Hydrostatic pressure in a tilted tube

[Jahnavi]

Homework Statement

Prove that the hydrostatic pressure difference between the liquid surface and the base of rightmost tube is ρgh (h is the vertical distance of the liquid surface from the base ) ?

Homework Equations

The Attempt at a Solution

I would like to prove this result just out of curiosity .

The tube is tilted at an angle θ with the vertical . Tube length is L .

Consider a small cuboidal shape fluid element of base area A and height dx at a distance x from the base of the tube .

Weight of the fluid element = (Adx)ρg

Since the fluid is in equilibrium , the net force along the length of the tube is zero .

If P (x) denotes pressure at a distance x from the base ,

P(x)A - P(x+dx)A - (Adx)ρg(cosθ) = 0

dP = ρg(cosθ)dx

P(0) - P(L) = ∫ ρg(cosθ)dx
= ρgL(cosθ)
= ρgh

Is this derivation correct ?

 

[Chestermiller]

That seems correct to me.

 

[Jahnavi]

Thanks !

 

[Jahnavi]

If we consider the rightmost tube to be a thin capillary tube such that the surface has a curved meniscus , will the pressure at the base still be P₀+ρgh ?

Actually I want to ask whether surface tension is relevant as far as hydrostatic pressure in the liquid is concerned ?

 

[Chestermiller]

If we consider the rightmost tube to be a thin capillary tube such that the surface has a curved meniscus , will the pressure at the base still be P₀+ρgh ?

Actually I want to ask whether surface tension is relevant as far as hydrostatic pressure in the liquid is concerned ?

Surface tension causes a pressure difference across the free surface. So, do you think this matters?

 

[Jahnavi]

Taking surface tension into account ,pressure at the base would be P₀ - 2T/r + ρgh . Right ?

 

[Chestermiller]

Taking surface tension into account ,pressure at the base would be P₀ - 2T/r + ρgh . Right ?

Yes, for a hemispherical surface contour.