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# Accelerated along inclined plane

Fluid masses subjected to acceleration along inclined plane

## 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).

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

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 $\inline&space;\phi$ 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

$F_H&space;=&space;F&space;\cos\phi$
and
$F_V&space;=&space;F&space;\sin\phi$

We know that the weight of the particle,

$W&space;=&space;mg$

Similarly, the accelerating force,

$F&space;=&space;ma$

Now resolving all the forces horizontally at A,

$P\sin\theta&space;=&space;F_H&space;=&space;F\cos\phi&space;=&space;ma\cos\phi$
$\Rightarrow&space;P\sin\theta&space;=&space;ma\cos\phi$

And resolving all the forces vertically at A,

$P\cos\theta&space;=&space;F_V&space;=&space;F\sin\phi&space;+&space;W&space;=&space;ma\sin\phi&space;+&space;mg$
$\Rightarrow&space;P\cos\theta&space;=&space;ma\sin\phi&space;+&space;mg$

Dividing equation (1) by (2)

$\frac{P\sin\theta}{P\cos\theta}&space;=&space;\frac{ma\cos\phi}{ma\sin\phi&space;+&space;mg}&space;=&space;\frac{a\cos\phi}{a\sin\phi&space;+&space;g}$
$\tan\theta&space;=&space;\frac{a_H}{a_V&space;+&space;g}$

where,
• aH = horizontal component of the acceleration
• aV = 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,
• $\inline&space;a$ = 3 m/s2
• $\inline&space;\phi$ = $\inline&space;30&space;^{\circ}$

Let, $\inline&space;\theta$ = Angle which the water surface will make with the horizontal.

Horizontal component of the acceleration,
$a_H&space;=&space;a\cos&space;30&space;^{\circ}&space;=&space;3\times&space;0.866&space;=&space;2.598\;m/s^2$

vertical component of the acceleration,
$a_V&space;=&space;a\sin&space;30&space;^{\circ}&space;=&space;3\times&space;0.5&space;=&space;1.5\;m/s^2$

$\tan\theta&space;=&space;\frac{a_H}{a_V&space;+&space;g}&space;=&space;\frac{2.598}{1.5&space;+&space;9.81}&space;=&space;&space;0.2297$
$\therefore&space;\theta&space;=&space;12.94&space;^{\circ}$
Solution
Slope of the free liquid surface = $\inline&space;12.94&space;^{\circ}$