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3 eso technology mechanism To print 3 eso technology mechanism To print Document Transcript

  • 11/26/11 Unit 2. Mechanism and machines Unit 2. Machines and mechanisms 1.  Introduction 2.  Rectilinear movement into an equivalent: Levers. Pulleys and Hoist. Sloping flat. Wedge. Screw 3.  Circular movement into i.  Rectilinear: Rack and Pinion , handle-winch ii.  An equivalent: gears, wheels, pulleys and strap. iii.  An alternative rectilinear: Crank-connecting rod, cam 4. Thermal machines i.  Steam engine ii.  Explosion engine iii.  Reaction engineWhich one of 2.1 Introductionthese objectsis a mechanism andwhich one is a structure? Structures and mechanisms resists forces and transmit them, but mechanism can transform these forces and movement in our benefit. Unit 2. Machines and mechanisms 2.2 Rectilinear into an equivalentA machine is a group of elementsthat help us do a job. Inside we canfind, mechanism, engines andstructures Rectilinear Rectilinear 1
  • 11/26/11 2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalent Lever: It is a mechanism made up of a rigid bar and a point of support which is also called a fulcrum.In this group we will find machines thattransform a rectilinear movement intoanother rectilinear movement. Thesimplest one is the lever 2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalent Lever elements Resistance (R) is a force (normally the weight of an Archimedes said once: object) that has to be overcome by the use of the applied Force (F). Give me a place to stand Resistance Force on, and I will move the Earth dRarm dFarm Fulcrum The point of support, or fulcrum, is the point on which the lever swings. The arms correspond to the distance between the fulcrum and the applied force or the resistance. 2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalent In physics we define mechanical work as Lever elements the amount of energy transferred by a Resistance Force force acting through a distance d= distance between A and B W= F•d F= Force applied to move the dRarm object dFarm Fulcrum F d The levers Law RdR=FdF 2
  • 11/26/112.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentLevers behave according to a law of physics, Therefore, if we apply the Newtons law, wecalled the LAW OF THE LEVER, that is get that, when there is an equilibrium, allderived from the forces and works applied to an object are Newton’s second Law. equal to ceroEquilibrium means that all forces applied to Equilibrium ∑ Fd=∑W=0an object are neutralized ∑W= Wr+Wf=0 ∑ F=o Wr= Wf2.2 Rectilinear into an equivalent 1º Ex 2.2 Rectilinear into an equivalent Exercise: Calculate the weight of the man to be able to raise the old lady. Wr= Wf Units: Data: R,F= [N] Man’s distance to fulcrum= 1 m D=[m] Lady’s distance to fulcrum= 2 m W=[Nm]=[j] Lady’s weight= 90 Kg The levers Law 1Kg= 9,8N RdR=FdF SolutionHow to do an exercise How to do an exercise1.  We read the text 4.  We write all the data that we need to solve any exercise. Change all units to IUPAC: Meters, Kg, Newtons, etc..2.  We identify the mechanism, and write Distance Mass Force Time all the related formulas Meters Kilograms Newtons Seconds Lever RdR=FdF 5.  We read the text again and write the value of the magnitudes needed.3.  We draw the diagram of this Force Resistance F=? R= 882N DR=2m DF=1m mechanism 4.  We calculate the magnitude dRarm dFarm Fulcrum 3
  • 11/26/112º Ex 2.2 Rectilinear into an equivalent 3º Ex 2.2 Rectilinear into an equivalentExercise: Calculate the force that has to be Exercise: What must the distance be applied to break this nut. between the ant and the fulcrum in orderExtra data: to rise an elephant that weights 1 ton.•  Dstance between the nut and fulcrum =2cm Extra data:•  Nut weight= 15gr •  Distance between elephant and fulcrum =1cm•  Nut Break limit Resistance= 1 N •  Ant weight= 1gr•  Force distance to fulcrum= 15cm •  Fulcrum weight= 30kg Solution•  Resistance distance to fulcrum= 5cm •  Ant height= 1m Force Resistance Solution2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentThere are three classes of levers and each First Class lever: Fulcrum is situatedclass has a fulcrum, load and effort which between the Force and Resistancetogether can move a heavy weight. Force Resistance Arm Arm2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentSecond Class lever: the Resistance is Third Class lever: the Force is situatedsituated between the Force and the Fulcrum between the Resistance and the Fulcrum Force Resistance Resistance Force Arm Arm Arm Arm Resistance Force Force Resistance 4
