the top, calculate the angle of the water surface to the horizontal.

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Accelerating Rectangular Tank Problem

Problem Statement

A rectangular tank is moving horizontally in the direction of its length with a constant acceleration of 2.4 m/s². The length, width and depth of the tank are 6 m, 2.5 m and 2 m respectively. If the depth of water in the tank is 1 m and the tank is open at the top, calculate: (i) the angle of the water surface to the horizontal, (ii) the maximum and minimum pressure intensities at the bottom, (iii) the total force due to water acting on each end of the tank.

Given Data

  • Constant acceleration, \( a = 2.4 \, \text{m/s}^2 \)
  • Length of tank, \( L = 6 \, \text{m} \)
  • Width of tank, \( b = 2.5 \, \text{m} \)
  • Initial depth of water, \( h_{initial} = 1 \, \text{m} \)

Diagram of Accelerating Tank

Diagram of a rectangular tank under constant horizontal acceleration

Solution

(i) Angle of the Water Surface to the Horizontal (\(\theta\))

The angle of the water surface is determined by the ratio of horizontal acceleration to gravitational acceleration.

$$ \tan \theta = \frac{a}{g} $$ $$ \tan \theta = \frac{2.4}{9.81} \approx 0.2446 $$ $$ \theta = \tan^{-1}(0.2446) \approx 13.74^\circ $$

(ii) Maximum and Minimum Pressure Intensities

First, we find the depths at the rear (maximum depth, \(h_2\)) and front (minimum depth, \(h_1\)) ends of the tank. The change in depth occurs over half the length of the tank (\(L/2 = 3\) m).

$$ \text{Depth at rear end, } h_2 = h_{initial} + \left(\frac{L}{2}\right) \tan \theta $$ $$ h_2 = 1 + 3 \times 0.2446 = 1.7338 \, \text{m} $$
$$ \text{Depth at front end, } h_1 = h_{initial} - \left(\frac{L}{2}\right) \tan \theta $$ $$ h_1 = 1 - 3 \times 0.2446 = 0.2662 \, \text{m} $$

The maximum pressure intensity occurs at the bottom rear (point A):

$$ P_{max} = \rho g h_2 $$ $$ = 1000 \times 9.81 \times 1.7338 $$ $$ \approx 17008.5 \, \text{N/m}^2 $$

The minimum pressure intensity occurs at the bottom front (point B):

$$ P_{min} = \rho g h_1 $$ $$ = 1000 \times 9.81 \times 0.2662 $$ $$ \approx 2611.4 \, \text{N/m}^2 $$

(iii) Total Force on Each End

Force on the front end (\(F_1\)):

$$ F_1 = \rho g A_1 \bar{h}_1 $$ $$ A_1 = h_1 \times b = 0.2662 \times 2.5 = 0.6655 \, \text{m}^2 $$ $$ \bar{h}_1 = h_1 / 2 = 0.2662 / 2 = 0.1331 \, \text{m} $$ $$ F_1 = 1000 \times 9.81 \times 0.6655 \times 0.1331 \approx 869 \, \text{N} $$

Force on the rear end (\(F_2\)):

$$ F_2 = \rho g A_2 \bar{h}_2 $$ $$ A_2 = h_2 \times b = 1.7338 \times 2.5 = 4.3345 \, \text{m}^2 $$ $$ \bar{h}_2 = h_2 / 2 = 1.7338 / 2 = 0.8669 \, \text{m} $$ $$ F_2 = 1000 \times 9.81 \times 4.3345 \times 0.8669 \approx 36862 \, \text{N} $$
Final Result:

(i) Angle of water surface is \( \theta \approx 13.74^\circ \).

(ii) Maximum pressure is \( P_{max} \approx 17008.5 \, \text{N/m}^2 \), Minimum pressure is \( P_{min} \approx 2611.4 \, \text{N/m}^2 \).

(iii) Total force on the front end is \( F_1 \approx 869 \, \text{N} \).

(iii) Total force on the rear end is \( F_2 \approx 36862 \, \text{N} \).

Explanation of Concepts

Tilted Water Surface: When a container of fluid is subjected to constant horizontal acceleration, the free surface of the fluid tilts. It is no longer horizontal. The surface forms an angle where the inertial force (due to acceleration) and the gravitational force on any fluid particle are in equilibrium. The tangent of this angle (\(\theta\)) is the ratio of the horizontal acceleration (\(a\)) to the vertical acceleration due to gravity (\(g\)).

Pressure in Accelerated Fluid: The pressure at any point within the fluid still depends on the vertical depth, but this depth is measured from the new, inclined free surface. This causes the pressure at the bottom of the tank to be non-uniform, being highest at the deeper (rear) end and lowest at the shallower (front) end.

Control Volume for Verification

Control volume of the liquid in the accelerating tank

Verification via Newton's Second Law: The net force on the body of water in the horizontal direction must be equal to the mass of the water multiplied by its acceleration (\(F_{net} = ma\)). The net force is the difference between the hydrostatic force on the rear end (\(F_2\)) and the front end (\(F_1\)).

$$ F_{net} = F_2 - F_1 \approx 36862 - 869 = 35993 \, \text{N} $$ $$ \text{Volume of Water} = \left( \frac{h_1 + h_2}{2} \right) \times L \times b = \left( \frac{0.2662 + 1.7338}{2} \right) \times 6 \times 2.5 = 15 \, \text{m}^3 $$ $$ \text{Mass of Water, M} = \rho \times V = 1000 \times 15 = 15000 \, \text{kg} $$ $$ M \times a = 15000 \times 2.4 = 36000 \, \text{N} $$
The calculated net force (\(35993 \, \text{N}\)) is approximately equal to the required force to accelerate the mass (\(36000 \, \text{N}\)), confirming the results.

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