Engineering Mechanics

Chapter 35 : Hydrostatics „„„„„ 719

Mathematically, if P is the total pressure (or force) acting on area a, then the intensity of
pressure,

Total pressure

Area

P

p

a

==

It will be interesting to know that the intensity of pressure, on any surface of the container, is
not uniform. But it increases linearly with the depth.

35.3.PASCAL’S LAW

It states, “The intensity of pressure at any point, in a fluid at rest, is the same in all directions.”
Consider a very small right-angled triangular element ABC
of a liquid at rest as shown in Fig. 35.1.

Let W= Weight of the element,

pX= Intensity of horizontal pressure on the

element of the liquid,

pY= Intensity of vertical pressure on the

element of the liquid,

pZ= Intensity of pressure on the diagonal of

the triangular element of the liquid, and

θ= Angle of the triangular element of the

liquid.

We know that the total pressure on the vertical side AC of the liquid element.

PX=pX × AC

Similarly, PY=pY × BC

and PZ=pZ × AB
Since the liquid element is at rest, therefore sum of the horizontal and vertical components
of the liquid pressure should be zero.
Resolving the forces horizontally,
PZ sin θ=PX
or pZ AB sin θ=pX AC ...(Q PZ = pZ AB and PX = pX AC )

From the geometry of the figure, we find that
AB sin θ=AC
∴ pZ AC=pX AC
or pZ=pX ...(i)

Now resolving the forces vertically,
PZ cos θ + W=PY ...(where W is weight of the liquid element)
Since we are considering very small triangular element of the liquid, therefore neglecting
weight of liquid (W)
PZ cos θ=PY

or pZ AB cos θ=pY BC ...(PY = pY AB)
From the geometry of the figure, we find that
AB cos θ=BC
∴ pZ BC=pY BC

or pZ=pY ...(ii)

From equations (i) and (ii), we find that

pX=pY = pZ

Fig. 35.1. Proof of Pascal’s law.