  • 11/26/112.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentPulleys: A pulley is a A pulley is a group of mechanisms forming awheel with a slot. It machine. And as a machine a lever is able tomakes easy to do workovercome aresistance offeredfrom an object But what is work?2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentPulleys: A pulley is a Axle: it holds the wheel Fixed Pulleys: they have only onewheel with a slot. Wheel wheel therefore they only changeThere is a rope, the direction of the Forcechain or strap thatgoes around it’s axle Slot: gap where the It is used to raise and lower weight easily. For rope goes example in wells around Resistance Force2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentIf we analyze the Fixed pulley we see that is a Mobile Pulleys: It is grouplever with equal distance to the fulcrum, so we of two pulleys, one of themcan apply the Levers law is fixed and the other one can move linearly.RRdR=FFdFSince dR=dF R=F In this case we only have to apply half of the resistance to get the balance balance 5
  • 11/26/11 2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentMultiple Mobile Hoist: It has multiple mobilePulleys: If we can wheels that decreasehave several exponentially the Forcecombinations of needed to achieve thethis mechanism. balanceIn this case, this is the formula used to define Where n is the numberthe equilibrium (where n is the number of of mobile wheelsmobile wheels) 2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalent Exercise: Data: We want to rise a fixed pulley that has a Water mas=5L x 1kg/L=5Kg water bucket hanging from the hook. What R=5kg x 9,8N/kg= 49N is the force that we have to apply to get F=? balance? Data: R=F Water volume: 5l 49N=F Wheel diameter: 30cm Well depth: 15m 1L=1kg 1kg=9,8N 2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalent Exercise: Sloping flat: It’s a flat that forms an angle that We have this hoist and we want helps to raise an object. to raise a heater. What is the force needed to get at least The smaller the balance? angle is, less force will be needed to raise Data: the object and Heater weight: 50kg the distance will Heater volume: 39L be longer Heater Brand: Fagor 6
  • 11/26/112.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalentThe formula is obtained using the trigonometry Wedge: It’s a double Sloping flat. Thelaws force applied is proportional to the faces length. b α a Fα b2.2 Rectilinear into an equivalent 2.3i Circular into RectilinearScrew: It’s a multiple Sloping flat rolled up.The force applied is proportional to thenumber of teeth. Circular Rectilinear2.3i Circular into Rectilinear 2.3i Circular into RectilinearHandle-winch: A handle is a bar joined to This mechanism is equal to a lever, sothe axle that makes it turn. A winch is a we can apply the same lever’s law:cylinder with a rope around it that is used toraise an object DR DF F R RDR=FDF 7
  • 11/26/11 2.3i 1º Ex Circular into Rectilinear 2.3i Circular into Rectilinear Calculate the force needed to raise a water bucket that has 10L of fresh water. Name the mechanism, draw its diagram and the Rack and Pinion: This mechanism is formulas applied used to transmit high efforts like a car transmission or a lift: solution Extra data: Handle size Df =30cm Bar radius Dr= 15 cm Water density 1kg/L 1Kg= 9,8N Bucket material: iron Bucket color: Black 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent GEARS: Wheels with “teeth” that fit into each other, so that, each wheel moves the other one. Used in cars, toys, drills, mixers, industrial machines, etc…Circular Circular 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent Gears with chain system: It consists of two Both wheels turn in the opposite direction. gears placed at a certain distance that turn simultaneously in the same direction thanks to a chain that joins them. Both gears turn driver gear simultaneously in driven gear the same All the teeth must have the same shape and direction size. The most common use is in bicycles and motorbikes. 8
