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Accelerated horizontally

Fluid masses subjected to horizontal acceleration

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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, originally at rest and containing some liquid. We know that the liquid, at rest, maintains its surface level as shown in fig-1(a). Now let the tank move towards the right side with a uniform acceleration.

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As the tank starts moving, we find that the liquid surface does not remain level any more. But the liquid surface fall down on the front side and rises up on the back side of the of the tank as shown in fig-1(b). The static pressure on the back side and front side due to liquid are shown in fig-1(c).

Let,

$\inline \theta$ = Angle, which the liquid surface makes with the horizontal, and
$\inline a$ = Horizontal acceleration of the tank

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Now consider any particle A on the inclined liquid surface as shown in fig-2. We know that the force acting on the liquid particle are :

1. Weight of the particle (W) acting vertically down,

2. Accelerating force (F) acting horizontally towards right, and

3. Pressure (P) exerted by the liquid particles normal to the free surface.

We know that the weight of the particle,

$W = mg$

where, m = Mass of liquid particle, g = Gravitational acceleration

Similarly, accelerating force,

$F = ma$

Now resolving the forces horizontally at A,

$P\sin\theta = F = ma$

(2)

And resolving the forces vertically at A,

$P\cos\theta = W = mg$

(3)

Dividing equation (2) by (3)

$\frac{P\sin\theta}{P\cos\theta} = \frac{ma}{mg} = \frac{a}{g}$

$\Rightarrow \tan\theta = \frac{a}{g}$

(4)

Now consider the equilibrium of the entire mass of the liquid. Let,

P₁ = Hydrostatic pressure on the back side of the tank, and
P₂ = Hydrostatic pressure on the front side of the tank

Therefore, Net pressure,

$P = P_1 + P_2$

Now as per Newton's Law of Motion, this net pressure,

$P = ma$

$\Rightarrow (P_1 - P_2) = ma$

(5)

Example:

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Example - Fluid masses subjected to horizontal acceleration

Problem

An open rectangular tank 3m long. 2.5m wide and 1.25m deep is completely filled with water. If the tank is moved with an acceleration of 1.5m/s², find the slope of the free surface of water and the quantity of water which will spill out of the tank.