A tank containing water up to a depth of 500 mm is moving vertically upward with a constant acceleration of 2.45 m/s². The width of the tank is 2 m. Find the force exerted by the water on the side of the tank. Also calculate the force on the side of the tank when (i) the tank is moving vertically downward with a constant acceleration of 2.45 m/s², and (ii) the tank is not moving at all.

Vertically Accelerating Tank Problem

Problem Statement

Given Data

Depth of water, $ h = 500 \, \text{mm} = 0.5 \, \text{m} $
Width of tank, $ b = 2 \, \text{m} $
Vertical acceleration, $ a = 2.45 \, \text{m/s}^2 $

Solution

Case 1: Moving Vertically Upward

Pressure distribution for upward acceleration.

Pressure diagram for upward acceleration

First, we calculate the pressure at the bottom of the tank ($p_B$).

$$ p_B = \rho g h \left(1 + \frac{a}{g}\right) $$ $$ p_B = 1000 \times 9.81 \times 0.5 \left(1 + \frac{2.45}{9.81}\right) $$ $$ p_B = 4905 \times (1 + 0.25) = 6131.25 \, \text{N/m}^2 $$

The total force on the side is the area of the triangular pressure distribution multiplied by the width of the tank.

$$ F_{up} = \left(\frac{1}{2} \times \text{depth} \times p_B\right) \times \text{width} $$ $$ F_{up} = \left(\frac{1}{2} \times 0.5 \times 6131.25\right) \times 2 $$ $$ F_{up} = 3065.625 \, \text{N} $$

(i) Case 2: Moving Vertically Downward

Pressure distribution for downward acceleration.

Pressure diagram for downward acceleration and stationary

The pressure at the bottom of the tank is calculated as:

$$ p_B = \rho g h \left(1 - \frac{a}{g}\right) $$ $$ p_B = 1000 \times 9.81 \times 0.5 \left(1 - \frac{2.45}{9.81}\right) $$ $$ p_B = 4905 \times (1 - 0.25) = 3678.75 \, \text{N/m}^2 $$

The total force on the side is:

$$ F_{down} = \left(\frac{1}{2} \times \text{depth} \times p_B\right) \times \text{width} $$ $$ F_{down} = \left(\frac{1}{2} \times 0.5 \times 3678.75\right) \times 2 $$ $$ F_{down} = 1839.375 \, \text{N} $$

(ii) Case 3: Tank is Stationary

When the tank is stationary, the pressure is standard hydrostatic pressure.

$$ p_B = \rho g h $$ $$ p_B = 1000 \times 9.81 \times 0.5 = 4905 \, \text{N/m}^2 $$

The total force on the side is:

$$ F_{stationary} = \left(\frac{1}{2} \times \text{depth} \times p_B\right) \times \text{width} $$ $$ F_{stationary} = \left(\frac{1}{2} \times 0.5 \times 4905\right) \times 2 = 2452.5 \, \text{N} $$

Final Results:

Force on the side when accelerating upward: $ F_{up} \approx 3065.6 \, \text{N} $.

Force on the side when accelerating downward: $ F_{down} \approx 1839.4 \, \text{N} $.

Force on the side when stationary: $ F_{stationary} = 2452.5 \, \text{N} $.

Explanation of Concepts

Effect of Vertical Acceleration on Pressure: When a fluid container is accelerated vertically, the pressure inside the fluid changes. This is due to the apparent change in the force of gravity.

Upward Acceleration: When accelerating upwards, the inertial force acts downwards, adding to the force of gravity. This increases the effective gravitational acceleration ($g_{eff} = g+a$), leading to a higher pressure at any given depth compared to a stationary fluid.
Downward Acceleration: When accelerating downwards, the inertial force acts upwards, opposing the force of gravity. This decreases the effective gravitational acceleration ($g_{eff} = g-a$), resulting in a lower pressure at any given depth.

The force on any submerged surface is directly proportional to this pressure.