  • 11/26/11 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent The gear that provides the energy is called Friction wheels: System with two or more wheels that are in direct contact. driver gear and the one that receives driven gear driver gear driven gear ωForce is applied in this gear These wheels cant transmit high forces but they can resist vibration and movements 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent Pulleys and strap system: Group of pulleys placed at a Pulleys and strap system are used also to change certain distance that turn simultaneously thanks to a movement direction in many mechanism like motor engines, strap that joins them industrial mechanism, etcThese wheels cant either transmit high forcesbut they can resist vibration and movements 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent Pulleys and strap system shown in this The speed of the wheel is measured in rpm picture has driven pulley A and five driven (revolutions per minute) that describe the pulleys. Indicate each wheel movement angular speed ω direction. ω = angular speed r= radio v =rw v= linear speed 9
  • 11/26/112.3ii Circular into an Equivalent 2.3ii Circular into an EquivalentGears are used to increase or decrease theangular speed. To describe the equilibriumwe have to know the number of teeth andangular speed WS= Z S=E= driver S=driven WS= Z S=2.3ii Circular into an Equivalent 2.3ii Ex 1 Circular into an EquivalentIn these mechanisms the ratio between the Exercise:speed of the driven wheel and speed of the We have a pulley and strap system formeddriver wheel is called transmission ratio i by two wheels as you can see in the picture. Which is the angular speed of the driver wheel? DriveN DriveR DriveR Sol DriveR DriveN DriveN2.3ii Ex 2 Circular into an Equivalent 2.3ii Circular into an EquivalentExercise: The transmission ratio I indicates if theWe have a gear system formed by two gears gear increase or decrease the driven gearwith 20 and 40 gears teeth (driven and speeddriver wheels respectively). Calculate:•  hich is the transmission ratio? W• If the driver gear is moving at 300 r.p.m.,how fast is the driven gear moving? Sol 10
  • 11/26/112.3ii Circular into an Equivalent 2.3ii Circular into an EquivalentI>1 indicates that the mechanism increases I<1 indicates that the mechanismthe driven gear speed, but decreases its decreases the driven gear speed, butpower decreases its power F driven driven driver driver2.3ii Ex 3 Circular into an Equivalent 2.3ii Ex 3 Circular into an EquivalentExercise. This pulley and strap system it’s used to modify a. Which positions allows us to get thethe speed of a drill, changing the pulleys combination. maximum speed on the drill?. b. If the engine speed is 1400 rpm, What is the smallest speed of the drill? Si el motor gira a 1400 rpm ¿Cuál es la mínima velocidad que se puede obtener en la broca? Si se elige la posición que aparece representada en la figura ¿A qué velocidad girará la broca? idad de giro en la broca? Si el motor gira a 1400 rpm ¿Cuál es la mínima velocidad que se puede obtener en la broca? Solution Si se elige la posición que aparece representada en la figura ¿A qué velocidad girará la broca?2.3ii Circular into an Equivalent 2.3ii Circular into an EquivalentGears are also used to raise heavy objects Therefore, we canapplying a low force at a low speed. apply the lever’s law This mechanism is also a lever, if we RDR=FDF want to raise something heavy we need a small driver gear and a big driven gear 11
