Pharos University in Alexandria Faculty of Allied Medical Science

1. Pharos University in Alexandria Faculty of Allied Medical Science Biomedical Physics (GRBP-101) Prof. Dr. Mostafa. M. Mohamed Vice Dean Dr. Mervat Mostafa Department of Medical Biophysics Pharos University Dr. Mervat Mostafa

2. Chapter (3) Part (1) Fluids Dr. Mervat Mostafa

3. The differences in the physical properties of solids, liquids, and gases are explained in terms of the forces that bind the molecules. In a solid, the molecules are rigidly bound; a solid therefore has a definite shape and vol­ume. The molecules constituting a liquid are not bound together with suffi­cient force to maintain a definite shape, but the binding is sufficiently strong to maintain a definite volume. Dr. Mervat Mostafa

4. Force and Pressure in a Fluid Solids and fluids transmit forces differently. When a force is applied to one section of a solid, this force is transmitted to the other parts of the solid with its direction unchanged. Because of a fluid’s ability to flow, it transmits a force uniformly in all directions. Therefore, the pressure at any point in a fluid at rest is the same in all directions. The pressure in a fluid increases with depth because of the weight of the fluid above. In a fluid of constant density ρ, the difference in pressure, P2 - P1, between two points separated by a vertical distance h is P2 - P1 = ρ gh Dr. Mervat Mostafa

5. Fluid pressure is often measured in millimeters of mercury, or torr [after Evangelista Torricelli (1608-1674), the first person to understand the nature of atmospheric pressure]. One torr is the pressure exerted by a column of mer­cury that is 1 mm high. Pascal, abbreviated as Pa is another commonly used unit of pressure. The relationship between the torr and several of the other units used to measure pressure follows: 1 torr = 1mm Hg = 13.5 mm water = 1.33 x 103 dyn/cm2 = 1.32 x 10-3 atm = 1.93x10-2psi = 1.33 x 102 Pa (N/m2 ) Dr. Mervat Mostafa

6. Pascal’s Principle When a force F1 is applied on a surface of a liquid that has an area A1, the pressure in the liquid increases by an amount P, given by The ratio A2/ A1 is analogous to the mechanical advantage of a lever. Dr. Mervat Mostafa

7. An illustration of Pascal’s principle. Dr. Mervat Mostafa

8. Hydrostatic Skeleton Let us now calculate the hydrostatic forces inside a moving worm. Con­sider a worm that has a radius r. Assume that the circular muscles running around its circumference are uniformly distributed along the length of the worm and that the effective area of the muscle per unit length of the worm is Am. As the circular muscles contract, they generate a force fm, which, along each centimeter of the worm’s length, is fM = SAM The hydrostatic skeleton. Dr. Mervat Mostafa

9. Here S is the force produced per unit area of the muscle. (Note that fMis in units of force per unit length.) This force produces a pressure inside the worm. The magnitude of the pressure can be calculated with the aid, which shows a section of the worm. The length of the section is L. If we were to cut this section in half lengthwise,, the force due to the pressure inside the cylinder would tend to push the two halves apart. This force is calculated as follows. The surface area A along the cut midsection is A = L x 2r Dr. Mervat Mostafa

10. Calculating pressure inside a worm. Dr. Mervat Mostafa

11. and Dr. Mervat Mostafa

12. Archimedes’ Principle Archimedes’ principle states that a body partially or wholly submerged in a fluid is buoyed upward by a force that is equal in magnitude to the weight of the displaced fluid. The derivation of this principle is found in basic physics texts. We will now use Archimedes’ principle to calculate the power required to remain afloat in water and to study the buoyancy of fish. Power Required to Remain Afloat Whether an animal sinks or floats in water depends on its density. If its density is greater than that of water, the animal must perform work in order not to sink. We will calculate the power P required for an animal of volume V and density ρ to float with a fraction f of its volume submerged. but our approach to the problem will be different. Dr. Mervat Mostafa

13. Because a fraction f of the animal is submerged, the animal is buoyed up by a force FBgiven by FB = gf Vρw where ρwis the density of water. The force FBis simply the weight of the displaced water. The net downward force FBon the animal is the difference between its weight gVρ and the buoyant force; that is, Fd = gV ρ — gVfρ w = gV(ρ — fρ w) Dr. Mervat Mostafa

