Strand C. Forces. Unit 4. Pressure. Text. Density and the Archimedes Principle 2 Pressure 5 Hydraulics and Pascal s Law 8

Transcription

1 Strand C. Forces Unit 4. Pressure Contents Page Density and the Archimedes Principle 2 Pressure 5 Hydraulics and Pascal s Law 8

2 C.4.1 Density and the Archimedes Principle Which is heavier, a 10 kg bag of feathers or a 10 kg bag of lead? Of course, they both weigh the same amount since they each have a mass of 10 kg. However, the 10kg of feathers will require a much bigger bag than the 10 kg of lead. The property that describes how much volume a given mass of a substance occupies is the objects density, and as such it is a very important material property. The density of a substance is defined as the mass per unit volume and is given the symbol ρ (the Greek letter rho ). If a substance is uniform throughout (homogeneous) and it has a mass m, and volume V, then: dddddddddddddd = mmmmmmmm vvvvvvvvvvvv = ρρ = mm VV The SI unit of density is the kilogram per cubic meter (kg/m 3 ). Different objects can have different masses and different volumes, and still have the same density. For example, a steel spanner and a steel nail have the same density since the material they are made from is the same. Even though the mass and volume of each object is different, the ratio of mass over volume is the same for each object. Some typical material densities are shown in Table Table Material Density (kg/m 3 ) Air (1atm, 20 C) 1.2 Ice Water Steam Concrete Aluminium Gold White Dwarf Star Neutron Star Notice that the density of ice and water is not significantly different. A solid that changes state to a liquid does not significantly reduce in density because the spacing between the molecules making up the substance does not significantly increase. When a liquid changes state to a gas however, the distance between molecules greatly increases, and the density of the substance in gas form will be much less than that of the liquid or solid form. Steam for example has a density on the order of 1000 times less than that of water at 100 C. 2

3 Worked Example: An Olympic sized swimming pool is 50m long, 25m wide, and has a constant depth of 3m. Using Table 4.1.1, and assuming g = 10m/s 2, calculate; (a) The mass and then the weight of the water contained (b) The mass and weight of the air that would occupy the same volume Answer (a) The volume V of the pool is 50m 25m 3m = 3750m 3. From Table 4.1.1, the density of water is kg/m 3. ρρ = mm VV mm = ρρρρ = kkkkmm mm 3 = kkkk ww = mmmm = kkkk 10mmss 2 = 37500kkkk (b) The density of air is 1.2kg/m 3. Therefore ρρ = mm VV mm = ρρρρ = 1.2kkkkmm mm 3 = 4500kkkk ww = mmmm = 4500kkkk 10mmss 2 = 45kkkk You may find it surprising that the air occupying an empty swimming pool has the same weight as two large 4x4 cars! The average density of an object also determines its buoyancy. When immersed in a fluid (for example air or water), if an objects density is greater than the fluid the object will sink, whereas if the objects density is less than the fluid, the object will float. When an object is immersed in a fluid, it occupies space and therefore displaces some of the fluid. The amount of fluid displaced is equal to the volume of the object that is immersed. The displaced fluid puts an upward force (buoyant force) on the object that is equal to the weight of the fluid that is displaced. If the weight of the object (and therefore the density) is less than the weight of the fluid displaced, the upward force on the object is greater than the force of gravity and the object floats. If the weight of the object is greater than the fluid displaced it sinks. Archimedes famously summed up this result in 250BC while taking a bath; Archimedes principle: When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. 3

4 Worked Example Tom and June are aboard the RMS Queen Mary II, a transatlantic cruise liner and flagship for the Cunard Line. She is 345m long, 41m wide and when laden, 72886m 3 of the ship sits under the water line. If the density of seawater is 1029kg/m 3, what is the upward force of buoyancy acting? Answer: The density of the seawater is 1029kg/m 3, and 72886m 3 is displaced. The mass of the seawater displaced is given by; ρρ = mm VV mm = ρρρρ = 1029kkkkmm mm 3 = kkkk The upwards force of buoyancy equals the weight of the water displaced, therefore ww = mmmm = kkkk 10mmss 2 = 750MMMM Exercise C What can be deduced about the average densities of a shark that has to continuously swim to avoid sinking, and an Ocellaris Clownfish (AKA Nemo) that can remain at a constant level with virtually no effort? 2. Osmium is the densest naturally occurring element on Earth. A cube of Osmium measuring 2 cm 2 cm 2 cm has a mass of 0.18 kg. Calculate the density of Osmium. 3. The world record for the largest watermelon tipped the scales at 159 kg and had a volume of m 3. Would it float in water? (The density of water is kg/m 3 ). 4. A rectangular barge with a mass of 3000 kg floats on a saltwater estuary (ρ= 1029 kg/m 3 ). If the bottom area of the barge is 2 m 3 m, how far from the bottom of the barge is the waterline? Challenge Question 5. A gold sphere of radius 5cm and an aluminium sphere of identical mass rest on a table. Calculate the radius of the aluminium sphere (Hint; use the information in Table and the volume of a sphere = 4/3 πr 3 ). 4