  • 11/26/11 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent Mechanical associations Mechanical associations We can create a mechanical association When we analyze this mechanism we study how the connecting several elements. With this energy and the movement is transmitted in each step association we can decrease or increase the out speed or the force applied 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent Mechanical associations Mechanical associations When we analyze this mechanism we study how the When we analyze this mechanism we study how the energy and the movement is transmitted in each step energy and the movement is transmitted in each step 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent Mechanical associations Mechanical associations So, when we have a mechanical association, the transmission ratio between the first and Indicate in which the last one is: direction moves each wheel and D1 ⋅ D3 ⋅ D5 ⋅⋅⋅ WS where is applied itotal = = more force D2 ⋅ D4 ⋅ D6 ⋅⋅⋅ WE D or Z drivers WS itotal = = D or Z driven WE itotal = i1−2 ⋅ i3−4 Solution€ 12
  • 11/26/11 2.3ii Ex 4 Circular into an Equivalent Mechanical associations 2.3ii Circular into an alternative rectilinear Analyze the next mechanism and answer the following questions: Crank-connecting rod: are also used to 1.  hat’s the name of the system formed by 1 and 2? And 3 and 4? W raise heavy objects applying a low force at 2.  f 1 spins clockwise, how do the 2,3 and 5? I a low speed. 3.  f 1 is spinning at 6 rpm, what’s 2 and 3? speed? I 4.  f 3 is spinning at 90 rpm and it’s 10cm Ø, what’s 4 speed if it’s 2cm I Ø? 5. What is the global transmission ratio between 1 and 4 Data Z1 =4 Z2 =16 Solution video solution Final exercises2.3ii Circular into an alternative rectilinear Final 1 Complete these tableWe can create also an alternativerectilinear movement also with: Name Movement transformationExcentric: it’s a regular wheel that has it’saxle off-center Excentric video ……..into……………Cam: it’s a wheel with a oval shape. Video ……..into…………… ……..into…………… ……..into…………… ……..into…………… ……..into…………… Final exercises Final exercises Final 2 Complete these table Solution Final 2 Complete these tableSolution Object/Use Justification •  o a D Toy car engine MILL Cheating log ELEVATOR •  X5 cm 5 CAR ENGINE Bicycle 13
  • 11/26/11Final exercisesPag 72, 73 ex 13, 18, 5.5 Thermal machines 19, 20, 21, 22, 23, 26 New rubrics: • Thermal machines, how do they work? • Classification • Main applications 5.5 Thermal machines 5.5 Thermal machinesThe thermal machines transform the thermal We classify these machines according where the energy from combustion into mechanical combustion takes place: energy. External combustion: the fuel is burned outside the engine. For example the steam engine. Internal combustion: the fuel is burned inside the engine creating a explosion. For example a car engine5.5 Thermal machines. External combustion 5.5 Thermal machines. External combustion Steam engine Steam engineA steam engine transform heats water using fuel combustion in order to obtain high pressure steam that is used to move a mechanism. Watt used this engine to move trains, ships and the first industrial machines that created the first Industrial Revolution. 14
  • 11/26/11 5.5 Thermal machines. External combustion 5.5 Thermal machines. Internal combustion Steam engine Explosion engines The first explosion engine was a gas engine where the gas was introduced in two entrances. The high pressure gasses of the explosion move theThe pressure of the steam moves the piston in both piston that moves the directions thanks to the mechanism that changes the wheel attached. entrance of the high pressured steam.5.5 Thermal machines. Internal combustion The four stroke engine The four stroke engine it’s the most popular engine due to it’s economy and resistance. It needs only air, oil and an ignition source from an electric discharge. Oil and airpiston Cylinder At high Combust pressur piston ion e Rod chamberCrank1º Intake: The admission valve is opened and at the same time the piston 2.Compression: Both valves are closed and the goes down creating vacuum absorbing piston goes up compressing the mixture of air air and fuel and fuel. 15
