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4.4.1: Fluid in a Linearly Accelerated System

For example, in a two dimensional system, for the effective gravity  
\[
    g_{eff} = a\, \hat{i}  + g\, \hat{k}
    \label{static:eq:geff2d} \tag{122}
\]
where the magnitude of the effective gravity is
\[
    | g_{eff} | = \sqrt{g^2 + a^2}
    \label{static:eq:g_eff_mag} \tag{123}
\]
and the angle/direction can be obtained from
\[
    tan \beta = \dfrac {a} {g}  
    \label{static:eq:g_effAngle} \tag{124}
\]
Perhaps the best way to explain the linear acceleration is by examples. Consider the following example to illustrate the situation. 

Example 4.9

Effective gravity of Accelerated Box

Fig. 4.13 The effective gravity is for accelerated cart.

A tank filled with liquid is accelerated at a constant acceleration. When the acceleration is changing from the right to the left, what happened to the liquid surface? What is the relative angle of the liquid surface for a container in an accelerated system of \(a=5[m/sec]\)?

Solution 4.9

This question is one of the traditional question of the fluid static and is straight forward. The solution is obtained by finding the effective angle body force. The effective  angle is obtained by adding vectors. The change of the acceleration from the right to left is like subtracting vector (addition negative vector). This angle/direction can be found using the following  
\[
    \tan^{-1} \beta = \tan^{-1} \dfrac {a} {g} =  
    \dfrac{5}{9.81} \sim 27.01^{\circ}  \tag{125}
\]
The magnitude of the effective acceleration is  
\[
    \left|g_{eff}\right| = \sqrt{5^2 + 9.81^2} = 11.015 [m/sec^2]  \tag{126}
\]

Fig. 4.14 A cart slide on inclined plane.

Example 4.10

A cart partially filled with liquid and is sliding on an inclined plane as shown in Figure 4.14. Calculate the shape of the surface. If there is a resistance, what will be the angle? What happen when the slope angle is straight (the cart is dropping straight down)?

Solution 4.10

\(\mathbf{a}\)
The angle can be found when the acceleration of the cart is found. If there is no resistance, the acceleration in the cart direction is determined from  
\[
    a = \mathbf{g} \sin \beta
    \label{static:eq:cartAcc} \tag{127}
\]
The effective body force is acting perpendicular to the slope. Thus, the liquid surface is parallel to the surface of the inclination surface.
\(\mathbf{b}\)
In case of resistance force (either of friction due to the air or resistance in the wheels) reduces the acceleration of the cart. In that case the effective body moves closer to the gravity forces. The net body force depends on the mass of the liquid and the net acceleration is  
\[
    a = \mathbf{g} - \dfrac{F_{net} }{m}
    \label{static:eq:aCartFnet} \tag{128}
\]
The angle of the surface, \(\alpha < \beta\), is now
\[
    \tan \alpha = \dfrac {\mathbf{g} - \dfrac{F_{net} }{m}}
        {\mathbf{g}\, cos{\beta}}    
    \label{static:eq:FnetAngle} \tag{129}
\]

\(\mathbf{a}\)
The angle can be found when the acceleration of the cart is found. If there is no resistance, the acceleration in the cart direction is determined from  
\[
    a = \mathbf{g} \sin \beta
    \label{static:eq:cartAcc} \tag{127}
\]
The effective body force is acting perpendicular to the slope. Thus, the liquid surface is parallel to the surface of the inclination surface.
\(\mathbf{b}\)
In case of resistance force (either of friction due to the air or resistance in the wheels) reduces the acceleration of the cart. In that case the effective body moves closer to the gravity forces. The net body force depends on the mass of the liquid and the net acceleration is  
\[
    a = \mathbf{g} - \dfrac{F_{net} }{m}
    \label{static:eq:aCartFnet} \tag{128}
\]
The angle of the surface, \(\alpha < \beta\), is now
\[
    \tan \alpha = \dfrac {\mathbf{g} - \dfrac{F_{net} }{m}}
        {\mathbf{g}\, cos{\beta}}    
    \label{static:eq:FnetAngle} \tag{129}
\]

Forces Sliding Cart 

Fig. 4.15 Forces diagram of cart sliding on inclined plane.


\(\mathbf{c}\)
In the case when the angle of the inclination turned to be straight (direct falling) the effective body force is zero. The pressure is uniform in the tank and no pressure difference can be found. So, the pressure at any point in the liquid is the same and equal to the atmospheric pressure.

Contributors

  • Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.