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Accelerated horizontally

Fluid masses subjected to horizontal acceleration

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.

23547/fluid_masses_subjected_to_acceleration.png
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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,
  • \theta = Angle, which the liquid surface makes with the horizontal, and
  • a = Horizontal acceleration of the tank

23547/fluid_horizontal.png
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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,

And resolving the forces vertically at A,

Dividing equation (1) by (2)

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

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

Example:

[metric]
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.

23547/fluid_horizontal_examples.png
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Workings
Given,
  • l = 3 m
  • b = 2.5 m
  • d = 1.25 m
  • a = 1.5 m/s2

Slope of the free surface of water

Let, \theta = Angle which the free surface of water will make with the horizontal.

From equation (#3) we know that,
\tan\theta = \frac{a}{g} = \frac{1.5}{9.81} = 1.53
\Rightarrow \theta = 8.7 ^{\circ}

Quantity of water which will spill out of the tank

From the above figure we can see that the depth of water on the front side,

h = 3\tan\theta = 3\times 0.153 = 0.459\;m

\therefore Quantity of water which will spill out of the tank,

V = \frac{1}{2}\times 3\times 2.5\times 0.459 = 1.72\;m^3 = 1720\;litres
Solution
Slope = 8.7^\circ

Quantity of water = 1720 litres