Section 7.10 Contraction of Muscles
95
If the density of the insect is 1 g/cm3, then with T 72.8 dyn/cm, themaximum weight is
2/3
Wm .8 (980)1/3 or Wm 19.7 dyn The mass of the insect is therefore about 2 × 10−2 g. The corresponding linear size of such an insect is about 3 mm.
As is shown in Exercise 7-11, a 70 kg person would have to stand on a platform about 10 km in perimeter to be supported solely by surface tension.
(This is a disk about 3.2 km in diameter.)
7.10
Contraction of Muscles myofibrils.
Further, examination with an electron microscope reveals that the myofibrilis composed of two types of threads, one made of myosin, which is about160 ˚ A (1 ˚ A 10−8 cm) in diameter, and the other made of actin, which has a diameter of about 50 ˚ A. Each myosin-actin unit is about 1 mm long. The threads are aligned in a regular pattern with spaces between threads so that thethreads can slide past one another, as shown in Fig. 7.9.
FIGURE 7.9 Contraction of muscles.
Chapter 7 Fluids
Clearly, a force must act along the myosin-actin threads to produce such a contracting motion. The physical nature of this force is not fully understood.
It has been suggested by Gamow and Ycas [7-5] that this force may be dueto surface tension, which is present not only in liquids but also in jellylikematerials such as tissue cells. The motion of the threads is then similar tocapillary movement of a liquid. Here the movement is due to the attractionbetween the surfaces of the two types of thread. The surface attraction maybe triggered by a release of the Ca2+ ions. Let us now estimate the forceper square centimeter of muscle tissue that could be generated by the surfacetension proposed in this model.
If the average diameter of the threads is D, the number of threads N per square centimeter of muscle is approximately
N
1
(7.26)
π × D2
4
The maximum pulling force Ff produced by the surface tension on each fiber is, from Eq. 7.16, Ff πDT
(7.27)
The total maximum force Fm due to all the fibers in a 1-cm2 area of muscle is Fm NFf 4T
(7.28)
D D of the muscle fibers is about 100 ˚ A (10−6 cm). There fore, the maximum contracting force that can be produced by surface tensionper square centimeter of muscle area is Fm T × 4 × 106 dyn/cm2 A surface tension of 1.75 dyn/cm can account for the 7 × 106 dyn/cm2 measured force capability of muscles. Because this is well below surface tensionscommonly encountered, we can conclude that surface tension could be thesource of muscle contraction. This proposed mechanism, however, shouldnot be taken too seriously. The actual processes in muscle contraction aremuch more complex and cannot be reduced to a simple surface tension model(see [7-7 and 7-9]).
97
7.11
Surfactants
Surfactants are molecules that lower surface tension of liquids. (The word is anabbreviation of surface active agent.) The most common surfactant moleculeshave one end that is water-soluble (hydrophilic) and the other end water insoluble (hydrophobic) (see Fig. 7.10). As the word implies, the hydrophilic endis strongly attracted to water while the hydrophobic has very little attractionto water but is attracted and is readily soluble in oily liquids. Many differenttypes of surfactant molecules are found in nature or as products of laboratorysynthesis.
When surfactant molecules are placed in water, they align on the surface with the hydrophobic end pushed out of the water as shown in Fig. 7.11. Such
an alignment disrupts the surface structure of water, reducing the surface tension. A small concentration of surfactant molecules can typically reduce surface tension of water from 73 dyn/cm to 30 dyn/cm. In oily liquids, surfactantsare aligned with the hydrophilic end squeezed out of the liquid. In this casethe surface tension of the oil is reduced.
Hydrophobe Hydrophile FIGURE 7.10 Schematic of a surfactant molecule.
FIGURE 7.11 Surface layer of surfactant molecules.
Chapter 7 Fluids
(c) FIGURE 7.12 Action of detergents. (a) Oil drop on a wet spot. (b) The hydrophobic end of surfactant molecules enter the oil spot. (c) The oil spot break up into smallersection surrounded by hydrophilic ends.
The most familiar use of surfactants is as soaps and detergents to wash away oily substances. Here the hydrophobic end of the surfactants dissolvesinto the oil surface while the hydrophilic end remains exposed to the surrounding water as shown in Fig. 7.12. The aligned surfactant molecules reduce thesurface tension of the oil. As a result, the oil breaks up into small dropletssurrounded by the hydrophilic end of the surfactants. The small oil dropletsare solubilized (that is suspended or dissolved) in the water and can now bewashed away.
Surfactants are widely used in experimental biochemistry. In certain types of experiments, for example, proteins that are hydrophobic such as membraneproteins and lipoproteins must be dissolved in water. Here surfactants are usedto solubilize the proteins in a process similar to that illustrated in Fig. 7.12.
The hydrophobic ends of the surfactant molecules dissolve into the surface ofthe protein. The aligned hydrophilic ends surround the protein, solubilizing itin the ambient water.
99
Some insects such as the Microvelia not only stand on water but also utilize surface tension for propulsion. They secrete a substance from theirabdomen that reduces the surface tension behind them. As a result they arepropelled in a forward direction. Here the effect is similar to cutting a tautrubber membrane which then draws apart, each section moving away from thecut. This effect known as Marangoni propulsion can be demonstrated simplyby coating one end of a toothpick with soap, and placing it in water. Thesoap acting as the surfactant reduces the surface tension behind the coatedend resulting in the acceleration of the toothpick away from the dissolvedsoap.
