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Proof of Pascal's law

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Pascal's Law states,

" The intensity of pressure at any point in a fluid at rest, is the same in all direction."

Theorem Proof

Consider a very small right angled triangular element ABC of a liquid as shown in figure.

23547/pascal.png

Let:

$p_{x}$ = Intensity of horizontal pressure on the element of the liquid
$p_{y}$ = Intensity of vertical pressure on the element of the liquid
$p_{z}$ = Intensity of pressure on the diagonal of the triangular element of the liquid
$\theta$ = Angle of the triangular element of the liquid

Now total pressure on the vertical side AC of the liquid,

$P_{x} = p_{x}\times AC$

(1)

Similarly,total pressure on the horizontal side BC of the liquid,

$P_{y} = p_{y}\times BC$

(2)

and total pressure on the diagonal side AB of the liquid,

$P_{z} = p_{z}\times AB$

(3)

Since the element of the liquid is at rest, therefore sum of the horizontal and vertical components of the liquid pressure must be equal to zero.

Now using eqilibrium condition for horizontal pressure, $P_{z}\sin \theta = P_{x}$

$\Rightarrow p_{z}.AB.\sin \theta = p_{x}.AC$

From the geometry of the figure, we find that, $AB\sin \theta = AC$

$\therefore p_{z}.AC = p_{x}.AC$

$\Rightarrow p_{z} = p_{x}$

(4)

Now using equilibrium condition for vertical pressure, i.e., $p_{z} \cos \theta = p_{y}-W$ (where W = Weight of the liquid)

As the triangular element is very small, the weight of the liquid W is neglected, so, $p_{z} \cos \theta = p_{y}$

$\therefore p_{z}.AB.\cos \theta = p_{y}.BC$

From the geometry of the figure, we find that $AB\cos \theta = BC$

$\therefore p_{z}.BC = p_{y}.BC$

$\Rightarrow p_{z} = p_{y}$

(5)

Now from equation (#4) and (#5), we find that $p_{x} = p_{y} = p_{z}$

Thus the intensity of pressure at any point in a fluid, at rest, is the same in all direction.