Pascal’s Law Derivation
Let us consider a right-angled triangle(with sides p, q, and r) prism (height s) submerged in the liquid of density ρ, also assume the size of the submersed element is negligible with compare to the volume of the liquid, and all the points on the element experience the same gravitational force.
Now, the area of the faces PQRS, PSUT, and QRUT of the prism is ps, qs, and rs respectively. Also, assume the pressure applied by the liquid on these faces is P1, P2, and P3 respectively.
Exerted force by this pressure to the faces in the perpendicular inward direction is F1, F2, and F3.
Thus, F1 = P1 × Area of PQRS = P1 × ps
F2 = P2 × Area of PSUT = P2 × qs
F3 = P3 × Area of QRUT = P3 × rs
Now, in triangle PQT,
sin θ = p/r and cos θ = q/r
The net force on the prism will be zero since the prism is in equilibrium.
F3 sin θ = F1 and F3 cos θ = F2 (putting values of F1, F2, and F3 from the above values)
⇒ P3 × rs × p/r = P1 × ps and P3 × rs × q/r = P2 × qs
⇒ P3 = P1 and P3 = P2
Thus, P1 = P2 = P3
Therefore, pressure throughout the liquid remains the same.
Pascal’s Law
Pascal’s law establishes the relation between pressure and the height of static fluids. A static fluid is defined as a fluid that is not in motion. When the fluid is not flowing, it is said to be in hydrostatic equilibrium. For a fluid to be in hydrostatic equilibrium, the net force on the fluid must be zero. This law can be applied to a wide range of real-life applications and hydraulic machines are one of the widely used applications of this law. These hydraulic systems allow us to design shockers and heavy-lifting machines.
Table of Content
- What is Pascal’s Law?
- Pascal Law Formula
- Difference of Pressure in Column
- Pascal’s Law Definition
- Example of Pascal’s Law
- Pascal’s Law Derivation
- Applications of Pascal’s Law
- Sample Problems on Pascal’s Law