A magic glass

Classical Mechanics Level 3

A typical science question young people have is "Why can a glass can contain water?" and a typical answer is "Because the distance between molecules that make up the glass is smaller than the size of each water molecule." This isn't quite right though. Imagine making a small hole in the bottom of a bottle that is full of water. If the hole is small enough, the water will not come out unless you squeeze the bottle a bit. So, answering that question with molecular distances and sizes is science overkill -- a glass can contain water even if there are holes in it. However, there's a limit on how big the holes can be.

Consider a glass with full of water of mass density $\rho=1,000~\mbox{kg/m}^3$ and height $h=20~\mbox{cm}$ . There's a circular hole in the bottom of the glass of radius $r$ . The maximum pressure that pushes the water back into the hole is roughly (on the order of) $p=\sigma/r$ , where $\sigma=0.072 ~\mbox{N/m}$ is the water's surface tension. This extra pressure comes from the curvature of the water surface, and it tends to flatten out the surface.

Estimate the largest possible radius of the hole in $\mu \mbox{m}$ such that water doesn't drip out of the glass.

Details and assumptions

The gravitational acceleration is $g=-9.8~\mbox{m/s}^2$ and the glass is placed vertically.
Neglect any other effects that can influence the pressure from other external sources.

The answer is 36.7.

5 solutions

Nishant Sharma
May 20, 2014

First of all we should draw the free body diagram of the glass with water. Let F $_{up}$ denote the net upward force and F $_{down}$ denote the net downward force on the water surface of the hole.

F $_{up}$ = P $_{0}$ + ( $\sigma$ /r) ,

F $_{down}$ = P $_{0}$ + $\rho$ gh , where the symbols have meanings as stated in the question.

Since the problem requires largest value of radius of hole, it implies that F $_{up}$ $\geq$ F $_{down}$

Now plugging in the values of the respective forces we find that

( $\sigma$ /r) $\geq$ $\rho$ gh

$\implies$ r $\leq$ ( $\sigma$ / $\rho$ gh)

Substituting the values of $\sigma$ , $\rho$ , g and h as given in the problem we find that r $\leq$ 36.73 $\times$ ${10}^{-6}$ m.

Finally as required by the problem largest possible value of r(in $\mu$ m) is $\boxed{36.7}$ .

Maria Fransiska
May 20, 2014

For the water not to drip out of the glass, the pressure pushing the water out must equal the pressure keeping the water in.

P(out) = P(in)

Hydrostatic pressure = Surface tension pressure

ρ.g.h=σ/r 1000(9.8)(0.20)=0.072/r r = 1000(9.8)(0.20)/0.072 = 36.7

Ravi Yadav
May 20, 2014

let the density of water be 'p', 'h' be the height and 'r' be the radius. pressure due to height = pgh pressure at bottom which pushes water back = 0.072/r

at equilibrium( as we want the max radius), pressure due to height = pressure at the bottom which pushes water back. pgh=0.072/r r=0.072/1000 * 9.8 * 0.2 r=(0.3673 * 10 pow -4) m r=36.73 um.

David Mattingly Staff
May 13, 2014

By balancing the pressure in and out of the hole, we get:

$\rho gh=\sigma/r \quad \Rightarrow \quad r=\sigma/\rho gh=3.67 \times 10^{-5}~\mbox{m}=36.7 ~\mu \mbox{m}$ .

So if the hole is about the size of a $\mu \mbox{m}$ , the glass can contain water (even if there are any strong external effects that increase the outward pressure).

Austin Horning
May 20, 2014

If you assume a column of water that is 20cm tall and has a 1cm^2

A magic glass

The answer is 36.7.

5 solutions

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