1 Mass, Density Pressure Buoyant Forces & Archimedes Principle Pascal’s Principle 2 Introduction: Phases of matter Solid Liquid Matter Gas 3 Introduction: Fluids Solid • Fixed shape, fixed size Liquid • does not maintain fixed shape (take the shape of its container) • does not • also not readily readily change in compressible shape or volume (volume can when force is change with applied large force) Gas • neither fixed shape nor fixed volume (it’s expand to its container) • compressible Fluids 4 Introduction: Fluids Fluid’s definition: “ a substance which has no definite shape and has the ability to flow. Such as liquid and gas.” 5 Density Every matter occupies space around it Space is called “Volume (m3)” 6 Density Two objects made of the same material have the same density even though they may have different masses and different volumes. That’s because the ratio of mass to volume is the same for both objects 7 Density 8 Density Closely pack more mass Iron has more mass per unit volume The density ρ of a substance is its mass per unit volume: m v The SI unit for density is kg/m3. Density is also sometimes given in g/cm3. Density is a characteristic property of any pure substance. 9 Density Object made of a particular pure substance, such as gold can have any size or mass, but the density will the same for each. Why helium balloons float ? Sometimes, we use the concept of density to write mass: m v ..and hence the weight as: mg vg 10 Density Example 1 What is the mass of a solid iron wrecking ball of radius 18 cm? Given the density of iron is 7800 kg/m3. Volume of ball Answer: 4 3 m v r 3 3 3 4 2 7800 kg/m 18 10 m 3 190 kg 11 Density Exercise 1 1. What volume does o.4 kg of alcohol occupy? What is the weight of this volume? 2. What volume of water has the same mass of 100 cm3 of lead? What is the weight density of lead? Ans : (1)5.06 104 m3 ,3.92 N (2)1.13 103 m3 ,1.1105 N/m3 12 Density Exercise 2 1. Calculate the density of cubical solid if one of its side is 0.02 m and its mass is 0.50 kg? 2. Air has density of 1.29 kg m/3 under the room temperature. Calculate the mass of air in a cubical room if one of the sides of the room in 8m? Concept Pressure is defined as the force per unit area. Pressure is a scalar; the units of pressure in the SI system are Pascal: 1 Pa = 1 N/m2. 14 15 Concept Example 2 The two feet of a 60-kg person cover an area of 500 cm2. (a) Determine the pressure exerted by the two feet on the ground. (b) If the person stands on one foot, what will the pressure be under that foot? 16 Fluid pressure Now, lets consider how the pressure in a liquid of uniform density varies with depth; 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: 20 Fluid pressure 21 Fluid pressure However, for the everyday situation of a liquid in an open container – such as water in the glass, a swimming pool, an ocean and etc, there is a free surface at the top exposed to the atmosphere. Absolute Pressure Thus, P = P₀ + ρgh Atmospheric Pressure The atmospheric pressure at sea level is about 1.013 x 105 N/m2 and decreases slowly with altitude. 25 Fluid pressure Example 4 A swimming pool has a dimension of 28.0 m by 8.5 m whose uniform depth is 1.8 m. Calculate the absolute pressure and the total force on the bottom of the pool. 27 Fluid pressure Example 4: Answer (a) Absolute pressure at the bottom of the pool: P P gh 1.103 105 Pa 1000 kg/m3 9.8 m/s 2 1.8 m 1.18 105 Pa (b) Total force at the bottom of the pool: F PA 1.18 105 Pa 28.0 m 8.5 m 2.8 107 N 28 Exercise Exercise 2 1. A Golf shoe has 10 cleats, each having an area of 6.5 ×10-6 m2 in contact with the floor. Assume that in walking, there is one instant when all 10 cleats support the entire weight of an 80 kg person, what is the pressure exerted by the cleats on the floor? 2. A pipe contains water under a gauge pressure of 400 kPa. If you patch a 4mm diameter hole in the pipe with a piece of tape, what force must the tape be able to withstand? Ans :(1)1.206 107 N/m2 (2)5.03 N 29 Exercise Exercise 2 3. The water pressure in a certain house is 1.1×106 N/m2. How high must be the water level be above the point to of release in the house? 