Physics 1408-002 Principles of Physics Lecture 21 Chapter 13 April 2, 2009 Sung-Won Lee Sungwon.Lee@ttu.edu Announcement I Lecture note is on the web Handout (6 slides/page) http://highenergy.phys.ttu.edu/~slee/1408/ *** Class attendance is strongly encouraged and will be taken randomly. Also it will be used for extra credits. HW Assignment #8 will be placed on MateringPHYSICS, and is due by 11:59pm on Wednesday, 4/** Announcements II 3rd Exam 4/9 Thursday (Next week) 9: 30 am 10:50 am Chapters 10, 11, 12, 13 Rotation motion, Angular momentum, Statics, Fluids Announcement III SI session by Reginald Tuvilla Thursday 4:00-5:30pm - Holden Hall 106 Next week, test review will be on Monday 04/06 in the BA room 55 from 4:30-7:30. Chapter 13 Fluids 13.3 Pressure! The pressure P of the fluid at the level to which the device has been submerged is the ratio of the force to the area!density and Specific Gravity!Pressure in Fluids!Atmospheric Pressure and Gauge Pressure & Measurement!Pascal s Principle!Buoyancy and Archimedes Principle!Fluids in Motion; Flow Rate and the Equation of Continuity!Bernoulli s Equation & its Applications!! Pressure is a scalar quantity! Because it is proportional to the magnitude of the force! If the pressure P varies over an area A, evaluate!f(=df) on a surface of area!a(=da) as df = P da 1 Pa = 1 N/m 2! Unit of pressure: Pascal (Pa)
Pressure vs. Force! Pressure is a scalar and force is a vector.! The direction of the force producing a pressure is perpendicular to the area of interest. Measuring Pressure! The spring is calibrated by a known force. Example 13-2: Calculating pressure. The two feet of a 60-kg person cover an area of 500 cm 2. Determine the pressure exerted by the two feet on the ground.! The force due to the fluid presses on the top of the piston and compresses the spring.! The pressure on the piston is then measured. Pressure is the same in every direction in a static (i.e. non-moving) fluid at a given depth; if it were not true, the fluid would be flow in motion. If there were a component of force parallel to the solid surface of the container, the liquid would move in response to it. For a liquid at rest, there is no component of force parallel (i.e. F ll = 0) surface of container The pressure at a depth h below the surface of the liquid is due to the weight of the liquid above it. We can quickly calculate the pressure at a depth h in a liquid: F = Mg = "Ahg This relation is valid for any liquid whose density does not change with depth. 13.3 Variation of P with Depth h! Fluids have pressure that varies with depth.! If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium.! Examine the darker region, a sample of liquid within a cylinder! It has a cross-sectional area A! Extends from depth d to d + h below the surface! Three external forces (F = PA) act on the region! The liquid has a density!! Assume the density is the same throughout the fluid! The three forces are:! Downward (- sign) force on the top, P 0 A! Upward (+ sign) on the bottom, PA! Gravity acting downward, Mg! The mass can be found from the density:
13.3 Variation of P with Depth h! Since the net force must be zero (because the fluid is in static equilibrium)! This chooses upward as positive! Solving for the pressure gives P = P 0 +!gh! The pressure P at a depth h below a point in the liquid at which the pressure is P 0 is greater by an amount!gh! If the liquid is open to the atmosphere, and P 0 is the pressure at the surface of the liquid, then P 0 is atmospheric pressure! P 0 = 1.00 atm = 1.013 x 10 5 Pa (REMEMBER!!) = 0 Variation of pressure with depth Density = Mass/Volume "!= M / V Units = kg/m 3 Feel it in your ears in a plane, in a pool! The surface of the water in a storage tank is 30 m above a water faucet in the kitchen of a house. Calculate the difference in water pressure between the faucet and the surface of the water in the tank. Calculate the force due to water pressure exerted on a 1.0 m x 3.0 m aquarium viewing window whose top edge is 1.0 m below the water surface. 13-4 Atmospheric Pressure and Gauge Pressure At sea level the atmospheric pressure is about 1.013 x 10 5 N/m 2 ; this is called 1 atmosphere (atm). Another unit of pressure is the bar: 1 bar = 1.00 x 10 5 N/m 2. Standard atmospheric pressure is just over 1 bar. 13-4 Atmospheric Pressure and Gauge Pressure Most pressure gauges measure the pressure above the atmospheric pressure this is called the gauge pressure. The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.
