Attraction between glass plates

Classical Mechanics Level 5

Find the maximum force of attraction in Newtons between two glass plates separated by a distance h = 0.1 mm h=0.1~\text{mm} after a drop of water of mass m = 70 mg m=70~\text{mg} was introduced between them.

Details and assumptions

Assume that water's surface tension and density are σ = 73 × 1 0 − 3 N / m \sigma=73\times 10^{-3}~\mbox{N}/\mbox{m} and ρ = 1 0 3 kg / m 3 \rho=10^{3}~\text{kg}/\text{m}^3 , respectively.


The answer is 1.022.

David Mattingly by David Mattingly

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2 solutions

Suppose S S is the area of water between the plates, so m = ρ h S m=\rho h S and S = m ρ h S=\frac{m}{\rho h} .

According to Young–Laplace equation, Δ p = 2 σ h \Delta p=\frac{2\sigma}{h} (since S > > h \sqrt{S}>>h )

Therefore, the force of attraction between the plates is: F = Δ p S = 2 σ m ρ h 2 = 1.022 ( N ) F=\Delta p S=\frac{2\sigma m}{\rho h^2}=1.022(N) .

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Its simple, where is the condition for Maximum?

Md Zuhair - 3 years, 3 months ago
Divyansh Singhal
Nov 3, 2013

Pressure inside the surface P(in)=P(0)-T/r(c)=P(0)-T/{t/2}=P(0)-2T/t

So,net inwards force=P(0)A-P(in)A=[P(0)-2T/t]A-P(0)A=2TA/t

Here volume between the plates V=A ×T⇒ t=V/A

Putting the value of t ,

F=2A^2T/V=1.022N.

Hence max. force of attraction between the two glass plates =1.022 N

Here P(0) indicates atmospheric pressure.

T is surface tension

t is distance between the two glass plates

r(c) denotes the radius of the water droplet formed between two glass plates.

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