IGCSE & CSEC Physics Lecture 7 PDF

IGCSE & CSEC  Mr.Emile Wihby Lecturer/ Instructor  BSc. Contact info:  784-6377  [email protected]  CSEC & IGCSE Physics Syllabus 6) ELECTRICITY 1) MECHANICS 7)...

IGCSE & CSEC  Mr.Emile Wihby Lecturer/ Instructor  BSc. Contact info:  784-6377  [email protected]  CSEC & IGCSE Physics Syllabus 6) ELECTRICITY 1) MECHANICS 7) MAGNETISM 2) THERMAL PHYSICS 8) PHYSICS OF THE ATOM 3) KINETIC THEORY 9) ELECTRONS AND ATOMS 4) WAVES 10) RADIOACTIVITY 5) OPTICS SUBTOPICS IN MECHANICS 1) SCIENTIFIC METHOD 2) MEASUREMENT 3) VECTORS AND SCALARS 4) STATICS (FORCES) 5) DYNAMICS : MOTION IN A STRAIGHT LINE 6) ENERGY 7) HYDROSTATICS  Today’s Agenda DEFINE PRESSURE AND APPLY DEFINITION RELATE THE PRESSURE AT A POINT IN A FLUID TO ITS DEPTH AND THE DENSITY. APPLY ARCHIMEDES’ PRINCIPLE TO PREDICT WHETHER A BODY WOULD FLOAT OR SINK IN A GIVEN FLUID. RELEVANT EXAMPLES INCLUDE RAFTS, BOATS, BALLOONS, AND SUBMARINES Lecture 7 SECTION A: MECHANICS HYDROSTATICS “Teachers who makes physics boring are criminals” – Walter Lewin “Force” and “Pressure” together.  The idea of force is familiar to all of us but we are not so familiar with the concept of pressure. We know that a net force acting on a solid body will cause that body to move. It is not so obvious, however, that a difference in pressure could cause a liquid or a gas to move. This is what happens when we turn on our water taps to get water and our gas jets to obtain gas to light a flame. When we drink from a straw, we use a pressure difference to get the liquid up the straw. Many of the incidents that take place during a hurricane in which household property is lost or destroyed can be put down to differences in air pressure inside and outside buildings. These are only a few of many instances of the consequences of pressure difference that we meet from day to day. (Physics for CSEC Examinations 2014)  What is Pressure?  Pressure  Pressure may be exerted on a surface by a solid, by a liquid or by a gas, which applies a thrust to that surface. The pressure on the surface is that quantity which expresses the normal (perpendicular) thrust per unit area of contact between the agency (i.e. the solid, liquid or gas) and the surface. If the thrust is applied over a small area the pressure will be large; if it is applied over a larger area, the pressure will be smaller in value.  If I stand in my shoes on soft, muddy ground, I run the risk of sinking into the mud. But if I stand on a piece of board and distribute my weight over the area of contact between the board and the ground, then I am less likely to sink. In the second case I would be distributing my weight over a larger area (that of the board), and so the perpendicular thrust over unit area of the ground would be smaller than before, when the area of contact was only that of my soles. In other words, the pressure when I stand on the board would be smaller (Physics for CSEC Examinations, 2014)  Pressure  Note: Pressure depends on the force (thrust) acting perpendicular to the surface that experiences the pressure and the area over which this perpendicular force is distributed.  Definition of force: “Pressure is the force acting normally per unit area.” (DeFreitas 2016)  Formula: Pressure on surface = thrust acting normal (perpendicular) to surface area of contact between agency and surface Pressure = force OR P=F area A  The SI unit of pressure is N m–2 which has been assigned the name pascal (Pa).  PRESSURE  WORKED EXAMPLE : Determine the maximum pressure that can be exerted on a horizontal surface by a rectangular block of mass 40 kg and dimensions 30 cm × 40 cm × 1.2 m when resting on one of its faces. Note: Gravitational field strength = 10 N kg–1 P=F 1.2m A 0.3m 40kg P = mg (the force is the weight) A Pmax = 40 × 10 0.30 × 0.40 0.40m Pmax = 3333.3 Pa Figure 1  PRESSURE IN FLUIDS  The pressure at a point in a fluid increases with increased depth and with increased density of the fluid.  Pressure due to liquids at rest is called hydrostatic pressure In the last slide we saw that the pressure due to a solid, is due to the weight of that solid or, more generally, the thrust exerted by that solid on the surface. Since liquids also have weight, we can use a similar formula to calculate the pressure due to a liquid increase in pressure = depth × density × gravitational field strength ΔP = Δhρg  PRESSURE IN FLUIDS  Figure 2 shows a tall can of water with small holes in its sides at different depths. Waterspouts with equal strength from holes on the same level, indicating that the pressure is the same in all directions at a particular depth. Waterspouts with greater force from holes at greater depth, indicating that the pressure increases with depth. Note: Pressure in a fluid increases with de Figure 2  PRESSURE IN FLUIDS WORKED EXAMPLE: Gravitational field strength = 10 N kg–1 Figure 3 shows a fish at a depth of h = 40 m in a lake where the density of the water is 1000 kg m–3. Determine: a) the pressure due to the water on the fish b) the total pressure on the fish if the atmospheric pressure is 1.0 × 105 Pa.  