  • 11/26/11 spark plug Inertia wheelThis movement is created by an electric engine 3º Power: An electric discharge from the spark of the starting mechanism. After the first plug explode the mixture creating high complete process the inertia energy reboot the pressure gases that makes the piston goes process over again. down Admision Escape Piston Valve Valve movement Intake OK Compression Power4. Exhaust: The escape calve is opened and the combustion gases are expulsed by the piston that goes up again. We have reached the start position again Exhaust OK 5.5 Thermal machines. Internal combustion 5.5 Thermal machines. Internal combustion The two stroke engine The two stroke engine Two stroke engine. It is a explosion engine that does the four steps in only two phases. It creates less energy that the four times but is simpler and cheaper. 16
  • 11/26/11 5.5 Thermal machines. Internal combustion 5.5 Thermal machines. Internal combustion The diesel engine Jet engineThe diesel engines Newtons third law says: For every Newtons third law says: use gasoil force acting on a acting there body instead of For every force body on a is an gasoline . The equal reaction. there is an equal and opposite mixture explodes reaction. it self when is compressed thus it doesn’t need a spark plug 5.5 Thermal machines. Internal combustion 5.5 Thermal machines. Internal combustion Jet engine Jet engine 1600 years before any flying machine could Rocket ever fly and before Newton announced the It carries two deposits with oxigen and fuel third mechanical law, Hero of Alexandria that when they are mixtured explode invented the Aeolipile.” creating a high preassure gases that push up the rocket. 5.5 Thermal machines. Internal combustion 5.5 Thermal machines. Internal combustion Jet engine Jet engine Turbojet: the air is absorbed and compresed into the combustion chamber and mixtured with querosene. The high preasure gases flow creating a push force and moving the turbina conected to the initial compressor. 17
  • 11/26/115.5 Thermal machines. Internal combustion 5.5 Thermal machines. Internal combustionJet engine Jet engineTurbofan: used by Turboprop. In this most of the case, the main comercial planes shaft is directly because it coupled to a generate les gearbox at the noise. front of the jet The plane moves engine to drive a due to the gases It is  similar to a turbojet. It propeller. and the helix consists of a large fan with a smaller  turbojet engine movement. mounted behind, so we may may that a turbofan is a Fan + a turbo jet.1º Ex Solution 2.2 Rectilinear into an equivalent 2º Ex Solution 2.2 Rectilinear into an equivalentData: Data:Man’s distance to fulcrum= 1 mLady’s distance to fulcrum= 2 m RdR=FdF Dr= 2cm=0,02m RdR=FdF Df= 15cm=0,15m 882x2=Fx1 1x0,02=Fx0,15Lady’s weight= 90 Kg1Kg= 9,8N R= 1N 1764N=F 0,02=Fx0,15F=?R=90Kgx9,8N/Kg= 882N F=?DR=2m F=1764N Weight=180Kg F=0,02/0,15 F=0,133NDF=1m Force Force Resistance Resistance Exercise dRarm dFarm dRarm dFarm Exercise Fulcrum Fulcrum3º Ex solution 2.2 Rectilinear into an equivalent How to do an exercise1.  We read the text 4.  We write all the data that we need to solve any exercise. Change all units to IUPAC: Meters, Kg, Newtons, etc..2.  We identify the mechanism, and write all the related formulas 1 ton= 1000kg •  Distance between elephant and fulcrum =1cm= 0,01m •  Ant weight= 1gr= 0,001Kg≠0,001x9,8N 1º Grade Lever RdR=FdF •  Fulcrum weight= 30kg3.  We draw the diagram of this mechanism •  Ant height= 1m 5. We read the text again and write the value of the magnitudes needed Force F= 0,001Kgx9,8N/kg= 0,0098N DF=? Resistance R=1000kgx9,8N/Kg=9800N DR=0,01m dRarm dFarm Fulcrum 18