Experiments have shown that the surfactant excreted by insects reduces the surface tension of water from 73 dyn/cm to about 50 dyn/cm. Measurementsshow that during Marangoni propulsion, Microvelia can attain peak speeds of17 cm/sec. This value is in agreement with a simple calculation presented inExercise 7-12.
The importance of surfactants in the process of breathing is described in Chapter 9.
EXERCISES
7-1. Verify Eqs. 7.11 and 7.14.
7-2. With the nose above the water, about 95% of the body is submerged.
Calculate the power expended by a 50-kg woman treading water in thisposition. Assume that the average density of the human body is aboutthe same as water (ρ ρw 1 g/cm3) and that the area A of the limbsacting on the water is about 600 cm2.
7-3. In Eq. 7.14, it is assumed that the density of the animal is greater than the density of the fluid in which it is submerged. If the situation is reversed,the immersed animal tends to rise to the surface, and it must expendenergy to keep itself below the surface. How is Eq. 7.14 modified forthis case?
7-4. Derive the relationship shown in Eq. 7.15.
7-5. Calculate the pressure 150 m below the surface of the sea. The density of sea water is 1.026 g/cm3.
7-6. Calculate volume of the swim bladder as a percent of the total vol ume of the fish in order to reduce the average density of the fish from1.067 g/cm3 to 1.026 g/cm3.
Chapter 7 Fluids 7-7. The density of an animal is conveniently obtained by weighing it first in air and then immersed in a fluid. Let the weight in air and in the fluidbe respectively W1 and W2. If the density of the fluid is ρ1, the averagedensity ρ2 of the animal is
W1 ρ2 ρ1 W1 −W2 Derive this relationship.
7-8. Starting with Eq. 7.20, show that the pressure P required to withdraw the water from a capillary of radius R and contact angle θ is
P 2T cos θ
R
With the contact angle θ 0◦, determine the pressure required to withdraw water from a capillary with a 10−3 cm radius. Assume that thesurface tension T 72.8 dyn/cm.
7-9. If a section of coarse-grained soil is adjacent to a finer grained soil of the same material, water will seep from the coarse-grained to the finergrained soil. Explain the reason for this.
7-10. Design an instrument to measure the SMT. (You can find a description of one such device in [7-4].)
7-11. Calculate the perimeter of a platform required to support a 70 kg person solely by surface tension.
7-12. (a) Estimate the maximum acceleration of the insect that can be pro duced by reducing the surface tension as described in the text. Assumethat the linear dimension of the insect is 3 × 10−1 cm and its mass is3 × 10−2 g. Further, assume that the surface tension difference betweenthe clean water and surfactant altered water provides the force to accelerate the insect. Use surface tension values provided in the text.
(b) Calculate the speed of the insect assuming that the surfactantrelease lasts 0.5 sec.
The Motion of Fluids The study of fluids in motion is closely related to biology and medicine. Infact, one of the foremost workers in this field, L. M. Poiseuille (1799–1869),was a French physician whose study of moving fluids was motivated by hisinterest in the flow of blood through the body. In this chapter, we will reviewbriefly the principles governing the flow of fluids and then examine the flowof blood in the circulatory system.
8.1
Bernoulli’s Equation Bernoulli’s equation, which gives the relationship between velocity,pressure, and elevation in a line of flow. Bernoulli’s equation states that at anypoint in the channel of a flowing fluid the following relationship holds:
P + ρgh + 1 ρv2 Constant
(8.1)
2
Here P is the pressure in the fluid, h is the height, ρ is the density, and v isthe velocity at any point in the flow channel. The first term in the equationis the potential energy per unit volume of the fluid due to the pressure in thefluid. (Note that the unit for pressure, which is dyn/cm2, is identical to erg/cm3,which is energy per unit volume.) The second term is the gravitational potentialenergy per unit volume, and the third is the kinetic energy per unit volume.
Bernoulli’s equation follows from the law of energy conservation. Because the three terms in the equation represent the total energy in the fluid, in the
101
102
Chapter 8 The Motion of Fluids FIGURE 8.1 Flow of fluid through a pipe with two segments of different areas.
absence of friction their sum must remain constant no matter how the flow isaltered.
We will illustrate the use of Bernoulli’s equation with a simple example.
Consider a fluid flowing through a pipe consisting of two segments with crosssectional areas A1 and A2, respectively (see Fig. 8.1). The volume of fluidflowing per second past any point in the pipe is given by the product of thefluid velocity and the area of the pipe, A × v. If the fluid is incompressible, ina unit time as much fluid must flow out of the pipe as flows into it. Therefore,the rates of flow in segments 1 and 2 are equal; that is, A1v1 A2v2 or v2 A1 v1
(8.2)
A2 In our case A1 is larger than A2 so we conclude that the velocity of the fluidin segment 2 is greater than in segment 1.
Bernoulli’s equation states that the sum of the terms in Eq. 8.1 at any point in the flow is equal to the same constant. Therefore the relationship betweenthe parameters P, ρ, h, and v at points 1 and 2 is P1 + ρgh1 + 1 ρv2 P2 + ρgh2 + 1 ρv2
(8.3)
2
1
2
2
where the subscripts designate the parameters at the two points in the flow.
Because in our case the two segments are at the same height (h1 h2), Eq. 8.2can be written as P1 + 1 ρv2 P2 + 1 ρv2
(8.4)
2
1
2
2
Because v2 (A1/A2)v1, the pressure in segment 2 is
2
A1 P2 P1 − 1 ρv2 − 1
(8.5)
2
1
A2 This relationship shows that while the flow velocity in segment 2 increases,the pressure in that segment decreases.