PRESSURE IN FLUIDS Solution: a) Pwater = hρg Pwater = 40 × 1000 × 10 = 400 000 Pa b) Ptot = Patmos + Pwater Ptot = 1.0 × 10^5 + 4.0 × 10^5 = 5.0 × 10 ^5 Pa  Liquids connected to each other will find their own level Figure 4 shows two quantities of the same liquid in separate vessels, A and B, on a table, joined by a tube carrying a closed clip, C. The surfaces of the two liquids are at different heights above the table. As long as, the liquids are kept separate by the closed clip, C, their levels remain unchanged. If the clip is opened, their levels will change, the higher one falling and the lower one rising. Why does this happen? The liquid pressure at point X in vessel A is less than that at point Y in vessel B, because the depth of X in A is less than that of Y in B. Because of this pressure difference, the liquid in the joining tube will move in the direction Y to X from a point where the pressure is higher to another point where the pressure is lower. This is by no means a strange occurrence, since we witness the same principle when we turn on our garden hose. As this happens, the level of liquid in B will fall and that in A will rise. When there is no longer a pressure difference, the final level becomes L, and the two liquids have equal depths. Liquids connected to each other will find their own level Figure 4 Liquids connected to each other will find their own level  The vessels containing the liquids could be of any shape, since liquid pressure does not depend on the shape of the vessel containing the liquid, but only on the depth of the point at which the pressure is taken. The depths of liquid in the vessels shown in figure 5 will therefore all have to be the same if the bases are all on the same horizontal level. It follows, too, from this ‘law’ that all points on the surface of a liquid at rest will be in the same horizontal plane. Figure 5  UNCONFINED LIQUIDS AND PASCAL’S LAW  A liquid in a closed vessel behaves differently from an unconfined liquid. This difference is summed up in Pascal’s law (Pascal’s principle), which states that:  Pressure applied to any point of a fluid in a closed vessel is transmitted equally to every other point in the fluid Example: This principle forms the basis of many hydraulic systems, for example the braking system of cars, and hydraulic jacks for raising motor vehicles.  THE MANOMETER  Now that we have discussed Pascal’s principle, we can discuss how the manometer works for the measurement of pressures that are a little larger or a little smaller than atmospheric. A diagram of this is shown in figure 6. The manometer is essentially a simple U-tube containing water, oil or (sometimes) mercury. It is the best device to use if the pressures to be dealt with are small. If the pressure to be measured is just below or just above atmospheric, then either water or an oil can be used in the U-tube (figure 6 (a)). But for pressures above this value and up to about 1½ times atmospheric, it will be more convenient to use mercury in the manometer (figure 6 (b)). (Physics for CSEC Examinations 2014) Figure 6  THE MANOMETER  HOW DOES THE MANOMETER WORK ? If you blow down one limb of the manometer, the liquid level in that limb falls while the other one rises (see figure 7). The difference between the liquid levels is used to find the actual pressure being measured. For example, in the figure where the manometer is connected to a laboratory gas supply, the pressure of the supply is acting on the lower meniscus on the left, labelled X. By Pascal’s law, since the liquid is continuous and at rest, the pressure at the level L1 in the diagram is the same in both manometer limbs. Thus if the liquid meniscus on the left is denoted by X and that at the same level on the right is Y, we can say that pressure (due to the gas) at X, Pgas = pressure at Y (inside the tube) due to the column h of liquid on the right above the point Y + the pressure of the atmosphere above, Patmos or Pgas = hρg + Patmos  THE MANOMETER  HOW DOES THE MANOMETER WORK ? Figure 7 THE MANOMETER ARCHIMEDES’ PRINCIPLE  ARCHIMEDES’ PRINCIPLE  Archimedes' principle states that when a body is completely or partially immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced. Figure 8 shows how the principle can be verified. An object is first weighed in air and then again when immersed in a displacement can filled to the spout with water. The difference between these weights is the upthrust of the water on the immersed object. The displaced water is collected in a beaker and weighed. The weight of the empty beaker is subtracted from the weight of the beaker of water to find the weight of water displaced. The readings verify that the upthrust is equal to the weight of the fluid displaced. (DeFreitas 2016) ARCHIMEDES’ PRINCIPLE Figure 8 ARCHIMEDES’ PRINCIPLE and RELATIVE DENSITY  Archimedes’ principle provides a convenient method of finding, by experiment, the relative density of a solid and also of a liquid, water being used as the reference substance. Thus, the density of aluminium is 2.7 g cm–3 and that of water is 1.0 g cm–3. We would therefore say that the relative density (R.D.) of aluminium is 2.7. The definition of “Relative density” is therefore: Relative density of a substance = density of the substance density of water NOTE: In calculating a relative density, the individual densities must be expressed in the same unit. So relative density has no units since, in calculating its value, we divide two quantities that have identical units. The value of a relative density does not depend on the units used for the respective densities.  