Absolute vs. Gauge Pressure!P = P0 + "gh! P: the absolute pressure!!! P0: the atmospheric pressure!! The gauge pressure: P P0 (= "gh) This is what you measure in your tires 13.5 Pascal s Law! The pressure in a fluid depends on depth & on the value of P0! An increase in pressure at the surface must be transmitted to every other point in the fluid! This is the basis of Pascal s law 0 P = P +!gh! Fig: A large output force can be applied by means of a small input force! The volume (A1*!x1) of liquid pushed down on the left must equal the volume pushed up on the right (A2*!x2)! Since the volumes are equal A2/A1 = "x1/"x2! Combining the equations,! which means (using W = F!x), W1 = W2! This is a consequence of Conservation of Energy 13.6 Pressure Measurements: Barometer! Invented by Torricelli to measure atmospheric pressure.! A long closed tube is filled with mercury and inverted in a dish of mercury! The closed end is nearly a vacuum! He measures atmospheric pressure as!!hg = density of the mercury (see table)! h = the height of the mercury column! Let us determine the h for one atmosphere of pressure, p0 = 1 atm = 1.013 x 105 Pa: ==> h = p0 /!Hg g = 0.706 m 13-6 Measurement of Pressure; Gauges and the Barometer Pressure is measured in a variety of different units. This table gives the conversion factors. 13.6 Pressure Measurements: Manometer! A device for measuring the pressure of a gas contained in a vessel! One end of the U-shaped tube is open to the atmosphere! The other end is connected to the pressure to be measured! Pressure @ B = P0+!gh Reminder: P = P0 +!gh 13.7 Buoyant Force Q: Have you ever tried to push beach ball under water?! A: Extremely difficult to do because of the large upward force exerted by the water on the ball.!! The beach ball is in equilibrium, there must be an upward force to balance the downward force!! The upward force, B, must equal (in magnitude) the downward gravitational force, Fg!! The upward force is called the buoyant force!
This is an object submerged in a fluid. There is a net force on the object because the pressures at the top and bottom of it are different. Archimedes principle: The buoyant force, FB, is found to be the upward force on the same volume of water: The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object. Archimedes's Principle! Before we proceed with a few examples, it is instructive for us to discuss about two common situations! A 70-kg ancient statue lies at the bottom of the sea. Its volume is 3.0 x 104 cm3. How much force is needed to lift it?! A totally submerged object!! A floating (partly submerged) object Archimedes's Principle: Totally Submerged Object! When an object is totally submerged in a fluid of density! the magnitude of upward buoyant force is!! If the object has mass M and density, "obj, the downward gravitational force is!!! Fg = w = Mg =!! So, the net force: B - Fg =! volume of object! " Archimedes's Principle: Totally Submerged Object! If the density of the object is less than the density of the fluid, (light object) the unsupported object accelerates upward!! If the density of the object is more than the density of the fluid, (heavy object) the unsupported object sinks!! The motion of an object in a fluid is determined by the densities of the fluid and the object! "
Archimedes's Principle: Floating Object! Now consider an object of volume V obj and density " obj < " fluid in static equilibrium - partially submerged (see Fig)!! The upward buoyant force is balanced by the downward force of gravity: F g = B!! The following equation is tell us that the fraction of volume of a floating object is equal to the ratio of the density of the object to that of the fluid.! If an object s density is less than that of water, there will be an upward net force on it, and it will rise until it is partially out of the water. (a) The fully submerged log accelerates upward because F B > mg. It comes to equilibrium (b) when!f = 0, so F B = mg = (1200kg)g. Thus 1200 kg, or 1.2 m 3, of water is displaced.