Minimum force required to form sphere

[Chestermiller]

Do you mind showing how you would arrive at the result using above approach .

Well. Let me get you started. The force per unit area exerted by the atmosphere on the curved surface of the sphere is $-p_{atm}\vec{i}_r$, where $\vec{i}_r$ is the unit vector in the radial direction in spherical coordinates. Do you know how to determine the x component of this unit vector in terms of the spherical coordinate angles φ and θ? Do you know how to determine the differential element of area on the surface of the sphere in terms of R, dφ, and dθ?

Chet

 

[Tanya Sharma]

The external atmospheric force acting on one of the hemispherical curved surfaces of the sphere filled with water is independent of what fluid is inside the sphere.

The left figure below shows a region of the atmosphere that has the same shape as a hemisphere of the sphere filled with water. The arrows indicate the external atmospheric force acting on the flat and curved surfaces of this portion of the atmosphere. The figure on the right is the sphere filled with water and the arrow indicates the external atmospheric force acting on the curved portion of one of the hemispheres of the sphere. All three forces F are equal. The force on the far left is easy to calculate.

Thanks .

Here is what I had been thinking .

Let the flat surface be in the z plane . The force on a very small surface element will be $pd\vec{A}$ . This $d\vec{A}$ could be decomposed into three area elements along the three planes . Now we could find another element on the surface such that the two elements $d\vec{A_x}$ and $d\vec{A_y}$ get canceled . So that we are only left with $d\vec{A_z}$ . Now $\int d\vec{A_z} = \pi R^2$ .

I am not sure if I could express myself properly . Does it make any sense ?

 

[TSny]

Let the flat surface be in the z plane . The force on a very small surface element will be $pd\vec{A}$ . This $d\vec{A}$ could be decomposed into three area elements along the three planes . Now we could find another element on the surface such that the two elements $d\vec{A_x}$ and $d\vec{A_y}$ get canceled . So that we are only left with $d\vec{A_z}$ . Now $\int d\vec{A_z} = \pi R^2$ .

I am not sure if I could express myself properly . Does it make any sense ?

Yes. That sounds very good.

 

[Tanya Sharma]

Yes. That sounds very good.

Hard to believe that you liked my reasoning .

 

[Tanya Sharma]

Well. Let me get you started. The force per unit area exerted by the atmosphere on the curved surface of the sphere is $-p_{atm}\vec{i}_r$, where $\vec{i}_r$ is the unit vector in the radial direction in spherical coordinates. Do you know how to determine the x component of this unit vector in terms of the spherical coordinate angles φ and θ? Do you know how to determine the differential element of area on the surface of the sphere in terms of R, dφ, and dθ?

Chet

Thank you Chet .

Sorry , but as of now I do not possesses required mathematical skills .

 

[TSny]

Hard to believe that you liked my reasoning .

I know. Probably some aftereffect of the eclipse of the moon last night.

 

[Tanya Sharma]

I know. Probably some aftereffect of the eclipse of the moon last night.