ARCHIMEDES’ PRINCIPLE and RELATIVE DENSITY WORKED EXAMPLE: A lump of a certain substance placed on a kitchen balance is shown to have a mass of 240 g. A beaker of water placed on the same balance is seen to have a mass of 420 g. When the lump of substance is suspended in the water on the balance, however, the beaker seems to weigh 480 g. Calculate the relative density of the substance. The water in the beaker is replaced by a liquid, L, and the experiment is repeated. The mass of the lump of substance is the same as before, but the beaker with liquid now weighs 490 g and 556 g before and after the same substance is suspended in it. Calculate the relative density of the liquid, L. ARCHIMEDES’ PRINCIPLE and RELATIVE DENSITY SOLUTION: When the lump is placed in the liquid there will be an upthrust on the lump. In response to this upthrust on the lump there will be a down thrust of the lump on the liquid and this downthrust is transmitted through the liquid on to the base of the container. This downthrust will be registered by the balance. Hence the apparent increase in the reading of the balance. It is as if the beaker has suffered an increase in weight. Newton’s third law at work! We have upthrust in water = (480 – 420) × 10 N = 0.60 N 1000 upthrust in liquid L = (556 – 490) × 10 N = 0.66 N 1000 ARCHIMEDES’ PRINCIPLE and RELATIVE DENSITY The upthrust is the weight of fluid displaced. We use dW and dL to stand for the density of water and liquid L respectively. If the volume of the substance used is V cm3 , then (since mass = density × volume) upthrust in water = V × dW × 10 N = VdW × 10–2 N 1000 upthrust in liquid L = V × dL 1 × 10 N = VdL × 10–2 N 1000 ARCHIMEDES’ PRINCIPLE and RELATIVE DENSITY So taking the ratio we get upthrust in water upthrust in liquid L = VdW × 10–2 N VdL × 10–2 N = 0.60 N 0.66 N or V dW V dL = 0.60 0.66 (on cancelling) giving R.D. = dL dW = 0.60 0.66 = 1.1  ARCHIMEDES’ PRINCIPLE and FLOATING OBJECTS  A body floats if its own weight is equal to the upthrust on it (the weight of fluid displaced by it). A rectangular block of base dimensions 20 cm × 20 cm and weight 50 N floats as shown in Figure 9. Calculate the depth, h, to which it is submerged, given that the density of water is 1000 kg m–3. (Gravitational field strength = 10 N kg–1) ARCHIMEDES’ PRINCIPLE and SUBMARINES Figure 11 shows a toy submarine that has a ballast tank which can hold water. If water is taken into the tank until the submarine’s weight exceeds the weight of water displaced (upthrust), the submarine will accelerate downwards. The resultant force on it is downwards, and it will descend without the use of its engines. To accelerate upwards, air under pressure expels water from the ballast tank, decreasing the weight of the submarine. When decreased to a value less than the weight of water displaced (upthrust), the submarine will accelerate upwards. Floating occurs if the weight is equal to the weight of water displaced (upthrust).  ARCHIMEDES’ PRINCIPLE and BALLOONS Figure 11(a) shows a balloon containing air. If the air in the balloon is heated, it will become less dense and will expand. The upthrust on the balloon will increase since it will now displace more of the cooler surrounding air. It will accelerate upwards if its weight plus the weight of its contents is less than the weight of the air it displaces (upthrust). If the balloon contains a gas of low density such as helium, as in Figure 11(b), it will accelerate upwards even without its contents being heated. ARCHIMEDES’ PRINCIPLE and BALLOONS  Note that air becomes less dense at higher altitudes and therefore the weight of the air displaced (upthrust) becomes less. When the upthrust has decreased to the weight of the balloon and its contents, the resultant force is zero and acceleration ceases. The balloon continues upwards but there is then a net downward force since the upthrust is further reduced. The balloon thus begins to bob up and down until it settles at a height where the net force on it is zero. Class Work JANUARY 2013 CSEC Past pap Class Work JUNE 2012 CSEC Past paper Class Work JUNE 2012 CSEC Past paper Class Work JUNE 2012 CSEC Past paper  RECAP  Today we learnt : DEFINE PRESSURE AND APPLY DEFINITION RELATE THE PRESSURE AT A POINT IN A FLUID TO ITS DEPTH AND THE DENSITY. APPLY ARCHIMEDES’ PRINCIPLE TO PREDICT WHETHER A BODY WOULD FLOAT OR SINK IN A GIVEN FLUID. RELEVANT EXAMPLES INCLUDE RAFTS, BOATS, BALLOONS, AND SUBMARINES  Next Class  In our next session, we will be covering:  Thermal Physics  REFERENCES  DeFreitas, Peter. 2016. Physics - a Concise Revision Course for CSEC (R). London, England: Collins.  Forbes, Darren. 2013. Physics for CSEC CXC. Cheltenham, England: Nelson Thornes.  Kennett, Heather, and Tom Duncan. 2010. Physics for CSEC Examination + CD. 2nd ed. London, England: Hodder Education.  Physics for CSEC Examinations. 2014.  THE END Thank you for your attention……. Have a nice day !!!!!