Mechanical Properties of Fluids

PAscal's law states that, if some pressure is applied at any point of incompressible liquid then the same pressure is transmitted to all the points of liquid and on the walls of the container.

Let us imagine an arbitrary right angled prismatic triangle in the liquid of density ρ. This prismatic element is very small so, every part is considered at the same depth from the liquid surface. Therefore, effect of gravity is the same at all these points. That, the small element is in equilibrium. 

The area of faces ABFE, ABDC and CDFE are ad, bd and cd respectively. Let the pressure of liquid on faces ABFE, ABDC and CDFE be P₁, P₂ and P₃respectively.

The pressure of liquid exerts the force normal to the surface. Let us assume pressure P₁exerts the force F₁ on the surface ABFE, pressure P₂ exerts force F₂ on the surface ABDC and pressure P₃ exerts force on the surface CDFE. 

So, Force F₁ is given by, 

$$\begin{gathered} {\mathrm{F}}_1 = {\mathrm{P}}_1 \times \mathrm{Area} \mathrm{ABFE} = {\mathrm{P}}_1\mathrm{ad} \\ {\mathrm{F}}_2 = {\mathrm{P}}_2 \times \mathrm{Area} \mathrm{ABDC} = {\mathrm{P}}_2\mathrm{bd} \\ {\mathrm{F}}_3 = {\mathrm{P}}_3 \times \mathrm{Area} \mathrm{CDFE} = {\mathrm{P}}_3\mathrm{cd} \\ \mathrm{Also}, \sin \mathrm{\theta } = \raisebox{1ex}{$\mathrm{b}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right. \mathrm{and} \sin \mathrm{\theta } = \raisebox{1ex}{$\mathrm{c}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right. \end{gathered}$$ 

Since the prism is in equilibrium, so net force on the prism is zero. 

Thus, 

$$\begin{gathered} {\mathrm{F}}_1 \sin \mathrm{\theta } = {\mathrm{F}}_2 , \mathrm{and} \\ {\mathrm{F}}_1 \cos \mathrm{\theta } = {\mathrm{F}}_3 \\ \mathrm{Therefore}, \\ {\mathrm{P}}_1 \mathrm{ad} (\mathrm{b}/\mathrm{a}) = {\mathrm{P}}_2 \mathrm{bd} , \mathrm{and} ... \left(1\right) \\ {\mathrm{P}}_1 \mathrm{ad} (\mathrm{c}/\mathrm{a})\hspace{0.17em}= {\mathrm{P}}_3 \mathrm{cd} ... \left(2\right) \\ \mathrm{So}, \mathrm{from} \mathrm{equations} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{have} \\ {\mathrm{P}}_1 = {\mathrm{P}}_2 , \mathrm{and} \\ {\mathrm{P}}_1 = {\mathrm{P}}_3 \\ \therefore {\mathrm{P}}_1 = {\mathrm{P}}_2 = {\mathrm{P}}_3 \\ \mathrm{So}, \mathrm{Pascal}\text{'}\mathrm{s} \mathrm{law} \mathrm{is} \mathrm{proved}. \end{gathered}$$

Verification of Pascal's law giving an example: 

Given, a container having opening of different cross-section provided with frictionless piston containing incompressible liquid. 

 

Let the area of cross-section of pistons be A, 2A and 3A respectively. 

Now, applying force F on the piston of area A, keeping the other pistons in position, a force of 2F is required on piston of area 2A and 3F force is required on piston of area 3A.

Therefore,

Pressure on piston of area A is, 
                         $$\begin{gathered} \mathit{ }\mathit{ }\mathit{ }\mathit{ } \\ {P}_{\mathit{1}}\mathit{=}\frac{F}{A} \end{gathered}$$
                       
Pressure on piston of area 2A is, 
                   ${\mathrm{P}}_2=\frac{2\mathrm{F} }{2\mathrm{A}}=\frac{F}{A}$

Pressure on piston of area 3A is,

                 $\mathit{ }{P}_{\mathit{3}}\mathit{=}\frac{\mathit{3}F}{3A}\mathit{=}\frac{\mathit{F}}{\mathit{A}}$ 

The practical application of Pascal's law are:

(i) Hydraulic lift
(ii) Hydraulic press
(iii) Hydraulic brake.

Hydraulic lift is a machine which is based on Pascal's law. In these devices fluids are used for transmitting presure. 

Hydraulic lift consists of two cylinders of different cross-sectional areas connected with a pipe. The cylinders are filled with incompressible liquid and frictionless pistons are fitted in both the cylinders as shown in figure below. 



Let 'a' and 'A' be the area of cross-section of smaller piston and bigger piston respectively. The load to be lifted is placed on bigger cross-section and effort is applied on smaller piston.
Let a force 'f' be applied on the smaller piston.

The pressure exerted by f on piston is,

            $\mathrm{P}=\frac{\mathrm{f}}{\mathrm{a}}$    ... (1) 

Now, according to Pascal's law, the pressure transmitted to bigger piston is also P.

Thus, force on bigger piston is, given by, 
$$\begin{gathered} \mathrm{F} = \mathrm{PA} = \frac{\mathrm{A}}{\mathrm{a}}\mathrm{f} \\ \mathrm{We} \mathrm{have}, \mathrm{A} > \mathrm{a}. \\ \mathrm{Let}, \mathrm{A} = \mathrm{\eta a} \\ \mathrm{So}, \mathrm{F} = \mathrm{\eta f} \end{gathered}$$ 

That is, the transmitted force gets multiplied by a factor $\mathrm{\eta }$ of applied force.