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Accelerated vertically

Fluid masses subjected to vertical acceleration

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Contents

Overview
Page Comments

Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\inline \tau$ ) = $\inline I\omega\Omega$ (In the limit).

$\inline I$ = Moment of Inertia.
$\inline \omega$ = Angular velocity
$\inline \Omega$ = 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.

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

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

We know that the forces acting on this column are :

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

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

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

Now resolving the forces vertically,

$P = W + F$

$\Rightarrow p.dA = w(h.dA) + \frac{w}{g}(h.dA)a = wh.dA(1 + \frac {a}{g})$

$\therefore p = wh(1 + \frac {a}{g})$

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,