Fluid pressure at a point:-
Consider a small area dA in large mass of fluid if the stationary, then the force exerted by the surrounding fluids kn the area dA be always perpendicular to the surface area dA.
Let dF is force acting on an area dA.
P=dF/dA.
If force F is uniformly distributed on area A then,
P=F/A.
S.I. unit = N/ m^2
M.K.S. unit = kgf/m^2
C.G.S. unit = dyne/cm^2
Pascal's Law:-
It's state that the pressure and intensity of pressure at a point in static fluid is equals in all direction.
Consider an arbitrary fluid element of wage shape in a fluid mass at rest.
Let, the width of the element perpendicular to the plane of paper is unit and Px, Py and Pz are the pressure or intensity of pressure action on the face AB, AC and BC respectively.
Let, angle ABC= θ
Then the force acting on the element are:-
i. Pressure force normal to the surface.
ii. Weight of element in the vertical direction.
The two forces on the face are
Force on face AB = Pn × area of face AB
= P × dY × 1
Similarly,
Force on face AC = Py × dX × 1
Force on face BC = Pz × dS × 1
Weight of element= (Mass of element) × g
= (volume × ρ ) × g
= (AB×AC)/2 × ρg.
Where,
ρ = density.
Resolving the force X - direction.
Px × dY × 1 - Pz(dS ×1) sin(90- θ ) =0 Px × dY × 1 - Pz dS × 1 cos θ = 0
from fig.,
ds cos θ = AB = dy
Px × dY × 1 - Pz ×dY ×1 =0
Px = Pz ---------- i
Resolving the force Y- direction.
Py × dx × 1 - Pz × ds × 1 × cos(90- θ ) - (dX × dY)/2 ×1 × ρg = 0
Pydx - Pzds sinθ - (dx.dy)/2 ×ρg = 0.
But, ds sin θ = dx and also element is very small
Py dx- Pz dx =0
Py = Pz ------------- ii
from equation i and ii .
Px = Py = Pz .
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