4. What is the absolute pressure at the bottom of a lake that is 30 m deep? Ans : (3)112.24 m (4)3.95 105 Pa 30 Exercise Exercise 2 5. A living room floor has floor dimensions of 4.50 m and 3.20 m and a height of 2.40 m. The density of air is 1.29 kg/m3. What does the air in the room weigh? What force does the atmosphere exerts on the floor of the room? Ans : 437 N, 1.46 106 N 31 Buoyant force 32 Buoyant force Object submerged in a fluid appear to weigh less when its outside the fluid. For e.g.: a large rock that you would have difficulty lifting off the ground can often be easily lifted from the bottom of a stream. When its break through the surface of the water, suddenly its seems to be heavier. This is the e.g. of the buoyant force. 33 Buoyant force The buoyant force is an upward force exerted by the fluid. 34 Buoyant force The BF occurs because the pressure in a fluid increases with depth. Thus, the upward pressure on the bottom surface of a submerged object is greater than the downward pressure on its top surface. 35 Buoyant force Buoyant force weight of the fluid displaced FB mF g FVF g Where: F = density of the fluid VF = volume of displaced fluid / volume of submerged object 36 Buoyant force Consider: At the top cylinder: Pressure, P1 = ρFgh1 and Force, F1 = P1A = ρFgh1A (downward) At the bottom cylinder: Pressure, P2 = ρFgh2 and, Force, F2 = P2A = ρFgh2A (upward) 38 Buoyant force Thus, the net force on the cylinder exerted by the fluid pressure is known as the buoyant force, FB and act upward. The magnitude of FB is; Weight of the Fluid displaced Volume of the cylinder 39 Archimedes Principle This result is valid no matter what the shape of the object. Archimedes’s Principle: “ The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object.” 40 Archimedes Principle “Fluid displaced” mean a volume of the fluid equal to submerged volume of the object ( or that part of the object that is submerged ) For e.g.: if the object is placed in a glass or tube initially filled to the brim with water, the water that flows over the top represents the water displaced by the object. 41 Archimedes Principle Example 5 Consider two identical pails of water filled to the brim. One pail contains only water, the other has a piece of wood floating in it. Which pail has the greater weight? 42 Archimedes Principle Example 5 Consider two identical pails of water filled to the brim. One pail contains only water, the other has a piece of wood floating in it. Which pail has the greater weight? Answer: Both pail weigh the same. According to Archimedes’ principle, the wood displaced a volume of water with equal weight of the wood. 43 Archimedes Principle Example 6 A 70-kg ancient statue lies at the bottom of the sea. Its volume is 3.0 x 104 cm3. (a) What is the weight of the statue? (b) What is the buoyant force exerted on the statue? (c) How much force is needed to lift it? Given the density for seawater, ρ = 1.1025 x 103 Kg/m3 . 44 Archimedes Principle Example 6: Answer (a) Weight of the statue: w mg 70 kg 9.8 m/s 2 6.9 102 N (b) The buoyant force (equal to weight of displaced seawater): FB ms g s vs g Volume of seawater = volume of statue 1.1025 103 kg 3.0 102 m3 9.8 m/s 2 3.0 102 N 45 Archimedes Principle Example 6: Answer (c) Force to lift the statue: F mg FB 6.9 102 N 3.0 102 N 3.9 102 N 46 Archimedes Principle Example 7 A typical hot air balloon has a volume of 2200 m3. The density of air at temperature 20°C is 1.205 kg/m3. The density of the hot air inside the balloon at a temperature 100°C is 0.946 kg/m3. How much weight can the hot air balloon lift? 47 Archimedes Principle Example 7: Answer FB The weight of the extra load can be found by: (refer to the FBD) FB mB g wL wL FB mB g Thus, find the buoyant force exerted on the balloon: FB mB g AvB g 1.205 kg 2200 m3 9.8 m/s 2 2.6 104 N WL m Bg 48 Archimedes Principle Example 7: Answer The mass of the balloon (equal to mass of the hot air inside): FB mB H v 0.946 kg/m3 2200 m3 2081 kg and the weight: mB g 2081 