Chapter 3 Fluids Part II PDF

CHAPTER 3 FLUIDS Part II Dr. Mohamed Ali Basic Science Center / Faculty of Dentistry / Phys 101 Email: [email protected] Mobile: 01002836484 Ref.: Physics, James Walker, Chapter 15, page 513-521 ARCHIMEDES’ PRINCIPLE AND BUOYANCY  Buoyant force A fluid surround...

CHAPTER 3 FLUIDS Part II Dr. Mohamed Ali Basic Science Center / Faculty of Dentistry / Phys 101 Email: [email protected] Mobile: 01002836484 Ref.: Physics, James Walker, Chapter 15, page 513-521 ARCHIMEDES’ PRINCIPLE AND BUOYANCY  Buoyant force A fluid surrounding an object exerts a buoyant force in the upward direction. This is due to the fact that pressure increases with depth, and hence the upward force on the object, F2, is greater than the downward force, F1. Fb  F2  F1   gL3   gV  M L g W L F1  P1A  P1L 2 F2  P2 A   P1   gL  L 2  P1L 2   gL3  ARCHIMEDES’ PRINCIPLE  An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object. More generally, if a volume V of an object is immersed in a fluid of density ρf, the buoyant force can be expressed as follows: Fb  V  f g  Note that the buoyant force is equal to the weight of displaced fluid. It does not depend on the weight of the object that displaces the fluid.  CONCEPTUAL CHECKPOINT  how is the scale reading affected?  A flask of water rests on a scale. If you dip your finger into the water, without touching the flask, does the reading on the scale (a) increase, (b) decrease, or (c) stay the same? ANSWER The reading on the scale ……………….  APPLICATIONS OF ARCHIMEDES’ PRINCIPLE Complete Submersion  An interesting application of complete submersion can be found in an apparatus commonly used in determining a person’s body-fat percentage.  We consider the basic physics of the apparatus and the measurement procedure in the next Example. Following the Example, we derive the relation between overall body density and the body-fat percentage. EXAMPLE 3–4 MEASURING THE BODY’S DENSITY  A person who weighs 720.0 N in air is lowered into a tank of water to about chin level. He sits in a harness of negligible mass suspended from a scale that reads his apparent weight. He now exhales as much air as possible and dunks his head underwater, submerging his entire body. If his apparent weight while submerged is 34.3 N, find (a) his volume and (b) his density. SOLUTION  Apply Newton’s second law to the person. W a  Fb W  0 W W a W a  W V P g W  0 VP  W g 720.0 N  34.3N  1000 kg/m3 9.8 m/s2   6.99  102 m 3  EXAMPLE 3–4 MEASURING THE BODY’S DENSITY  (b) the density of the person ρP by using W   PV P g W 720.0 N P    1050 kg/m 3 V P g  6.99  102 m3  9.81 m/s 2   Notice that the density of the person is only slightly greater than the density of seawater THE PERCENTAGE OF BODY FAT  The percentage of body fat can be obtained by noting that: The body fat has a density of ρf = 9 × 102 kg/m3,  The lean body mass (muscles and bone) has a density of ρl = 1.10 × 103 kg/m3.  Suppose that the total body mass M  xf is a fraction of body fats, mf = xf M  (1- xf) the fraction of lean mass; ml = (1- xf) M  The total volume of the body is V = Vf + Vl. THE PERCENTAGE OF BODY FAT  Using the fact that V = m/ρ, we can write the total volume as V  m fat  ml  xf M  1  x f  M f l f l  Combining these results, the overall density of a person’s body, is M 1 P   V xf  1  x f  f l  Solving for the body-fat fraction and substituting the values for ρf and ρl, we find 4950 kg/m3 xf   4.5 P  This result is known as Siri’s formula. THE PERCENTAGE OF BODY FAT  For example, if ρP = 900 kg/m3 (all fat), we find xf = 1; if ρP = 1100 kg/m3 (no fat), we find xf = 0. In the case of Example 2–7 where ρP = 1050 kg/m3, we find that this person’s body-fat fraction is xf = 0.214, for a percentage of 21.4%. This is a reasonable value for a healthy adult male.  Next we consider a low-density object, such as a piece of wood, held down below the surface of a denser fluid.  EXAMPLE 3–5 FIND THE TENSION IN THE STRING  Apiece of wood with a density of 706 kg/m3 is tied with a string to the bottom of a water- filled flask. The wood is completely immersed, and has a volume of 8×10-6 m3. What is the tension in the string? SOLUTION Fb  T  mg  0 T  Fb  mg W  mg   PV P g   706 kg/m 3  8.00  106 m3  9.81 m/s 2   0.0554 N Fb  WV P g Since the wood floats in  1000 kg/m3  8.00  106 m3  9.81 m/s 2   0.0785 N water, its buoyant force when completely immersed is greater than its weight. T  Fb  mg  0.0785  0.0554  0.0231 N  FLOATATION An object floats when it displaces an amount of fluid whose weight (buoyant force) is equal to the weight of the object. Consider, then, a solid of density ρs floating in a fluid of density ρf as in Figure 3–9. If the solid has a volume Vs its total weight is W s  sV s g Similarly, the weight of a volume Vf of displaced fluid is Fb   f V f g Equating these weights, we find the following: W s  Fb sV s g  f V f g Canceling g, and solving for the volume of displaced fluid, Vf, we have   V f V s  s   f  Since, by definition, the volume of displaced fluid, Vf , is the same as the volume of the solid that is submerged, Vsub, we find V  sub  s Vs f CONCEPTUAL CHECKPOINT THE PLIMSOLL MARK  On the side of a cargo ship you may see a horizontal line indicating “maximum load.” When a ship is loaded to capacity, the maximum load line is at water level. The ship shown here has two maximum load lines, one for freshwater and one for salt water. Which line should be marked ANSWER “maximum load for salt water”: (a) the top line or (b) The ………… line (b) the bottom line? should be used in salt water. s Vsub  V f s EXAMPLE 3–7 THE TIP OF THE ICEBERG  What percentage of a floating chunk of ice projects above the level of the water? Assume a density of 917 kg/m3 for the ice and 1000 kg/m3 for the water. V sub s 917 kg/m3    0.917 Vs f 1.00  103 kg/m3  91.7%  Then the part above water surface is 1  0.917  0.083  8.3% CONCEPTUAL CHECKPOINT THE NEW WATER LEVEL I  A cup is filled to the brim with water and a floating ice cube. When the ice melts, which of the following occurs? (a) Water overflows the cup, (b) the water level decreases, or (c) the water level remains the same.  Since the ice cube floats, it displaces a volume of water equal to its weight. But when it melts, it becomes water, and its weight is the same. W s W f s V s  g  f  V  g   f ice water ANSWER  The water level ………………….. CONCEPTUAL CHECKPOINT THE NEW WATER LEVEL II  A cup is filled to the brim with water and a floating ice cube. Resting on top of the ice cube is a small pebble.  As for the pebble, when it floats on  the ice it displaces an amount of water  equal to its weight  When the ice melts, the pebble drops to the  bottom of the cup, where it displaces a  volume of water equal to its own volume. W s W f sV s g  f V f g  ANSWER The water level ………………