A tank contains water up to a height of 10 m. One of the sides of the tank is inclined. The angle between the free surface of water and the inclined side is 60°. The width of the tank is 5 m. Find: (i) the force exerted by water on the inclined side and (ii) the position of the centre of pressure.

Leave a Comment / Hydrostatic Forces on Surfaces / By
Inclined Tank Side Problem

Problem Statement

A tank contains water up to a height of 10 m. One of the sides of the tank is inclined. The angle between the free surface of water and the inclined side is 60°. The width of the tank is 5 m. Find: (i) the force exerted by water on the inclined side and (ii) the position of the centre of pressure.

Given Data

  • Vertical height of water, \( h = 10 \, \text{m} \)
  • Angle with free surface, \( \theta = 60^\circ \)
  • Width of tank side, \( b = 5 \, \text{m} \)
  • Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)

Solution

(i) Force Exerted by Water (\(F\))

First, find the length of the wetted inclined side (\(L\)).

$$ L = \frac{h}{\sin \theta} $$ $$ L = \frac{10}{\sin 60^\circ} $$ $$ L = \frac{10}{0.866} \approx 11.547 \, \text{m} $$

Next, calculate the area of the inclined side.

$$ A = L \times b $$ $$ A = 11.547 \times 5 = 57.735 \, \text{m}^2 $$

The vertical depth of the centroid (\(\bar{h}\)) is half the vertical water height.

$$ \bar{h} = \frac{h}{2} $$ $$ \bar{h} = \frac{10}{2} = 5 \, \text{m} $$

Now, calculate the total force on the inclined side.

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 57.735 \times 5 $$ $$ F = 2831800 \, \text{N} $$

(ii) Position of Centre of Pressure (\(h^*\))

The vertical depth of the centre of pressure for an inclined rectangular plane starting from the free surface is given by:

$$ h^* = \frac{2}{3} h $$ $$ h^* = \frac{2}{3} \times 10 $$ $$ h^* \approx 6.667 \, \text{m} $$

Alternatively, using the general formula:

$$ I_G = \frac{bL^3}{12} = \frac{5 \times (11.547)^3}{12} \approx 641.5 \, \text{m}^4 $$
$$ h^* = \frac{I_G \sin^2 \theta}{A \bar{h}} + \bar{h} $$ $$ h^* = \frac{641.5 \times (\sin 60^\circ)^2}{57.735 \times 5} + 5 $$ $$ h^* = \frac{641.5 \times 0.75}{288.675} + 5 $$ $$ h^* = 1.667 + 5 $$ $$ h^* = 6.667 \, \text{m} $$
Final Results:

(i) Force Exerted: \( F \approx 2.832 \, \text{MN} \).

(ii) Position of Centre of Pressure: \( h^* \approx 6.667 \, \text{m} \) (vertical depth from free surface).

Explanation of Concepts

Area of an Inclined Surface: The actual wetted area of the inclined side is larger than the vertical projection. We must first calculate the length of the incline (\(L\)) using trigonometry before finding the area.

Total Force: The total hydrostatic force is calculated using the vertical depth to the centroid (\(\bar{h}\)). Even though the area is larger due to the incline, the average pressure is determined by the average vertical depth, which is simply half the total vertical depth of the water.

Centre of Pressure: For a simple rectangular plane starting at the free surface, the centre of pressure is located at two-thirds of the total vertical depth. This can be proven with the general formula (\(h^* = \frac{I_G \sin^2 \theta}{A \bar{h}} + \bar{h}\)), which gives the same result and is applicable to surfaces that do not start at the free surface.

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top