Solutions to Theory Problems

S3‒1
Theory Problem 3: Water and Objects (10 points)
Part A. Merger of water drops (2.0 pt)
A.1

The surface energy per drop before the merger is
𝐸 = 4𝜋𝑎2 𝛾.

(S3.1)

Therefore, the surface energy difference becomes
Δ𝐸 = 4𝜋 (2 − 22/3 ) 𝑎2 𝛾.

(S3.2)The transfer of surface energy to kinetic energy is represented by
𝑀 𝑣2 /2 = 𝑘Δ𝐸,

(S3.3)

where 𝑘 = 0.06 and 𝑀 = 4𝜋𝑎3 𝜌/3 × 2 = 8𝜋𝑎3 𝜌/3 is the mass of the drop after the merger. The numericalcomputation gives𝑣=√

2𝑘Δ𝐸𝑘𝛾0.06 × (7.27 × 10−2 )= √3 (2 − 22/3 )= √3 (2 − 22/3 ) ×= 0.232 m/s.
𝑀𝜌𝑎(1.0 × 103 ) × (100 × 10−6 )(S3.4)▷ Note: We point out an interesting phenomenon related to this question. On a superhydrophobic surface, when small droplets merge, they release surface energy, causing the surface area to shrink. Thisenergy release propels the merged droplet to jump up. This phenomenon mirrors the natural mechanism seen in cicadas. Cicadas' wings, which possess superhydrophobic surfaces, facilitate the removalof water droplets upon coalescence. This process serves as a natural self-cleaning system, converting
surface energy to kinetic energy, as reported in the following paper: Wisdom et. al., Proc. Natl. Acad. Sci.
USA 110, 7992−7997 (2013).

A.1

2.0 pt𝑣 = 0.23 m/s

Part B. A vertically placed board (4.5 pt)
B.1
Consider a vertical upright column-shaped water block as shown in thehatched area of Fig. S3-1. The vertical force balance with respect to thisblock per unit area leads to 𝑃 + 𝜌𝑔𝑧 = 𝑃0 .

Fig. S3-1
B.1

0.6 pt𝑃 = 𝑃0 − 𝜌𝑔𝑧

B.2
Because the atmospheric pressure 𝑃0 exerts no net horizontal force on the water block, we have𝑧1𝑓𝑥 = ∫ (−𝜌𝑔𝑧)𝑑𝑧 =𝑧2

1
𝜌𝑔 (𝑧22 − 𝑧12 ) .
2(S3.5)

This force acts in the leftward direction.
1

Solutions to Theory Problems

S3‒2▷ Note: The reason why 𝑃0 exerts no net horizontal force is understood as follows. Consider a smallarea (infinitesimally divided piece) near the surface, which is regarded as a right-angled triangle (seeFig. S3-2).

Fig. S3‒2The horizontal component of the combined force exerted on the right-angled triangle ABC per unit
length along the 𝑦-axis by atmospheric pressure is𝑃0 × AB − 𝑃0 × AC ×

AB=0
AC

Integrating infinitesimal pieces over a finite domain yields a finite volume of water, while the net force
remains zero.

B.2

0.8 pt1𝑓𝑥 = 𝜌𝑔 (𝑧22 − 𝑧12 )2

B.3
The horizontal component of the surface tension acting on the water block is 𝛾 cos 𝜃2 − 𝛾 cos 𝜃1 . Thus,
the horizontal force balance is expressed as
𝑓𝑥 + 𝛾 cos 𝜃2 − 𝛾 cos 𝜃1 = 0.

(S3.6)

B.3

0.8 pt𝑓𝑥 = 𝛾 cos 𝜃1 − 𝛾 cos 𝜃2

B.4
From the results of B.2 and B.3, we have11𝜌𝑔𝑧12 + 𝛾 cos 𝜃1 = 𝜌𝑔𝑧22 + 𝛾 cos 𝜃2 .
22(S3.7)

Since this equation holds at an arbitrary point (𝑥, 𝑧) on the water surface, we conclude1
𝜌𝑔𝑧 2 + 𝛾 cos 𝜃 = constant,
2(S3.8)which is written as
1 𝑧 𝑎( ) + cos 𝜃(𝑥) = constant,2 ℓwith 𝑎 = 2 and ℓ = √(S3.9)𝛾.
𝜌𝑔▷ Note: The equaiton (S3.9) is a kind of conservation law. The constant ℓ is called the capillary length.

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Solutions to Theory Problems

S3‒3B.4

0.8 pt𝛾𝑎 = 2, ℓ = √𝜌𝑔

B.5
The derivative of the water surface coordinate 𝑧, denoted by 𝑧 ′ , is associated with the angle ofinclination 𝜃, given by the equation: 𝑧′ = tan 𝜃. This relation yields
cos 𝜃 =1,
√1 + (𝑧 ′ )2(S3.10)

which leads to1cos 𝜃 ≃ 1 − (𝑧 ′ )2 .
2(S3.11)

Plugging this into Eq. (S3.9), we obtain𝑧2− 𝑧 ′2 = const.
ℓ2(S3.12)

Taking the derivative of both sides with respect to 𝑥, we have𝑧″ =𝑧,ℓ2(S3.13)

which is the differential equation that determines the water surface form.
Its general solution is𝑧 = 𝐴𝑒𝑥/ℓ + 𝐵𝑒−𝑥/ℓ .

