Physics Tools and Applications Pressure, Fluids, and Archimedes’ Principle

1. OSE:  = m / V. Physics Tools and Applications Pressure, Fluids, and Archimedes’ Principle Density and Specific Gravity Density: how much mass there is in a given volume. Example: how much mass is in a cubic meter of air? Table 10-1 gives the density of air as 1.29 kg/m3. m =  V = (1.29 kg/m3) (1 m3) = 1.29 kg. You may encounter “specific gravity:” Specific gravity = ( of material)/( of water at 4C)

2. OSE: P = F / A. Pressure in Fluids In words: pressure is the perpendicular component of force on a surface divided by the surface area. I won’t trick you with a non-perpendicular force. The unit of pressure is the pascal: 1 Pa =1 n / 1 m3. Example: a waterbed mattress is about 2 m x 1.5 m x 0.3 m when filled with water. How much does it weigh? What pressure does it exert on the floor? OSE:  = m / V  m =  V w = m g  w =  g V

3. w =  g V = (1x103)(9.8)(2·1.5·0.3) = 8820 N = 1982 pounds (a ton!) OSE: P = F / A = W / A P=  g (L W H) / (L W) =  g H = (1x103)(9.8)(0.3) = 2940 Pa = 0.029 atm (we’ll discuss this later) = 0.43 pounds / in2 Of course, if the mattress rests on a frame which has only a few square inches of contact area on the floor, the pressure is enormous.

4. Interesting equation in the previous calculation: P = gH. The pressure exerted by a fluid depends only on the height of the fluid. Is this true in general? Important facts about fluids:  A fluid exerts pressure in all directions. • The force due to the pressure of a fluid at rest is always  to any surface it is contact with (otherwise the fluid would flow parallel to the surface and not be at rest).

5. OSE: Pbelow - Pabove =  g h, OSE: P = - g y (y-axis up). Back to my question aboutP = gH. Is it true that the pressure exerted by a fluid depends only on the height of the fluid? Most texts derive an equation P = gh for the pressure in a fluid a depth h below the surface, and another equation P = gh for the change in pressure on going down an amount h. Better ways of writing these equations are: where h is the magnitude of the distance between “below” and “above,” and

6. The above equations for P are true if the fluid is uniform (its density does not change with depth). One would expect  to vary with depth because pressure varies with depth. However, for a relatively incompressible fluid like water the variation is so small that the approximation is a good one. Note that the variation of pressure with depth does not depend on the shape of the container, only with the depth! Application: you go home, turn on the faucet, and water comes out. Why?

7. ywater = H+h H y yfaucet=h P = - g (yfaucet - ywater) = - g [h-(H+h)] = gH.

8. Atmospheric Pressure and Gauge Pressure Air pressure varies with height and weather conditions. The “average” atmospheric pressure at sea level is defined as 1 atm. It turns out that 1 atm = 1.013x105 n/m2 = 101.3x103 Pa = 101.3 kPa. Also, 1 atm = 14.7 pounds/in2 1 bar = 100 kPa (slightly less than 1 atm).

9. Ptire air Patmosphere OSE: P = PA + PG. You use a tire gauge to measure tire pressure. It records 28 psi. Is that the actual pressure in the tire? You measure the pressure PG, which is the gauge pressure, or the difference between the absolute pressure and atmospheric pressure. Thus P = 28 psi + 14.7 psi = 42.7 psi is the actual pressure inside.

10. F2 = ? F1 A1 A2 P2 = ? P1 = F1 / A1 fluid Pascal’s Principle Pressure applied to a confined fluid increases the pressure throughout by the same amount. Pressure is same throughout, so P2 = F2/A2 = P1 = F1/A1.Thus F2 = (A2 / A1) F1.

11. Buoyancy and Archimedes’ Principle Who remembers the story of Archimedes and the king’s crown?

12. http://www.engineering.usu.edu/jrestate/workshop/buoyancy.htm:http://www.engineering.usu.edu/jrestate/workshop/buoyancy.htm: “As the story goes, the king of Syracuse had given a craftsman a certain amount of gold to be made into an exquisite crown. When the project was completed, a rumor surfaced that the craftsman had substituted a quantity of silver for an equivalent amount of gold, thereby devaluing the crown and defrauding the king. Archimedes was tasked with determining if the crown was pure gold or not. The Roman architect Vitruvious relates the story: ‘While Archimedes was considering the matter, he happened to go to the baths. When he went down into the bathing pool he observed that the amount of water which flowed outside the pool was equal to the amount of his body that was immersed. Since this fact indicated the method of explaining the case, he did not linger, but moved with delight, he leapt out of the pool, and going home naked, cried aloud that he had found exactly what he was seeking. For as he ran he shouted in Greek: Eureka! Eureka! (eureka translated is "I have found it").’

13. “Although there is speculation as to the authenticity of this story, it remains famous. Probably no other tale in all of science combines the elements of brilliance and bareness quite so effectively. Whether the story is true or not, there is no doubt to the truth of Archimedes understanding of buoyancy.” Archimedes’ death also makes an interesting story: “In 212 BC Syracuse surrendered to Rome. Before sending his men to sack the city Marcellus told them "Spare that mathematician". Plutarch records what happened next: ‘As fate would have it, intent upon working out some problem by a diagram, and having fixed his mind alike and his eyes upon the subject of his speculation, he [Archimedes] never noticed the incursion of the Romans, nor that the city was taken. In this transport of study and contemplation, a soldier, unexpectedly coming up to him, commanded him to follow Marcellus; which he declining to do before he had worked out his problem to a demonstration, the soldier, enraged, drew his sword and ran him through.’ ”

14. Finally, you can go here to read why the traditional story of Archimedes and the king’s crown is probably not true: http://www.mcs.drexel.edu/~crorres/Archimedes/Crown/CrownIntro.html What is important for us, and is what Archimedes understood, is that an object immersed in a fluid experiences a buoyant force equal in magnitude to the weight of the fluid displaced. Why? An object submerged in a fluid experiences pressure. The pressure increases with depth. Because P = F / A, the force per unit area on the object also increases with depth.

15. Force on object by fluid increases with depth. Force is always  to surface. Horizontal forces cancel. Upward force on bottom is greater than downward force on top.

16. OSE: B = fluid g Vdisplaced. Summary: an object submerged in a fluid experiences a net upward force because the pressure in the fluid increases with depth. The net force is independent of the shape of the object, and depends only on the weight of the fluid displaced. The weight of the fluid is mg = (g)V. The buoyant force is If the object is completely submerged, Vdisplaced = Vobject. If the object is only partially submerged, Vdisplaced is the volume of the submerged part of the object.

17. B a y w = mg x Example: a log having a volume of 2.0 m3 and a density of 0.9 g/cm3 is held under water and released. What is the acceleration of the log at the instant it is released? First ask yourself: what kind of a problem is this? “What is the acceleration…” so it must be a kinematics or force problem. You are not given information about positions, velocities, or times, so it sounds like a force problem. Draw a free body diagram! OSE: Fy = may By + wy = may

18. B a y w = mg x Substitute for generic quantities: fluid g Vlog + (-mg) = +ma Solve: a = (fluid g Vlog - mg) / m Use the density and volume of the log to get its mass: m = log Vlog a = (fluid g Vlog - log Vlog g) / log Vlog a = (fluid - log) g / log a = ( 1.0 - 0.9) g / 0.9(using water = 1 g/cm3) (I admit--a bit sloppy with units here.) a = (1/9) g