  • 11/26/11 3º Ex 2.2 solution Rectilinear into 4º Ex solution 2.2 Rectilinear into an equivalent an equivalent 5 We write all the data that we need to solve any exercise. We read 1.  We read the text the text again and write the value of the magnitudes needed F= 0,001Kgx9,8N/kg= 0,0098N DF=? 2.  We identify the mechanism, and write all the related formulas R=1000kgx9,8N/Kg DR=0,01m 6. Calculate the magnitude 1º Fixed pulley R=F F ∗ DF = R ∗ DR 3.  We draw the diagram of this mechanism 0,0098 ∗ DF = 9800 ∗0,01 Force 9800 ∗0,01 Resistance DF = = 10000m exercise 0,0098 DF = 10000m Fulcrum € 3º Ex 2.2 solution Rectilinear into How to do an exercise an equivalent4.  We write all the data that we need to solve any exercise. Change all 5 We write all the data that we need to solve any exercise. We read units to IUPAC: Meters, Kg, Newtons, etc.. the text again and write the value of the magnitudes needed1 ton= 1000kg F= 0,001Kgx9,8N/kg= 0,0098N DF=?•  Distance between elephant and fulcrum =1cm= 0,01m R=1000kgx9,8N/Kg DR=0,01m 6. Calculate the magnitude•  Ant weight= 1gr= 0,001Kg≠0,001x9,8N•  Fulcrum weight= 30kg•  Ant height= 1m F ∗ DF = R ∗ DR5. We read the text again and write the value of the magnitudes needed 0,0098 ∗ DF = 9800 ∗0,01F= 0,001Kgx9,8N/kg= 0,0098N DF=? 9800 ∗0,01 DF = = 10000mR=1000kgx9,8N/Kg=9800N DR=0,01m exercise 0,0098 DF = 10000m € 2.3i 1º Ex Circular into Rectilinear 4º Ex solution 2.2 Rectilinear into an equivalent Calculate the force needed to raise a water bucket that has 10L of fresh water. Name the mechanism, draw its diagram and the 1.  We read the text formulas applied 2.  We identify the mechanism, and write all the related formulas Exercise Handle-winch that has the same structure as a 1º grade lever. RdR=FdF Extra data: Handle size Df =30cm 1.  We draw the diagram of this mechanism Bar radius Dr= 15 cm Water density 1kg/L Force Resistance 1Kg= 9,8N Bucket material: iron Bucket color: Black Fulcrum 19
  • 11/26/11 3º Ex 2.2 solution Rectilinear into How to do an exercise an equivalent4.  We write all the data that we need to solve any exercise. Change all 5 We write all the data that we need to solve any exercise. We read units to IUPAC: Meters, Kg, Newtons, etc.. the text again and write the value of the magnitudes neededHandle size DF= 30cm=0,3m F= ? DF=0,3mBar radius DR= 15 cm=0,15m Resistance=10kgx9,8N/Kg= 98N DR=0,15m 6. Calculate the magnitudeWater= 10L=10kg5. We read the text again and write the value of the magnitudes neededF= ? DF=0,3m F ∗ DF = R ∗ DRResistance=10kgx9,8N/Kg= 98N DR=0,15m F ∗0,3 = 98 ∗0,15 98 ∗0,15 14,7 F= = = 49N F = 49N 0,3 0,3 exercise € 2.3ii Ex 1 Sol Circular into an Equivalent 2.3ii Ex 1 Sol Circular into an Equivalent Exercise: 1.  We read the text We have a pulley and strap system formed 2.  We identify the mechanism, and write all the related formulas by two wheels as you can see in the picture. Which is the angular speed of the Pulley and Strap driver wheel? 1.  We draw the diagram of this mechanism WE WS DE DS 2.3ii Ex 1 Sol Circular into an Equivalent 2.3ii Ex 1 Sol Circular into an Equivalent4.  We write all the data that we need to solve any exercise. Change all 5 We write all the data that we need to solve any exercise. We read units to IUPAC: Meters, Kg, Newtons, etc.. the text again and write the value of the magnitudes neededDriver wheel DE= 20cm= 0,2m DE= 20cm= 0,2mDriveN wheel DS= 60 cm=0,6m DS= 60 cm=0,6m WS =250 rpmDriveN speed WS= 250 rpm WE =? rpm5. We read the text again and write the value of the magnitudes needed 6. Calculate the magnitudeDE= 20cm= 0,2m WS DEDS= 60 cm=0,6m = =i WE DSWS =250 rpm 250 0,2 250 • 0,6 WE = exercise = ⇒WE =? rpm WE 0,6 0,2 WE = 750rpm € 20
  • 11/26/11 2.3ii Ex 2 Sol Circular into an 2.3ii Ex 1 Sol Circular into an Equivalent Equivalent Exercise: 1.  We read the text We have a gear system formed by two gears 2.  We identify the mechanism, and write all the related formulas with 20 and 40 gears teeth (driver and driven). Calculate: Gears •  hich is the transmission ratio? W • If the driver gear is moving at 300 r.p.m., 1.  We draw the how fast is the driven gear moving? diagram of this mechanism Z=40 driver gear Z=20 driven gear 2.3ii Ex 1 Sol Circular into an Equivalent 2.3ii Ex 1 Sol Circular into an Equivalent4.  We write all the data that we need to solve any exercise. Change all 5 We write all the data that we need to solve any exercise. We read units to IUPAC: Meters, Kg, Newtons, etc.. the text again and write the value of the magnitudes neededDriver wheel ZE= 20 ZE= 20DriveN wheel ZS= 40 ZS= 40 WS =? rpmDriveN speed WE= 300 rpm WE =300 rpm5. We read the text again and write the value of the magnitudes needed 6. Calculate the magnitudeZE= 20 WS Z EZS= 40 = =i WS Z E WE ZSWS =? rpm = =i WS 20 20 • 300 WE ZS = ⇒ WE =WE =300 rpm 300 40 40 WE = 150rpm € € 2.3ii Ex 1 Sol Circular into an Equivalent 2.3ii Circular into an Equivalent 5 We write all the data that we need to solve any exercise. We read 1º Read the text again and write the value of the magnitudes needed ZE= 20 2º Identify the mechanism exe ZS= 40 It is a pulley and strap with three WS =? rpm different positionsl WE =300 rpm 6. Calculate the magnitude WS Z E = =i WE ZS 20 i= 40 ⇒ i = 0,5 exercise i = 0,5 € 21
