Physics (101) Properties of Matter Lecture Notes PDF

1st Grade chemistry Physics Department Faculty of Science Mansoura University Properties of Matter [email protected]  CH 3 Fluid Mechanics Fluids are divided into liquids and gases hard to compress (incompre...

1st Grade chemistry Physics Department Faculty of Science Mansoura University Properties of Matter [email protected]  CH 3 Fluid Mechanics Fluids are divided into liquids and gases hard to compress (incompressible easy to compress (compressible fluids), compressibility must be fluids), compressibility may be considered whenever they are highly disregarded when the change in pressurized pressure is small. ‫غير قابلة لالنضغاط ولكن اذا وقع السائل تحت تأثير ضغط قوى‬ ‫ومن الممكن ان يتم تجاهل‬. ‫لها قابلية االنضغاط‬ ‫من الممكن ان يكون قابل لالنضغاط‬ ‫صغيرا‬ ً ‫االنضغاطية عندما يكون التغير في الضغط‬ Water takes the shape of its container fully expands to fill its container with an upper free surface with no free surface. a fluid shows resistance whenever two layers slide over each other, (viscosity). ‫يتميز المائع بوجود قوى احتكاك بين طبقاته التى بدورها تقاوم حركة او سريان هذا المائع التى نسميها اللزوجة‬ fluid is treated as a continuous isotropic substance ‫تعتبر الموائع من المواد المتناحية التى التعتمد خواصها على االتجاه‬ Fluids do not sustain shearing stresses or tensile stresses. ‫ال تحافظ السوائل على إجهاد القص أو إجهاد الشد‬  ideal fluid or perfect fluid: a non-existent, assumed fluid without either viscosity or compressibility. A fluid with compressibility but without viscosity is occasionally discriminated and called a perfect Liquid. Pressure is‫لالنضغاط‬ ‫المائع المثالى هو مائع افتراضى غير موجود يفترض انه ليس له لزوجة وغير قابل‬ a scalar quantity because it is proportional to the magnitude of the force. 𝑴 𝞺 = = 𝐊𝐠. 𝐦 − 𝟑 𝑽  Example 3.1 The mattress of a water bed is 2.00 m long by 2.00 m wide and 30.0 cm deep. (A) Find the weight of the water in the mattress. (B) Find the pressure exerted by the water bed on the floor when the bed rests in its normal position. Assume the entire lower surface of the bed contacts the floor. Solution (A) The weight of the water filling the mattress: Volume Mass Weight Mg = (1.2 x 103 kg ) (9.8 m/ 𝑠 2 ) = 1.18 x 104 N B) When the water bed is in its normal position, the area in contact with the floor is 4.00 m2. Use Equation 4.1 to find the pressure: There are two methods used to express the pressure: one is based on the perfect vacuum; 👉 absolute pressure. and the other on the atmospheric pressure; 👉 gauge pressure. gauge pressure = absolute pressure- atmospheric pressure, gauge pressure = P - P0 In gauge pressure, a pressure under 1 atmospheric pressure is expressed as a negative pressure. Select a parcel of the liquid (uniform density ρ) contained within an imaginary block of cross-sectional area A extending from depth d to depth d + h. The upward force exerted by the liquid on the bottom face of the parcel is PA, and the downward force on the top face is P0 A. The mass of liquid in the parcel is M = ρ V = ρ A h; therefore; Mg = ρ A h g. Because the parcel is at rest, so that the net force acting on it must be zero. the pressure is the same at all points having the same depth, independent of the shape of the container. If the liquid is open to the atmosphere; P0 then P0 is atmospheric pressure=1 bar. the hydraulic press is an application of Pascal’s law. A force of magnitude F1 is applied to a small piston of surface area A1. The pressure is transmitted through an incompressible liquid to a larger piston of surface area A2. P = F1 / A1 = F2 / A2 , we can say that force F2 greater than F1 by A2 / A1 Without any change in the volume of the enclosed liquid, as the piston moves downward through a displacement ∆x1 the liquid pushed up the right piston through a displacement ∆x2. That is, A1 ∆x1 = A2 ∆x2; therefore, P = F2 / F1 = A2 / A1 = ∆x1 /∆x2 F1∆x1 = F2∆x2 ; 👉 is the work done by the force on its respective piston. Therefore, the work done by F1 equals the work done by F2, energy conservation. Example 3.2 In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of radius 5.00 cm. This pressure is transmitted by a liquid to a piston that has a radius of 15.0 cm. (A) What force must the compressed air exert to lift a car weighing 13300 N? (B) What air pressure produces this force? Solution This pressure is approximately twice atmospheric pressure.  