Magdeburg hemispheres

  • Thread starter kidsmoker
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  • #1
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Homework Statement



Simple problem but for some reason I get double the answer in the book and I can't understand why. Two hemispherical shells of diameter D are placed together and the air is pumped out so that the pressure inside is p. The pressure outside the shells is q. How much force would you have to exert on each shell to pull them apart?

Homework Equations



Force = pressure*area
Surface area of a sphere = 4*pi*r^2

The Attempt at a Solution



If I just concentrate on one shell, then the force from the outside is 2*pi*(D/2)^2*q = 0.5*pi*D^2*q . The force on the inside is then 0.5*pi*D^2*p, so the difference between these would be the force required I would have thought? This gives

F = 0.5*pi*D^2*(q-p).

However the answer in the book gives it as F = 0.25*pi*D^2*(q-p) and I don't understand why.

Thanks.
 

Answers and Replies

  • #2
mgb_phys
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hint - Does the shape of the hemispheres matter?
Think about the plane (circle) where they join, suppose you changed the shape of one of the spheres but kept the same pressure, would the pressure on the interface change?

It's possibly easier to think about one sphere stuck on flat wall.
 
  • #3
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Aha gotcha. So the pressure that matters is the one pushing the two spheres apart/together, and the area over which this acts is the inner join which is 2*pi*r^2, not 4*pi*r^2. This then gives the correct answer.

This means that if I changed the shape of one of the spheres but kept it so they still joined perfectly together then the pressure on the interface would not change and neither would the force required.

Thanks.
 
  • #4
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Sorry just to go back to this quickly. When you're talking about the force acting from the outside, do you look at the hemisphere face-on and so consider the pressure to be acting over the area of a circle? Is that an okay way to look at it?
 
  • #5
mgb_phys
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Yes, mathematically it's equivalent to looking at the force on each point on the hemisphere * the direction of the force compared to where they join.

A simpler picture is looking at the total solar energy from the sun on the earth.
You could calculate the amount on each square meter and multiply by the angle of the sun at noon (to allow for their being less light near the poles ) or you can consider a disc the same diameter as the earth just in front of the earth and square-on to the sun. Obviously exactly the same amount of light must hit the disc as hits the earth.
With calculus you can easily prove that they are equivalent.
 
  • #6
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Okay cool thanks. I will try it with calculus.
 

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