  • 11/26/11 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent 1º Read 4º Formulas and data related 2º Identify the mechanism 1º Position 2º Position 3º Position It is a pulley and strap with three DE 80cm 70cm 60cm DS 100cm 120cm 140cm different positionsl WE 1400rpm 1400rpm 1400rpm 3º Draw the mechanism diagram WS ? ? ? i ? ? ? Driver Driven 1º Position X3 2º Position 3º Position Drill Engine 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent 5º Calculate the missing magnitudes 5º Calculate the missing magnitudes 1º Position 2º PositionDE 80cm DE 70cmDS 100cm DS 120cmWE 1400rpm WE 1400rpmWS 0,8 WS 0,583i 1120rpm i 816,7rpm 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent 5º Calculate the missing magnitudes 4º Formulas and data related 3º Position 1º Position 2º Position 3º PositionDE 60cm DE 80cm 70cm 60cmDS 140cm DS 100cm 120cm 140cmWE 1400rpm WE 1400rpm 1400rpm 1400rpmWS 0,583 WS 1120rpm 816rpm 600rpmi 816,7rpm i 0,8 0,583 0,428 a.  Which positions allows us to get the maximum speed on the drill?. Position A b.  If the engine speed is 1400 rpm, What is the smallest speed of the drill? 600rpm back 22
  • 11/26/11 2.3ii Circular into an Equivalent 2.3ii Ex 4 sol Circular into an Equivalent Mechanical associations Mechanical associations Analyze the next mechanism and answer the following questions:Indicate in which 1.  hat’s the name of the system formed by 1 and 2? And 3 and 4? Wdirection moves 2.  f 1 spins clockwise, how do the 2,3 and 5? Ieach wheel and 3.  f 1 is spinning at 6 rpm, what’s 2 and 3? speed? Iwhere is applied 4.  f 3 is spinning at 90 rpm and it’s 10cm Ø, what’s 4 speed if it’s 2cm Ø? Imore force 5. What is the global transmission ratio between 1 and 4 Data Z1 =4 Z2 =16 Back 1.  What’s the name of the system formed by 1 and 2? 2.3ii Circular into an Equivalent And 3 and 4? Gears with chain and Pulley and strap Mechanical associations Analyze the next mechanism and answer the following 2.  If 1 spins clockwise, how do the 2,3 and 5? questions: Data 3.  f 1 is spinning at 6 rpm, what’s 2 and 3 speed? I Z1 =4 Data Driver Driven Z2 =16 Z1 =4 Z2 =16 W1=6rpm W2=? W3=? W2=W3=1,5rpm 2.3ii Circular into an Equivalent 2.3ii Circular into an Equivalent Mechanical associations Mechanical associations 4. If 3 is spinning at 90 rpm and it’s 10cm Ø, what’s 4 5. What is the global transmission ratio between 1 and 4 speed if it’s 2cm Ø? Data Data D3 =10cm D3 =10cm D4 =2cm D4 =2cm W3=90rpm W4=? W4=450rpm Back 23
  • 11/26/11 exercise Final exercises Final exercises Final 1 solution Complete these table Final 2 Complete these table. Solution Exercise Name Movement transformation Object/Use Justification Friction wheels Circular into equivalent Thanks to the strap we can change the movement Toy car engine direction and adapt the mechanism to the car shape. Not high efforts are applied Pulley and strap Circular into equivalent The vibrations from the mill are absorbed by the MILL wheels while they transmit high efforts at slow Mobile pulley Circular into equivalent spped Cranck-connecting The high effort from the engine is transformed into Circular into linear ELEVATOR rod a high speed linear movement to rise the elevator. Lever Linear into equivalent The high effort from the explosion is transformed CAR ENGINE into circular movement transmitting all the energy created Rack and pinion Circular into linear The chain transmit high effort to the wheel separated Bicycle and allows to absorb vibrations Unit 2. Machines and mechanismsA machine is a group of elementsthat help us do a job. Inside we canfind, mechanism, engines andstructures 24