How Does Surface Tension Affect the Force Needed to Separate Glass Plates?

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Hello ,


A Problem I found In a book , although I have the answer , I can't understand it


the problem is


Calculate the force required to separate two glass plates of area 10 cm^2 each , with a film of water .5mm thick between them . take the coefficient of surface tention (T)to be 72 dyne/cm and theta = 0

The Answer written

since theta = 0 so the radius of curvature of the cylindrical surface between the plates is r = d/2 =.025 cm

((note : I Understand this but what If theta was another angle ??))


Delta P = T/r = 72/.025 = 2880 dyne/cm^2

the Normal force required = Delta P * Area = 2880*10 =2.88*10^4 dyne = .288 N #

I nead some explanation for this


Another _Identical problem_ says

two circular plates of radus 5 cm with 5 mm thick film of water betweem them , If the surface tention = 72.7*10^3 N/m , the normal force required to pull them Apart is what ?

it is the same I know , but in the solution ((the solution this time is not in the book , it is an assistant Engineer at the college)) said that :

Delta P = T(1/R1-1/R2) , R2 = infinity , so Delta P = T/R1

My problem is Why did he write a negative sign ?? I know it won't change the answer but why is it minus or it is just a writing error ??
 
on Phys.org
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Read http://en.wikipedia.org/wiki/Surface_tension

Zero contact angle means that the wall is tangent to the liquid surface. There are two parallel walls at distance d, therefore the radius of curvature is r=d/2. This is not true for other angles.

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Yes I do know that and I wrote that above , I asked what would be (mathematically) the radius curvature if theta was 30 degrees for example

Any way it is not my basic question here :D
 
I found some answers , but what I can't well understand is

how the pressure difference between the water and the air due to the curviture is multiplies by the (whole area) of the sheet , why ?
 
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