5 C.4.2. Pressure Pressure describes the perpendicular force per unit area that the surface of an object experiences when a force is applied. Pressure (P) = Perpendicular Force (F) per Unit Area (A) PP = FF AA Since pressure is force per unit area the unit of pressure is Nm -2, or the Pascal (Pa) (1Pa = 1Nm -2 ). In many physical situations it is pressure as opposed to force that is the most important and convenient quantity to use when describing an interaction. For example, for the same force F applied, a hammer striking a piece of wood might not pass through the wood, but if it is striking a nail, the nail does pass through the wood. The force per unit area (pressure) the nail exerts is much greater than that of the hammer because the surface area is smaller. Worked Example A lead block is at rest on a flat and level surface. The block has a weight of 200N. The block is 25cm long, 5cm wide and 5cm in breadth. Calculate the pressure exerted on the surface when the block is (a) Lying flat length ways (b) Stood on its end. Answer 5cm (b) (a) 5cm 25cm w = 200N Figure (a) The force F acting is the weight of the block, which acts perpendicular to the surface. Therefore, F = 200N. The area A in contact with the surface is the bottom face of the block, A = 25cm 5cm = 0.25m 0.05m = m 2. PP = FF AA = 200NN mm 2 = 16000PPPP (b) Force F remains the same but the area of the block in contact with the surface is now 5 cm 5 cm = 0.05 m 0.05 m = m 2. PP = FF AA = 200NN mm 2 = 80000PPPP 5

6 We see from the above example that the pressure experienced by the surface is very different depending on the orientation of the block, even though the force remains the same. The greater the surface area the less force per unit area experienced by the surface because the force is spread out over a larger contact area. This is why caterpillar tracks on a tank allow it to travel over soft surfaces without sinking, why snow shoes (tennis racket like objects strapped to the bottom of your feet) stop you sinking into the snow and make it easier to walk longer distances, and why the force you put on the round end of a drawing pin with your thumb is enough to push the pointy end into a notice board. When a fluid (either liquid or a gas) is at rest, it exerts a force perpendicular to any surface in contact with it, such as the container wall or an object immersed in the fluid. This is the force that you feel pressing on you when you dive to the bottom of a swimming pool. While the fluid as a whole is at rest, the molecules that make up the fluid are always in motion; the force exerted by the fluid is due to molecules colliding with their surroundings. Atmospheric pressure (p a) is the pressure of the earth s atmosphere, the pressure we experience at the bottom of this sea of air in which we live and which is constantly pressing down on us. This pressure varies with weather changes and with height above sea level. Normal atmospheric pressure (at sea level) is 1 atmosphere (atm), defined to be exactly 101,325 Pa. 1 atmosphere (atm) = the pressure of the Earth s atmosphere at sea level. 1 atm = 101,325 Pa The weight of a fluid causes pressure to change with depth. The deeper you go within a fluid, the greater the pressure you experience, due to the increase in the amount of fluid (and therefore its weight) above you. This is why atmospheric pressure is greater at sea level than at the top of Mt Everest, or why your ears let you know that the pressure is increasing the deeper you dive in the pool. The pressure in a liquid at a depth h below the surface is given by PP = PP 0 + ρρρρh = PP 0 + mmmmh VV Where ρ is the density (m/v) of the liquid and P0 is the pressure at the surface. However, the above equation is ONLY valid for liquids. 6

