A rectangular tank of length 6 m, width 2.5 m and height 2 m is completely filled with water when at rest. The tank is open at the top. The tank is subjected to a horizontal constant linear acceleration of 2.4 m/s² in the direction of its length. Find the volume of water spilled from the tank.

Spilling Water from an Accelerating Tank

Problem Statement

Given Data

Length of tank, $ L = 6 \, \text{m} $
Width of tank, $ b = 2.5 \, \text{m} $
Height of tank, $ H = 2 \, \text{m} $
Initial condition: Tank is completely filled
Constant acceleration, $ a = 2.4 \, \text{m/s}^2 $

Diagram of Spilling Water

The new water surface tilts downwards, causing a triangular volume of water to spill.

Diagram of a full tank accelerating and spilling water

Solution

1. Calculate the Angle of the Water Surface

The slope ($\tan\theta$) of the new free surface is determined by the acceleration.

$$ \tan \theta = \frac{a}{g} $$ $$ \tan \theta = \frac{2.4}{9.81} \approx 0.2446 $$

2. Determine the Drop in Height

The drop in height ($BC$) at the front of the tank over its full length ($L$) is calculated from the slope.

$$ BC = L \times \tan \theta $$ $$ BC = 6 \times 0.2446 \approx 1.4676 \, \text{m} $$

3. Calculate the Volume of Spilled Water

The spilled water forms a triangular prism. Its volume is the area of the triangular cross-section (ABC) multiplied by the width of the tank.

$$ \text{Volume}_{spilled} = \text{Area}_{ABC} \times \text{width} $$ $$ \text{Area}_{ABC} = \frac{1}{2} \times AB \times BC $$ $$ \text{Area}_{ABC} = \frac{1}{2} \times 6 \times 1.4676 = 4.4028 \, \text{m}^2 $$ $$ \text{Volume}_{spilled} = 4.4028 \times 2.5 \approx 11.007 \, \text{m}^3 $$

Final Result:

The volume of water spilled from the tank is $ V \approx 11.007 \, \text{m}^3 $.

Explanation of Concepts

When a tank that is full of liquid is subjected to horizontal acceleration, the free surface of the liquid tilts. The new surface is an inclined plane, lower at the front end and higher at the rear end. Since the tank was initially full, the volume of liquid that is now "above" the original top edge of the tank spills out. This spilled volume corresponds to a triangular prism, with the base of the triangle being the length of the tank and the height being the vertical drop of the water level at the front of the tank.