A reversed funnel problem (static fluid mechanics)

[fluidistic]

Hi,
I've been thinking a while on the problem but I don't know how to proceed and if the method I'm thinking of is the most efficient.

Homework Statement

A reversed funnel (see the figure I posted) is on a table. We fill the funnel with water until the water reaches the height h. Because when it does so, the water starts to flow from the bottom of the funnel. (that is, it raises the funnel a little bit in order to flow).
Find the mass of the funnel.

2. The attempt at a solution
I realize I must find what net force the water create on the walls of the funnel when the water is filled up to an height of h.
But it seems very complicated since the shape of the funnel is relatively complicated and I think I must integrate the force exerted starting from the table level up to the height H.
If you could help me to start the problem I'd be happy.

 

[tiny-tim]

I realize I must find what net force the water create on the walls of the funnel when the water is filled up to an height of h.
But it seems very complicated since the shape of the funnel is relatively complicated and I think I must integrate the force exerted starting from the table level up to the height H.

Hi fluidistic!

I can't see the diagram yet, but I assume the funnel is axially symmetric.

So cut the funnel into slices of height dh, calculate the force on each slice, and integrate.

 

[fluidistic]

Hi tiny-tim, (thanks for the input)
I get it but I have some problems. For example for the first slice (the lower possible one), I know that the water exerts a pressure of [tex]-\rho \cdot h[/tex]. So the net force exerted on the first slice is [tex]A\cdot (-\rho \cdot h)[/tex] but I don't know how to calculate the area. It's a portion of a cone.

 

[tiny-tim]

… but I don't know how to calculate the area. It's a portion of a cone.

Well, the area is 2π times the radius times dh times a factor allowing for the slope …

but you don't need the area, you need the fluxy thingy

 

[fluidistic]

Well, the area is 2π times the radius times dh times a factor allowing for the slope …

but you don't need the area, you need the fluxy thingy

What do you mean by fluxy thingy? (flux thing? I still don't understand if it is so.)
I must calculate the net force on the slice... I have the pressure but it isn't enough. The area is not important? I'm lost!

 

[tiny-tim]

Force is a vector.

Pressure is normal force per area, and you only want the vertical component.

 

[fluidistic]

Force is a vector.

Pressure is normal force per area, and you only want the vertical component.

Ok I get the picture. So the vertical component of the net force equals to [tex]\sin (\theta ) \cdot \text{Net force}[/tex] where [tex]\theta[/tex] is the angle that the funnel forms with the table. I don't see how to calculate the net force without passing by calculating the area of a slice. I'm mixed up, the formula I just wrote is only valid for a point on the total area of the slice, but as I'm dealing with differentials, a point might be the whole slice. (I don't understand well what I'm doing).

 

[rl.bhat]

Consider any in between slice. Pressure on that slice is ρgh. This pressure is perpendicular to the slanted side of the funnel. Force at that point is ρgh*2πrdh. Its vertical component is ρgh*2πrdh*cosθ. Now you have to convert it into single variable h and integrate from
h = h to h - H

 

[fluidistic]

Consider any in between slice. Pressure on that slice is ρgh. This pressure is perpendicular to the slanted side of the funnel. Force at that point is ρgh*2πrdh. Its vertical component is ρgh*2πrdh*cosθ. Now you have to convert it into single variable h and integrate from
h = h to h - H

I don't understand why the vertical component of the force is not ρgh*2πrdh*sinθ.
I've formed the integral [tex]\int _h^{h-H} \rho gh2\pi R \cos \theta dh = \rho g2 \pi R \cos \theta (-hH+\frac{H^2}{2})[/tex] so the mass of the funnel would be this but without the "g" variable.

 

[tiny-tim]

Force on slice dh = pressure times area = ρ times 2πrdh/sinθ,

so vertical component = … ?