Force Needed to pull apart hemispheres of an evacuated sphere

  • #1

Homework Statement



Prove that the horizontal force needed to pull two weightless, thin hemispheres apart has magnitude

[tex] F = \pi*r^2 * \Delta p [/tex]

where [tex] \Delta p [/tex]

is the pressure difference between the inside and outside of the spheres.

Homework Equations



[tex] surface area = 4\pir^2
F = pA [/tex]


The Attempt at a Solution



I can see that the force needed to pull the hemispheres apart will be equal to the net force on one of the hemispheres along the X-axis created by the pressure (which is uniform).

I am not sure how to calculate the net force along the X axis on either hemisphere. If this were a ring in a plane and there were a simple force along the circumference towards the center of the ring, then I would integrate the x-components of the force elements dF along the circumference of a half ring from 0 to pi. But I am not sure what kind of integration I can do since I'm dealing with pressure and 3 dimensions. Thanks.

EDIT:
I apologize for the poor formatting, can someone tell me how to write this properly in Tex? You should be able to see my tex code by mousing over.
 

Answers and Replies

  • #2
gneill
Mentor
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It doesn't really matter if the spheres have horizontal or vertical orientation for purposes of analysis.

Think of the hemisphere as being made up of very thin horizontal slices. The surface area of a 'differential slice' depends upon the axial radius (determining its circumference) and the small change in radius angle from its bottom edge to its top edge (this radius being the radius of the hemisphere from the center of its flat bottom).

Pressure will always be perpendicular to this differential bit of surface area, whose surface normal will be radial. You want to find the component of the total force that acts towards the flat bottom.
 

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