kg 9.8 m/s 2 2.0 104 N WL m Bg Hence, the extra load: wL FB mB g 2.6 104 N 2.0 104 N 6000 N 49 Archimedes Principle Case 1: Totally submerged object An object is totally submerged in a fluid if its density is greater than the density of the fluid (ρ₀ > ρF). When an object is totally submerged, the upward FB is: FB Fluid vFluiddispl g Fluid vobject g 50 Archimedes Principle If the object has a mass, m and density,ρ₀ thus, Fg mobject g object vobject g Hence, the net force, FB Fg Fluid vobject g object vobject g Fluid object vobject g When ρ₀ > ρF , then FB < Fg and the supported object sink! 51 Archimedes Principle Case 2: Floating object In general, an object floats on a fluid if its density is less than that of fluid (ρ₀ < ρF). In this case, the upward buoyant force is balanced by the downward gravitational force, FB Fg Fluid vF g object vobject g displ vFdispl vobject object Fluid 52 Archimedes Principle Example 8 A 100 g block of wood has a volume of 120 cm3. Will it float in water? Will it float in gasoline? 53 Archimedes Principle Example 8: Answer Lets compare the buoyant force with the weight: Weight of the block: mB g 100 103 kg 9.8 m/s 2 0.98 N Buoyant force by the water: FB wvB g 1000 kg/m3 120 106 m3 9.8 m/s 2 1.176 N Since FB > mg, it will float in water. 54 Archimedes Principle Example 8: Answer Buoyant force by the gasoline: FB g vB g 680 kg/m3 120 106 m3 9.8 m/s 2 0.78 N Since FB < mg, it will sink in gasoline. 55 Archimedes Principle Example 9 A scuba diver and her gear displace a volume of 65.0 L and have a total mass of 68.0 kg. (a) What is the buoyant force on the diver in seawater? (b) Will the diver sink or float? Given the density for seawater, ρ = 1.025 x 103 Kg/m3 . 56 Archimedes Principle Example 9: Answer (a) Buoyant force on the diver: FB s vd g 1.025 103 kg/m3 0.065 m3 9.8 m/s 2 653 N (b) Will the diver sink/float? md g 68 kg 9.8 m/s 2 666 N Thus, since the weight > buoyant force, the diver will sink! 57 Archimedes Principle Exercise 3 A crane lift the 16,000 kg steel hull of a sunken ship out of the sea. Determine: (a) the buoyant force exerted by seawater on the hull (b) the tension in the crane’s cable when the hull is fully submerged in the water (c) the tension when the hull is completely out of the water? [Given the density for seawater, ρ = 1.1025 x 103Kg/m3 and density of steel, ρ = 7.8 x 103 Kg/m3] Ans : (a) 2.2 104 N (b) 1.34 105 N (c) 1.56 105 N 58 Archimedes Principle Exercise 4 A tin coffee can floating in water has an internal volume of 180 cm3 and a mass of 112 g. How many grams of metal can be added to the can without causing it to sink in the water? Ans : 68 g 59 Pascal’s Principle 60 Pascal’s Principle In fact that the pressure in a fluid depends on depth and the atmospheric pressure P₀, any increase in pressure at the surface must be transmitted to every other points in fluid. Pascal’s principle: “If an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount”. A change in pressure at any point in an enclosed fluid is transmitted equally throughout the fluid. 61 Pascal’s Principle Some application of Pascal’s Principle: a) Hydraulic lift which a small input force is used to exert large output force 62 Pascal’s Principle The pressure must be same for both piston. Pin Pout Since the pressure is define as force per unit area: Fin Fout Ain Aout The mechanical advantage: Fout Aout Fin Ain 63 Pascal’s Principle Example 7 In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a radius of 5.0 cm. This pressure is transmitted by a liquid to a piston that has a radius of 15.0 cm. What force must the compressed air exert to lift a car weighing 13,300 N? What air pressure produced this force? 65 Pascal’s Principle Example 7: Answer input force: Fin Fout Ain Aout 5 102 m 2 1.33 104 N 2 15 102 m 1.48 103 N pressure due this force: Fin 1.48 103 N 5 Pin 1.88 10 Pa 2 Ain 5 102 m 66 Pascal’s Principle Exercise 5 A force of 400 N is applied to a small piston of hydraulic press whose diameter is 4 cm. What must be the diameter of the large piston if it is to lift a 200 kg load? 0.088 m 67