6. More on Fluid statics
Last time, we saw that the balance of curvature and hydrostatic pressures requires
−ηxx
.
−ρgη = σ∇ · n = σ (1+η
2 )3/2
x
We linearized, assuming ηx ≪ 1, to find η(x). Note: we can integrate directly
ηx ηxx
d η2
d
1
ρg ⇒
⇒
ρgηηx = σ
=σ
3/2
2
dx (1 + ηx2 )1/2
dx 2
(1 + ηx )
Z ∞
1
d
1
1
ρgη 2 =
dx = 1 −
= 1 − sin θ
1/2
1/2
2
2σ
dx
x
(1 + ηx )
(1 + ηx2 )
1
σ sin θ + ρgη 2 = σ
2
(6.1)
Figure 6.1: Calculating the shape and maximal rise height of a static meniscus.
Maximal rise height: At z = h we have θ = θe , so from (6.1) 12 ρgh2 = σ(1 − sin θe ), from which
h=
√
2ℓc (1 − sin θe )1/2
where ℓc =
p
σ/ρg
(6.2)
Alternative perspective: Consider force balance on the meniscus.
Horizontal force balance:
1
σ
+
ρgz 2
= |{z}
σ
| sin
{z θ}
2
| {z }
T2
horiz. pr oj ection of T1
(6.3)
hydrostatic suction
Vertical force balance:
σ
| cos
{z θ}
=
vert. proj. of T1
Z
|
∞
x
ρgzdx
{z }
(6.4)
weight of f luid
At x = 0, where θ = θe , gives σ cos θe = weight of fluid displaced above z = 0.
Note: σ cos θe = weight of displaced fluid is +/− according to whether θe is smaller or larger than π2 .
Floating Bodies Without considering interfacial effects, one anticipates that heavy things sink and light
things float. This doesn’t hold for objects small relative Rto the capillary length.
Recall: Archimedean force on a submerged body FA = S pndS = ρgVB .
In general,
the hydrodynamic force acting on a body in a fluid
R
Fh = S T · ndS,
R where T = −
R pI + 2µE = −pI for
R static fluid.
HereR Fh = − S pndS = − S ρgzndS = −ρg V ∇z dV by divergence theorem. This is equal to
−ρg V dV ẑ = −ρgV ẑ = weight of displaced fluid. The archimedean force can thus support weight
of a body M g = ρB gV if ρF > ρB (fluid density larger than body density); otherwise, it sinks.
22
6.1. Capillary forces on floating bodies
Chapter 6. More on Fluid statics
Figure 6.2: A heavy body may be supported on a fluid surface by a combination of buoyancy and surface
tension.
6.1
Capillary forces on floating bodies
• arise owing to interaction of the menisci of floating bodies
• attractive or repulsive depending on whether the menisci are of the same or opposite sense
• explains the formation of bubble rafts on champagne
• explains the mutual attraction of Cheerios and their attraction to the walls
• utilized in technology for self-assembly on the microscale
Capillary attraction Want to calculate the attractive force between two floating bodies separated by
a distance R. Total energy of the system is given by
Etot = σ
f
dA(R) +
1
∞
dx
−∞
1
h(x)
ρgzdz
(6.5)
0
where the first term in (6.5) corresponds to the total surface energy when the two bodies are a distance
R apart, and the second term is the total gravitational potential energy of the fluid. Differentiating (6.5)
yields the force acting on each of the bodies:
F (R) = −
dEtot (R)
dR
(6.6)
Such capillary forces are exploited by certain water walking insects to climb menisci. By deforming the
free surface, they generate a lateral force that drives them up menisci (Hu & Bush 2005).
MIT OCW: 18.357 Interfacial Phenomena
23
Prof. John W. M. Bush
MIT OpenCourseWare
http://ocw.mit.edu
357 Interfacial Phenomena
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.