(S3.14)The boundary condition, 𝑧(∞) = 0, leads to 𝐴 = 0.
The boundary condition, 𝑧′ (0) = tan 𝜃0 , leads to 𝐵 = −ℓ tan 𝜃0 .
B.5

1.5 pt𝑧 (𝑥) = −ℓ tan 𝜃0 𝑒−𝑥/ℓPart C. Interaction between two rods (3.5 pt)
C.1The horizontal component of the force due to the pressure is𝑧b

∫ (𝜌𝑔𝑧) 𝑑𝑧 =𝑧a1
𝜌𝑔 (𝑧b2 − 𝑧a2 )2(S3.15)

Let 𝑧bottom be the 𝑧-coordinate at the bottom of the rod. Then, we have𝑧bottom𝑧b𝑧b𝑧a𝐹𝑥 = ∫(−𝜌𝑔𝑧)𝑑𝑧 + (− ∫𝑧bottom(−𝜌𝑔𝑧)𝑑𝑧) = ∫ (𝜌𝑔𝑧)𝑑𝑧(S3.16)

𝑧a▷ Note: The fact that the contribution due to the pressure does not depend on the shape of the crosssection can be demonstrated as follows. The pressure at the point 𝑠 on the contour 𝐶 along the crosssectional boundary is(S3.17)−𝑃 𝑛𝑑𝑠̂ = (−𝑃0 +𝜌𝑔𝑧)𝑛𝑑𝑠.
̂3

Solutions to Theory Problems

S3‒4Let 𝑥̂ be the unit vector pointing the positive 𝑥-direction and noting 𝑥̂ ⋅ 𝑛𝑑𝑠
̂ = 𝑑𝑧 (see Fig. S3-3), we obtainits horizontal component as(S3.18)−𝑃 𝑛̂ ⋅ 𝑥𝑑𝑠
̂ = −𝑃0 𝑑𝑧+𝜌𝑔𝑧𝑑𝑧.
Integrating along the contour 𝐶.1 We obtain𝑧b∫ (𝜌𝑔𝑧)𝑑𝑧 =𝑧a1
𝜌𝑔 (𝑧b2 − 𝑧a2 )2(S3.19)

Fig. S3‒3
C.1

C.2

1.0 pt1𝐹𝑥 = 𝜌𝑔 (𝑧b2 − 𝑧a2 ) + 𝛾 (cos 𝜃b − cos 𝜃a )2By applying the boundary conditions to Eq. (S3.8), we obtain1
1
𝜌𝑔𝑧a2 + 𝛾 cos 𝜃a = 𝜌𝑔𝑧02 + 𝛾⏟22⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝑥=𝑥a(S3.20)

𝑥=01
𝜌𝑔𝑧b2 + 𝛾 cos 𝜃b = 𝛾⏟2⏟⏟⏟⏟⏟⏟⏟
𝑥→∞(S3.21)𝑥=𝑥b

1Then, 𝐹𝑥 = − 𝜌𝑔𝑧02 is obtained by subtracting (S3.20) from (S3.21).
2▷ Note: The physical background of this problem is as follows. When a single rod is placed on the water
surface, the shape of the water surface on both sides of the rod remains the same. In other words, therod is placed in an environment that exhibits the left-right symmetry. Then, there is no force acting on
the rod. On the other hand, when two rods are placed on the water surface, the left-right symmetry of
the water surface is broken from the perspective of each rod. As a result, an attractive force is generated.
The displacement of the water surface at the midpoint between the two rods differs from that of a horizontal water surface. This deviation is represented by 𝑧0 . That is to say, 𝑧0 plays the role of a symmetrybreaking parameter (see Fig. S3-4).
The fact that the attractive force between the two rods is determined solely by this parameter suggeststhat the symmetry breaking directly becomes the origin of the force. This corresponds to the fundamental principle in physics that relates symmetry breaking to force generation.

1 This integral is symbolically written as ∮ (−𝑃 𝑛̂ ⋅ 𝑥𝑑𝑠).
̂𝐶

4

Solutions to Theory Problems

S3‒5Fig. S3‒4
C.2

C.3

1.5 pt1𝐹𝑥 = − 𝜌𝑔𝑧02
2The general solution for the water surface height is given by the equation𝑧(𝑥) = 𝐴𝑒𝑥/ℓ + 𝐵𝑒−𝑥/ℓ .

(S3.22)

By considering the left-right symmetry, we find(S3.23)𝐴 = 𝐵.
Applying the boundary condition 𝑧(0) = 𝑧0 , we obtain(S3.24)

𝐴 + 𝐵 = 𝑧0 .
We thus have(S3.25)𝐴 = 𝑧0 /2, 𝐵 = 𝑧0 /2.

C.3

1.0 pt2𝑧𝑧0 = 𝑥 /ℓ a −𝑥 /ℓ𝑒 a +𝑒 a5