Pascals Law and its Application - Definition, Formula, Applications, FAQs

Pascal Law Definition

Definition: Pascal’s Law states that “When pressure is applied to a confined fluid, the pressure is transmitted equally and undiminished in all directions throughout the fluid.”

Static pressure acts perpendicular to any surface in contact with the fluid. Pascal also found that at a point, the pressure would be the same across any arbitrary planes passing through that point within that fluid for the case of a static fluid. Pascal's law is also referred to as Pascal's principle or principle of transmission of fluid pressure. This law was formulated in 1653 by French mathematician Blaise Pascal.

Pascal Law Formula

Pascal's law is expressed as follows:

$
P=\frac{F}{A}
$

where
$P=$ Pressure, $F=$ Force, $A=$ Area.

Derivation of Pascal Law

Consider a right-angled triangle in a liquid of density. Due to the small size of the element, each point is presumed to be at the same depth as the liquid surface. Gravity has the same effect at all of these locations.

Let $a b$, bd, and cd represent the cross-sectional areas of the ABFE, ABDC, and CDFE faces, respectively.

Let P₁, P₂, and P₃ represent the pressures communicated on the faces $A B F E, A B D C$, and CDFE, respectively.

The pressure exerts a force parallel to the surface. Allow P₁ to apply force F₁ to the surface ABFE, P₂ to apply force F2 to the surface ABDC, and P₃ to apply force F₃ to the surface CDFE.

Here,
$F_1=P_1 \times$ area of $A B F E=P_1 \mathrm{ab}$
$F_2=P_2 \times$ area of $A B D C=P_2$ bd
And, $F_3=P_3 \times$ area of $C D F E=P_3 c d$
We also know that,

$
\begin{aligned}
& \sin \theta=\frac{b}{a} \\
& \cos \theta=\frac{c}{a}
\end{aligned}
$
The prism's net force will be zero because the prism is in equilibrium.

$
\begin{aligned}
& F_1 \sin \theta=F_2 \\
& F_1 \cos \theta=F_3
\end{aligned}
$

Pressure is expressed as,

$
\begin{aligned}
& P_1 a d b / a=P_2 b d(\text { equation-i) } \\
& P_1 a d c / a=P_3 c d \text { (equation-ii) }
\end{aligned}
$
From (i) and (ii),

$
\begin{gathered}
\mathrm{P}_1=\mathrm{P}_2 \text { and } \mathrm{P}_1=\mathrm{P}_3 \\
\therefore \mathrm{P}_1=\mathrm{P}_2=\mathrm{P}_3
\end{gathered}
$

Applications of Pascal's law

1. Fluid Lift

It has many applications in everyday life. Many devices, such as hydraulic lifting and pressure brakes, are based on Pascal's law. The liquid is used to transfer pressure to all these devices. In a hydraulic lift, the two pistons are separated by a space filled with liquid. The cross-section piston at the small cross A is used to apply force F directly to that liquid. The pressure P = F / A is transferred across the liquid to a larger cylinder fitted with a large area piston B, resulting in a higher force of (P × B. ). Platform B can be moved up or down. Therefore, the force used is increased by the B / A factor

2. Brake Fluid

In automobiles, pressure brakes also serve the same purpose. When using less force on the foot, the master piston moves inside the master cylinder, and the resulting pressure is transferred by the oil brakes to form a larger area piston. A high force was then applied to the piston and it was pulled down, stretching the braces on the brake line. As a result, the small force at the base produces excessive force returning to the wheel. The main advantage of the system is that the pressure, which is stopped by pressing the pedal, is transmitted evenly across all cylinders, attached to four wheels to make the braking effort equal to all the wheels.

Example Problems on Pascal's Law

Example 1:
A hydraulic press has two pistons. The area of the small piston is $\mathrm{2 0} \mathrm{c m}^{\mathrm{2}}$, and the area of the large piston is $\mathrm{4 0 0} \mathrm{c m}^{\mathrm{2}}$. If a force of $\mathrm{1 0 0} \mathrm{N}$ is applied on the small piston, find the force exerted by the large piston.

Given:

$
A_1=20 \mathrm{~cm}^2, \quad A_2=400 \mathrm{~cm}^2, \quad F_1=100 \mathrm{~N}
$

By Pascal's Law:

$
\begin{gathered}
\frac{F_1}{A_1}=\frac{F_2}{A_2} \\
F_2=\frac{A_2}{A_1} \times F_1 \\
F_2=\frac{400}{20} \times 100=20 \times 100=2000 \mathrm{~N}
\end{gathered}
$

Answer: The large piston exerts a force of $\mathbf{2 0 0 0}$ N.

Example 2:
In a hydraulic brake system, the area of the master piston is $\mathrm{5 c m}^{\mathrm{2}}$, and the area of each wheel piston is $\mathrm{2 0} \mathrm{c m}^{\mathrm{2}}$. If a force of $\mathrm{5 0} \mathrm{N}$ is applied on the master piston, calculate the force on each wheel piston.

Given:

$
\begin{gathered}
A_1=5 \mathrm{~cm}^2, \quad A_2=20 \mathrm{~cm}^2, \quad F_1=50 \mathrm{~N} \\
\frac{F_1}{A_1}=\frac{F_2}{A_2} \\
F_2=\frac{A_2}{A_1} \times F_1 \\
F_2=\frac{20}{5} \times 50=4 \times 50=200 \mathrm{~N}
\end{gathered}
$

Answer: Each wheel piston experiences a force of $\mathbf{2 0 0 ~ N}$.

Related Topics

• Archimedes Principle
• Buoyant Force
• Bernoulli's Principle
• Density of Water
• Poissons Ratio