14. Surface Tension The molecules constituting a liquid exert attractive forces on each other. A molecule in the interior of the liquid is surrounded by an equal number of neighboring molecules in all directions. Therefore, the net resultant intermolecular force on an interior molecule is zero. The situation is different, however, near the surface of the liquid. Because there are no molecules above the surface, a molecule here is pulled predominantly in one direction, toward the interior of the surface. This causes the surface of a liquid to contract and behave somewhat like a stretched membrane. This contracting tendency results in a surface tension that resists an increase in the free surface of the liquid. Dr. Mervat Mostafa

15. Surface tension. total force FTproduced by surface tension tangential to a liquid surface of boundary length L is Ft = TL Dr. Mervat Mostafa

16. When a liquid is contained in a vessel, the surface molecules near the wall are attracted to the wall. This attractive force is called adhesion. At the same time, however, these molecules are also subject to the attractive cohesive force exerted by the liquid, which pulls the molecules in the opposite direction. If the adhesive force is greater than the cohesive force, the liquid wets the container wall, and the liquid surface near the wall is curved upward. If the adhesion is greater than the cohesion, a liquid in a narrow tube will rise to a specific height h, which can be calculated from the following considerations. The weight W of the column of the supported liquid is Dr. Mervat Mostafa

17. Fm = 2πRT The upward component of this force supports the weight of the column of liquid that is, Dr. Mervat Mostafa

18. Angle of contact when (a) liquid wets the wall and (b) liquid does not wet the wall. (a) Capillary rise. (b) Capillary depression. Dr. Mervat Mostafa

19. Therefore, the height of the column is If the adhesion is smaller than the cohesion, the angle Өis greater than 90◦. In this case, the height of the fluid in the tube is depressed. Equation still applies, yielding a negative number for h. These effects are called capillary action. Dr. Mervat Mostafa

20. Soil Water Most soil is porous with narrow spaces between the smallm particles. These spaces act as capillaries and in part govern the motion of water through the soil. When water enters soil, it penetrates the spaces between the small particles and adheres to them. If the water did not adhere to the particles, it would run rapidly through the soil until it reached solid rock. Plant life would then be severely restricted. Because of adhesion and the resulting capillary action, a significant fraction of the water that enters the soil is retained by it. For a plant to withdraw this water, the roots must apply a negative pressure, or suction, to the moist soil. Dr. Mervat Mostafa

21. Fine-grained soil (a) holds water more tightly than coarse-grained soil (b). saturated with water. As the amount of water in the soil decreases, the SMT increases. In loam, for example, with a moisture content of 20% the SMT is about 0.19 atm. When the moisture content drops to 12%, the SMT increases to 0.76 atm. Dr. Mervat Mostafa

22. Contraction of Muscles An examination of skeletal muscles shows that they consist of smaller mus­cle fibers, which in turn are composed of yet smaller units called myofibrils. Further, examination with an electron microscope reveals that the myofibril is composed of two types of threads, one made of myosin, which is about 160 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 the threads can slide past one another Contraction of muscles.

23. If the average diameter of the threads is D, the number of threads N per square centimeter of muscle is approximately The maximum pulling force Ffproduced by the surface tension on each fiber The total maximum force Fmdue to all the fibers in a 1-cm2 area of muscle is Dr. Mervat Mostafa

24. The average diameter D of the muscle fibers is about 100 A˚ (10-6 cm). There­fore, the maximum contracting force that can be produced by surface tension per square centimeter of muscle area is Fm = T x 4 x 106 dyn/cm2 A surface tension of 1.75 dyn/cm can account for the 7 x 106 dyn/cm2 mea­sured force capability of muscles. Because this is well below surface tensions commonly encountered, we can conclude that surface tension could be the source of muscle contraction. Dr. Mervat Mostafa

25. Surfactants Surfactants are molecules that lower surface tension of liquids. (The word is an abbreviation of surface active agent.) The most common surfactant molecules have one end that is water-soluble (hydrophilic) and the other end water insol­uble (hydrophobic) Surface layer of surfactant molecules. Dr. Mervat Mostafa

26. Exercises (Problems For Part (1)) • 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 expend energy to keep itself below the surface. How is Eq. 7.14 modified for this case? Dr. Mervat Mostafa

27. Calculate the perimeter of a platform required to support a 70 kg person solely by surface tension. Dr. Mervat Mostafa