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Accelerated along inclined plane

Fluid masses subjected to acceleration along inclined plane

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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 open at top, containing a liquid and moving upwards along inclined plane with a uniform acceleration as shown in fig-1(a).

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We know that when the tank starts moving, the liquid surface falls down on the front side and rises up on the back side of the tank as shown in fig-1(b).

Let,

$\inline \phi$ = Inclination of the plane with the horizontal
$\inline \theta$ = Angle, which the liquid surface makes with the horizontal, and
$\inline a$ = Acceleration of the tank

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Now consider any particle A on the liquid surface. We know that the forces acting on the liquid particle are :

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

2. Accelerating force (F) on an angle $\inline \phi$ with the horizontal.

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

Now resolving the accelerating force (F) horizontally and vertically, we get

$F_H = F \cos\phi$

and

$F_V = F \sin\phi$

We know that the weight of the particle,

$W = mg$

Similarly, the accelerating force,

$F = ma$

Now resolving all the forces horizontally at A,

$P\sin\theta = F_H = F\cos\phi = ma\cos\phi$

$\Rightarrow P\sin\theta = ma\cos\phi$

(2)

And resolving all the forces vertically at A,

$P\cos\theta = F_V = F\sin\phi + W = ma\sin\phi + mg$

$\Rightarrow P\cos\theta = ma\sin\phi + mg$

(3)

Dividing equation (2) by (3)

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

$\tan\theta = \frac{a_H}{a_V + g}$

where,

a_H = horizontal component of the acceleration
a_V = vertical component of the acceleration

Example:

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Example - Fluid masses subjected to acceleration along inclined plane

Problem

A rectangular box containing water is accelerated at 3m/s² upwards on an inclined plane 30 degree to the horizontal. Find the slope of the free liquid surface.

Workings

Given,