Problem 9: Force Exerted by Drop on Plates
Analysis:
The problem describes a liquid drop in a zero-gravity environment forming a cylinder of diameter $D$ between two parallel plates. The contact angle is $90^\circ$, which results in straight vertical sides (as shown in the diagram).
1. Excess Pressure Inside the Drop:
The pressure difference across a curved liquid surface is given by the Young-Laplace equation: $$ \Delta P = P – P_0 = \sigma \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $$ For a cylinder:
- One principal radius of curvature is the radius of the cylinder: $R_1 = D/2$.
- The other principal radius (along the vertical straight side) is infinite: $R_2 = \infty$.
2. Forces Acting on the Top Plate:
There are two opposing forces acting on the plate due to the liquid:
- Force due to Excess Pressure (Repulsive): The pressure inside the liquid ($P$) is higher than the atmospheric pressure ($P_0$) outside. This pushes the plates apart. $$ F_{pressure} = (P – P_0) \times \text{Area} = \left( \frac{2\sigma}{D} \right) \left( \frac{\pi D^2}{4} \right) = \frac{\pi D \sigma}{2} $$
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Force due to Surface Tension (Attractive): Surface tension acts along the circumference of the contact circle ($\pi D$). Since the contact angle is $90^\circ$, the surface tension pulls the plate boundaries inward, but more importantly, the tension in the flat liquid surface acts to minimize the area, effectively pulling the plates together.
The net force is the Pressure Force minus the Surface Tension Force acting on the perimeter. $$ F_{net} = F_{pressure} – F_{tension} $$ $$ F_{tension} = \sigma \times \text{Circumference} = \sigma (\pi D) $$
3. Net Force Calculation:
$$ F_{net} = \frac{\pi D \sigma}{2} – \pi D \sigma $$ $$ F_{net} = -\frac{\pi D \sigma}{2} $$
Answer: The force is attractive with magnitude $F = \frac{\pi D \sigma}{2}$
Note: The negative sign indicates attraction (the drop pulls the plates together).