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# Accelerated vertically

Fluid masses subjected to vertical acceleration
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Contents

### Key Facts

Gyroscopic Couple: The rate of change of angular momentum ($\inline&space;\tau$) = $\inline&space;I\omega\Omega$ (In the limit).
• $\inline&space;I$ = Moment of Inertia.
• $\inline&space;\omega$ = Angular velocity
• $\inline&space;\Omega$ = Angular velocity of precession.

Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.

Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.

Henry Philibert Gaspard Darcy (1803-1858) was a French engineer who made several important contributions to hydraulics.

## Overview

Consider a tank open at top, containing a liquid and moving vertically upwards with a uniform acceleration. Since the tank is subjected to an acceleration in the vertical direction only, therefore the liquid surface will remain horizontal.

##### MISSING IMAGE!

Now consider a small column of the liquid of height h and area dA in the tank as sown in fig-1.

Let, $\inline&space;p$ = Pressure due to vertical acceleration

We know that the forces acting on this column are :

1. Weight of the liquid column $\inline&space;[W&space;=&space;w(h.dA)]$ acting vertically downwards,

2. Acceleration force, $\inline&space;F&space;=&space;ma&space;=&space;\frac{w}{g}(h.dA)a$

3. Pressure $\inline&space;(P&space;=&space;p.dA)$ exerted by the liquid particles on the column.

Now resolving the forces vertically,

$P&space;=&space;W&space;+&space;F$
$\Rightarrow&space;p.dA&space;=&space;w(h.dA)&space;+&space;\frac{w}{g}(h.dA)a&space;=&space;wh.dA(1&space;+&space;\frac&space;{a}{g})$
$\therefore&space;p&space;=&space;wh(1&space;+&space;\frac&space;{a}{g})$

Example:
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##### Example - Fluid masses subjected to vertical acceleration
Problem
An open rectangular tank 4m long and 2.5m wide contains an oil of specific gravity 0.85 up to a depth of 1.5m. Determine the total pressure on the bottom of the tank, when the tank is moving with an acceleration of of g/2 m/s2 (i) vertically upwards (ii) vertically downwards.
Workings
Given,
• $\inline&space;l$ = 4 m
• $\inline&space;b$ = 2.5 m
• $\inline&space;d$ = 1.5 m
• $\inline&space;a$ = g/2 m/s2
• Specific gravity of liquid = 0.85

(i) Total pressure on the bottom of the tank, when it is vertically upwards

Specific weight of oil, $\inline&space;w&space;=&space;0.85\times&space;9.81&space;=&space;8.34\;KN/m^3$

Intensity of pressure at the bottom of the tank,
$p_1&space;=&space;wh(1&space;+&space;\frac{a}{g})$
$\Rightarrow&space;p_1&space;=&space;8.34\times&space;1.5(1&space;+&space;\frac{g}{2g})$
$\therefore&space;p_1&space;=&space;18.765\;KN/m^2$

Total pressure on the bottom of the tank,
$P_1&space;=&space;p_1.A&space;=&space;18.765\times&space;(4\times&space;2.5)&space;=&space;187.65\;KN$

(i) Total pressure on the bottom of the tank, when it is vertically downwards

Intensity of pressure at the bottom of the tank,
$p_2&space;=&space;wh(1&space;+&space;\frac{a}{g})$
$\Rightarrow&space;p_2&space;=&space;8.34\times&space;1.5(1&space;+&space;\frac{g}{2g})$
$\therefore&space;p_2&space;=&space;6.255\;KN/m^2$

Total pressure on the bottom of the tank,
$P_2&space;=&space;p_2.A&space;=&space;6.255\times&space;(4\times&space;2.5)&space;=&space;62.55\;KN$