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Pascal’s Law : It states that intensity of pressure for a fluid at rest is equal in all directions. This is proved as : The fluid element is of very small dimensions i.e. dx, dy, ds. Consider a arbitrary fluid element of the wedge shape in a fluid mass at rest. Let the width of the element perpendicRead more

Pascal’s Law : It states that intensity of pressure for a fluid at rest is equal in all directions. This is proved as :

The fluid element is of very small dimensions i.e. dx, dy, ds.

See lessConsider a arbitrary fluid element of the wedge shape in a fluid mass at rest. Let the width of the element perpendicular to the plane of paper is unity and

px, py , pz are the pressure intensity acting on the face AB, AC, BC respectively.

Then the forces acting on the element are :

i. Pressure forces normal to the surfaces ,

ii. Weight of element in the vertical direction ,

The forces on the faces are :

Forces on the face AB = px × Area of face AB = px × dy × 1

Forces on the face AC = py × area of face AC = py × dx × 1

Forces on the face BC = pz × area of face BC = pz × ds × 1

Weight of element = ( Mass of element ) × g = ( Volume × rho ) × g

= [ ( AB × AC ) / 2 × 1 ] × rho × g

where, rho = density of fluid

Resolving the forces in x – direction, we have

px × dy × 1 – pz.ds × 1 cos ( thita ) = 0

px × dy × 1 – pz × dy × 1 = 0

px = pz ……(i)

{ ds cos ( thita ) = AB = dy }

Similarly, Resolving the forces in y – direction, we get

py × dx × 1 – pz × ds × 1 cos ( 90 ° – thita ) – ( dx × dy ) / 2 × 1 × rho × g = 0

py × dx – pz ds sin ( thita ) – ( dx . dy ) / 2 × rho × g = 0

But [ ds . Sin ( thita ) ] = dx and also the element is very small and hence the weight is negligible.

Therefore, py.dx – pz × dx = 0

py = pz ……(ii)

From (i) and (ii) ; we have

px = py = pz

This equation shows that the pressure at any point in x, y and z directions is equal.