7 Worked Example. Matt is Scuba diving off the coast of Komodo Island, hoping to see a manta ray. He floats at the surface cleaning his mask where the pressure is 1 atmosphere (101325Pa). When his dive master gives the signal, he deflates his buoyancy control device (BCD) to start his decent. He is careful to regularly (every 1 meter) equalize his ears. He reaches the bottom and kneels on the seabed, 36m below the surface (where, luckily for him, four manta ray are visiting a cleaning station). If the density of water is 1030kg/m 3 and the pressure at the surface was 1 atmosphere, what pressure is Matt experiencing at 36m? Why does he need to equalize his ears? Assume g = 10ms -2. Answer 1 atmosphere = Pa, and h = 36m. Therefore PP = PP 0 + ρρρρh = PPPP + (1030kkkkmm 3 10mmss 2 36mm) =472125Pa Since 1atm = Pa, Matt is experiencing / = 4.66atm Scuba divers must regularly equalize their ears during a decent since the ear canal is pressurized (to 1 atmosphere). As a diver descends the pressure increases, pushing on the eardrum. If the diver does not equalize (by swallowing or blowing) so that there is the same pressure on each side of the eardrum, it will implode. Exercise C James and Alison are visiting Alison s mum. Alison s dad has just laid new wooden flooring in the lounge and dining room. He happily lets James in who weighs 96kg and is wearing his size 11 work boots but won t let Alison in even though she only weighs 60kg, unless she takes off her high heels. Why is Alison s dad worried about her shoes and not James s boots? 2. Mr Sparks puts a teacher vs. student rugby match notice up in the teacher staff room using drawing pins. The head of a drawing pin has a circular area with a radius of 4mm, whereas the pointy end has a circular area with a radius of only 0.05mm. If Mr Sparks applies a force of 10N, (a) what is the pressure experienced by Mr Sparks thumb? (b) what is the pressure experienced by the notice board? (the area of a circle = πr 2 ) 3. A 10 story stack of encyclopaedias sit on a table in the library. Each book weighs 25 N. If the pressure on the table is Pa, what is the surface area of one encyclopaedia? 4. A polar bears feet are on average 31 cm across and are roughly circular, to help them spread their weight when walking in the snow. Considering that a 7

8 polar bear has four feet, and when he stands still his weight is distributed evenly, with each foot applying 3726 Pa of pressure on the snow, how heavy is the average polar bear? 5. A British diver, John Bennett set a SCUBA world record by diving to 308m in The density of seawater is 1030 kg/m 3 and the pressure at the surface is 1 atm ( Pa). Assuming g = 10m/s 2, (a) What pressure did John Bennett withstand in Pascals? (b) A blue whale can withstand 52 atmospheres. What is its maximum diving depth? Challenge Question. 6. Scientists now believe that the planet Mars could once have had an ocean of depth 0.5km. If the gravitational field strength g on Mars is 3.71m/s 2, the Martian atmosphere at sea level is 600Pa, and the density of seawater (which depends on mass and not weight) is the same on Earth as it is on mars (1030kg/m 3 ) calculate the pressure at the bottom of the Martian Ocean. To what depth would we have to dive in the Atlantic Ocean to experience the same pressure? C.4.3. Hydraulics and Pascal s Law The pressure in a liquid (note the term liquid does not include gases) acts equally in all directions. In addition, liquids are incompressible. This means that when a liquid is put under pressure by a compressing force, for example by a piston that fits tightly inside the container and pushes down on the fluid, the volume of the fluid does not change. Instead, the pressure at any depth increases by the amount of pressure supplied by the downward force, a fact summarized by French scientist Blaise Pascal in 1653, and now called Pascal s Law: Pascal s Law states that the pressure applied to a fluid is transmitted to every point within that fluid and to the walls of the container equally and without loss. Pascal s law is the principle behind hydraulics, which are the mechanical muscles of machines like diggers, car jacks, and robotic arms. A basic hydraulic ram is a metal cylinder that is capable of withstanding high pressure, filled with an oil, that acts as the incompressible fluid as shown in Figure When oil is forced into the ram at high pressure via line 1, that extra pressure is transmitted to the piston. As a result the piston is driven downwards, since the pressure in the top half of the ram is greater than the pressure in the Piston Hydraulic oil Figure

