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

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Homework Help Overview

The discussion revolves around a physics problem involving surface tension and the force required to separate two glass plates with a thin film of water between them. The problem includes calculations related to pressure differences caused by surface tension and the geometry of the setup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the calculations related to pressure differences and the impact of contact angle on the radius of curvature. They also question the reasoning behind the use of a negative sign in a related problem's solution.

Discussion Status

Participants are exploring the implications of varying contact angles on the radius of curvature and discussing the mathematical reasoning behind pressure calculations. Some guidance has been offered regarding the relationship between surface tension and pressure difference, but there is no explicit consensus on the interpretation of the negative sign in the equations.

Contextual Notes

Participants are working within the constraints of specific problem setups and assumptions about surface tension and geometry. The original poster expresses confusion about certain aspects of the solutions provided, indicating a need for further clarification.

ahmeeeeeeeeee
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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 ??
 
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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.

ehild
 

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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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