Q1 In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross-section having a radius of 3.11 cm. This pressure is transmitted by a liquid to a second piston of radius 18.3 cm. What force must the compressed air exert in order to lift a car weighing 11200 N? Solution Aim: barometer used to measure atmospheric pressure. Construction: A long tube closed at one end is filled with mercury and then inverted into a dish of mercury. The closed end of the tube is nearly a vacuum, Idea: pressure at (A), due to the column of mercury equal pressure at (B), due to the atmosphere. Therefore, P0 = ρHg g h, where ρHg is the density of the mercury and h is the height of the mercury column. Conclusion: As atmospheric pressure varies, the height of the mercury column varies, so the height can be calibrated to measure atmospheric pressure. P0 = 1 atm = 1.013 x105 Pa: one atmosphere of pressure is defined to be the pressure equivalent of a column of mercury that is exactly 0.760 m in height at 0oC. 2: Manometer Aim: A device which measures the fluid pressure by the height of a liquid column. Construction: U-tube filled with a liquid with density ρ, one branch connected with a bulb enclosed gas of unknown pressure p and the other branch exposed to the atmospheric pressure po Idea: Pressure of the gas p can be obtained by measuring the height of liquid h coming upwards into a manometer as shown in the figure. P0 is the atmospheric pressure, ρ is the density; Note: when the pressure p is large, this is inconvenient because h is too high. So, 👉 a U-tube manometer, containing a high-density liquid such as mercury is used. Archimedes’ principle 👉 the magnitude of the buoyant force on an object equals the weight of the fluid displaced by the object.  the pressure Pbot at the bottom of the cube is greater than the pressure Ptop at the top by an amount ρfluid g h, where h is the height of the cube and ρfluid is the density of the fluid. The upward force = Pbot A, where A is the area of the bottom face. The pressure at the downward force = Ptop A. The resultant of these two forces is the buoyant force  Vdisp = Ah is the volume of the fluid displaced by the cube. ρfluid V disp is equal to the mass of fluid displaced by the object, B = Mg, where Mg is the weight of the fluid displaced by the cube. If an object (Mass M, Volume Vobj and density ρobj) totally submerged in fluid; the volume Vdisp of the displaced fluid = the volume Vobj of the object; so, the upward buoyant force is B = ρfluid g Vobj,. If the object has a mass M and density ρobj, its weight is equal to Fg = Mg = ρobj gVobj, and the net force on the object is B – Fg = Mg = (ρfluid - ρobj) g Vobj.  If ρobj < ρfluid,👉 Fg < B and the object accelerates upward (Fig. a) if ρobj > ρfluid, 👉 Fg > B and the object sinks (Fig. b). If the density of the object ρobj = ρfluid,👉 Fg = B, and the object remains in eqm. Therefore, the direction of motion of an object submerged in a fluid is determined only by the densities of the object and the fluid.  In this case, the upward buoyant force is balanced by the downward gravitational force acting on the object. If Vdisp is the volume of the fluid displaced by the object (this volume is the same as the volume of that part of the object beneath the surface of the fluid), the buoyant force B = ρfluid g Vdisp. Because the weight of the object is Fg = Mg = ρobj g Vobj , and Fg = B, So, ρfluid g Vdisp = ρobj g Vobj, or Example 3.3 Wood floats. A block of oak wood 5.00 cm high, 5.00 cm wide, and 10.0 cm long is placed into a tub of water. The density of the wood is 7.20 × 102 kg/m3. How far will the block of wood sink before it floats? Solution: Wwood = (mwood) (g) = (ρwood ) (Vwood ) (g ) = (7.2×102 kg/m3) (0.05x0.05x0.1 m3) (9.80 m/s2) = 1.76 N The buoyant force = the weight of the water displaced Wwater Thus, the block sinks to a depth of 3.59 cm. At this point the buoyant force becomes equal to the weight of the wooden block and the wooden block floats. Example 3.4 Iron sinks. Repeat example 3.3 for a block of iron of the same dimensions. (The density of iron is 7860 kg/m3.) Solution: The weight of the iron block is given by But the block is only 10 cm high. Hence, the buoyant force is not great enough to lift an iron block of this size, and the iron block sinks to the bottom. calculate the buoyant force on this piece of iron. The buoyant force on the iron, the upward buoyant force is B = ρfluid g Vobj, Thus, the net force upward on a block of iron of this size is 2.45 N. But the block weighs 19.3 N. Hence, Wiron > B Why should the block sink and not the ship? If this same weight of iron is made into thin slabs, these thin slabs could be welded together into a boat structure of some kind. By increasing the size and hence the volume of this iron boat, a greater volume of water can be displaced. An increase in the volume of water displaced increases the buoyant force. If this can be made equal to the weight of the iron boat, then the boat floats.  Example 3.5 An iron boats. The iron block of example 3.4 is cut into 16 slices, each 5.00 cm by 10.0 cm by 5/16 cm. They are now welded together to form a box 20.0 cm wide by 10.0 cm long by 10.0 cm high. Will this iron body now float, or will it sink? (The density of iron is 7860 kg/m3.) Solution: the weight of the box is also equal to 19.3 N. Because the iron box is 10 cm high, it sinks to a depth of 9. 48 cm and it then floats.