9 lower half, below the piston. This causes the mechanical arm to contract. When fed with Line 2, the piston is forced upwards, extending the arm. In this way the hydraulic ram acts similarly to the bicep muscle of the arm. A hydraulic machine can also be used as a force multiplier. For a small force (effort) applied over a small surface area is transmitted through the hydraulic fluid to a larger surface area creating a large force (load). This is extremely useful for applications such as car braking systems where a small force applied by the driver s foot results in a force at all four wheels that is large enough to stop a car. Consider the hydraulic ram from a car jack as shown in Figure F2 Oil Area A1 F1 Area A2 Figure As the effort is applied to the jack, force F1 acts on the piston in the small chamber of cross sectional area A1. This creates a pressure in the oil at A1 given by; PP = FF 1 AA 1 (1) From Pascal s law, this pressure is transmitted to every point within the oil without loss, and acts on the piston with larger cross sectional area A2. PP = FF 2 AA 2 (2) Since the pressure at A2 is the same as the pressure at A1, we can replace P with F1/A1: FF 1 AA 1 = FF 2 AA 2 (3) Thus the force F2 at the larger piston is given by; FF 2 = FF 1 AA 1 AA 2 (4) 9

10 From (4) we see that the force at the large piston is equal to the force per unit area applied at the small piston multiplied by the area of the large piston. The hydraulic system using different sized areas therefore acts as a force multiplier. When a hydraulic ram has a cross sectional area A1 on the effort side and a different cross sectional area A2 on the load side, FF 1 AA 1 = FF 2 AA 2 Worked Example Bob puts his motorcycle of weight 750N on his motorcycle lift, and starts to lift the entire bike off the floor so that he can change the oil. The small circular cross section piston at the jack handle side of the lift has a radius of 2cm and the large circular cross section piston lifting the load has a radius of 10cm. Calculate the force that bob needs to apply to the small piston to lift the bike. Answer Let the small piston have an area A1 and a radius r1 = 2cm = 0.02m. Then the large piston has an area A2 and r2 = 10cm = 0.1m. Since the area of a circle is given by πr 2, AA 1 = ππrr 2 = ππ = mm 2 AA 2 = ππrr 2 = ππ = mm 2 Since the force F2 required to lift the bike is 750N (the bike s weight), FF 1 AA 1 = FF 2 AA 2 FF 1 = FF 2 AA AA 1 = 750NN mm mm 2 = 30.1NN We can see from the worked example that thanks to Pascal s law in the form of the piston of the motorcycle jack as a force multiplier, Bob managed to lift a 750N motorcycle by supplying a force of only 30.1N. Of course, most car jacks use a long lever at the small piston, and so the force supplied to produce a moment of 30.1N at the small piston is even less. 10

11 Exercise C Identify the following statements as either true, or false. (a) Fluids are incompressible (b) Pressure in a fluid is transmitted equally in all directions (c) A hydraulic ram with two pistons of identical cross sectional area is a force multiplier (d) The piston in a car lift that lifts the car is forced upwards by the molecules of hydraulic oil colliding with the piston 2. Two syringes are connected with a rubber tube. When a force is applied to plunger of the small syringe, oil flows through the tube to the large syringe. The large syringe has a cross sectional area of 0.003m 2 and the small syringe a cross sectional area of m 2. When a 24N force is applied to the small syringe; (a) What is the pressure of the oil? (b) What is the force acting on the plunger of the large syringe? 3. A mechanics car lift has to raise a van of weight 12100N so that the mechanic can check the exhaust pipe for holes. The force supplied to the effort end is a maximum of 100N. What is the minimum cross sectional area needed at the load side of the lift in order to lift the van if the effort side has a cross sectional area of 0.002m 2? 4. A set of hydraulic scales at a vehicle weighbridge allows lorries to be driven on and weighed. The platform the lorries drive on to weighs 8000N and sits on a 0.1m 2 piston. When a 40,000kg artic lorry is parked on the weighbridge, what is the pressure on the hydraulic fluid at the piston? (Assume g = 10m/s 2 ) 5. For the weighbridge in question 4, the bridge piston supporting the 40,000kg lorry is connected to a smaller piston of area 0.02m 2 inside the weighbridge office. What is the force at this smaller piston? Challenge Question 6. A lorry is fitted with a tail lift driven by a single hydraulic piston. The area of the piston is 0.005m 2. The pressure in the system must not exceed Pa. If the weight of the tail platform is 1000N, (a) what is the maximum load the tail lift can lift? (b) what would be the pressure in the system if the tail lift was loaded with a pallet of weight w = 1500N? 11