Pascal's Law for Static Fluids
As shown in the figure three coordinate axes x, y, and z are arbitrarily selected for the point O.
Consider the point being surrounded by a small tetrahedron OQRS, having three surfaces as ORS, QOS and QRO with areas δAx, δAy and δAz normal to the axes x, y and z respectively.
Also let the pressure on these surfaces be as px, py and pz respectively.
The fourth surface of the tetrahedron which is given as QRS has an area δAn with pressure pn acting normally on it.
Here we won’t consider the weight of the fluid inside the tetrahedron because it’s infinitesimally small and the axes are arbitrarily introduced.
Now, the limits may be represented as,
δAx → 0, δAy → 0, δAz → 0, δAn → 0
The referred pressures px, py, pz and pn are towards the point O.
Now, under the above pressures, for equilibrium of the fluid, the following equations are met:
Ʃ Fx = 0, Ʃ Fy = 0 and Ʃ Fz = 0
Also for the direction x,
pxδAx - pnδAn (n.i) = 0,
Where δAn (n.i) is the area of δAn which is shown projected and is normal to the x-axis, can now be represented as δAx.
Therefore the above equation becomes, pxδAX - pnδAx = 0
δAx can be cancelled out from both the sides, producing
px = pn
py = pn and pz = pn indicating equilibrium along the directions of the y and the z axes.
Thus finally we can write,
px = py = pz = pn
The above conditions prove that the pressure over all the sides of the tetrahedron or the planes of the enclosed static fluid is equal, confirming Pascal's law.