Force needed to unstick a semi sphere on the ground

[fluidistic]

0. Homework Statement

What is the force required to unstick a semi sphere from the ground if the pressure inside the semi sphere is lesser than the one outside it?
Assume the semi sphere has a radius of r.


1. The attempt at a solution

\Delta P = \frac{F_{\text {net}}}{A} where A is the surface area of the semi sphere, that is A=2\pi r^2 and P is the pressure.
So I get that the net force (pointing upward) is worth 2\pi r^2 \Delta P.
I wonder if I'm correct.

 

[kuruman]

Not so fast. I assume that one pulls in a direction perpendicular to the flat face. Now suppose that instead of pulling on the hemisphere, you pulled on the flat face. Then the area would be just πr² giving half as much force. This is the correct answer because the force has the same magnitude and direction over the flat face so it is fair to say that "force is pressure times area" in this case. For the hemisphere, you have to do an integral in which case you will get the same answer. Area elements dA near the "equator" generate little force to oppose the direction of the pull along the "pole."

 

[fluidistic]

Ok thanks kuruman. I over simplified the problem, now I know how it is.

 

[fluidistic]

This problem is haunting my brain.
I'm trying to do it via the integration method.

I can use the fact that the surface area of the hemisphere is 2\pi r^2 and it remains to find the force acting on a differential of area (dA) and integrate it. But I don't success in it.
I realize that only the vertical component of the force matters, that is F\sin \theta but I have to replace F by \Delta PA. I have some difficulties when it comes to find the A (more precisely the dA). So I'd have dF=\Delta P dA because the pressure is constant. Hence F=\Delta P \int dA.

As you can see I'm confused.

Can someone help me a bit more?

 

[kuruman]

Take the axis of symmetry of the hemisphere to be z and assume that you pull along the z-axis. Use standard spherical angles θ and φ. Note that θ is measured from the z-axis. An area element on the surface of the sphere is

dA = r²sinθ dθ dφ

The force on this element is dF = p dA in a direction radially in and perpendicular to the surface. The z- component of this is

(dF)_z = r²sinθ cosθ dθ dφ

Integrate. It is best to double the force over half a hemisphere because of the unconventional limits of integration for θ.

 

[fluidistic]

Take the axis of symmetry of the hemisphere to be z and assume that you pull along the z-axis. Use standard spherical angles θ and φ. Note that θ is measured from the z-axis. An area element on the surface of the sphere is

dA = r²sinθ dθ dφ

The force on this element is dF = p dA in a direction radially in and perpendicular to the surface. The z- component of this is

(dF)_z = r²sinθ cosθ dθ dφ

Integrate. It is best to double the force over half a hemisphere because of the unconventional limits of integration for θ.

Thanks infinitely. Although I did calculus III (didn't ace it) I feel that was a bit over my head. I'll think about it.