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3 ESO Mechanism Unit

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Here you have all we have been working with through the mechanism unit

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3 ESO Mechanism Unit

1. 1. Unit 2. Mechanism and machines
2. 2. Unit 2. Machines and mechanisms1.  Introduction2.  Rectilinear movement into an equivalent:Levers. Pulleys and Hoist. Sloping flat. Wedge. Screw3.  Circular movement intoi.  Rectilinear: Rack and Pinion , handle-winchii.  An equivalent: gears, wheels, pulleys and strap.iii.  An alternative rectilinear: Crank-connecting rod, cam4. Thermal machinesi.  Steam engineii.  Explosion engineiii.  Reaction engine
3. 3. Which one ofthese objectsis a mechanism andwhich one is a structure?
4. 4. 2.1 IntroductionStructures and mechanisms resists forcesand transmit them, but mechanism cantransform these forces and movement inour benefit.
5. 5. Unit 2. Machines and mechanismsA machine is a group of elementsthat help us do a job. Inside we canfind, mechanism, engines andstructures
6. 6. 2.2 Rectilinear into an equivalentRectilinear Rectilinear
7. 7. 2.2 Rectilinear into an equivalentIn this group we will find machines thattransform a rectilinear movement intoanother rectilinear movement. Thesimplest one is the lever
8. 8. 2.2 Rectilinear into an equivalentLever: It is a mechanism made up of arigid bar and a point of support which isalso called a fulcrum.
9. 9. 2.2 Rectilinear into an equivalentArchimedes said once:Give me a place to standon, and I will move theEarth
10. 10. 2.2 Rectilinear into an equivalentLever elementsResistance (R) is a force (normally the weight of anobject) that has to be overcome by the use of theapplied Force (F). Force Resistance 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.
11. 11. 2.2 Rectilinear into an equivalentLever elements Force Resistance dRarm dFarm Fulcrum The levers Law RdR=FdF
12. 12. 2.2 Rectilinear into an equivalentIn physics we define mechanical work asthe amount of energy transferred by aforce acting through a distance d= distance between A and B W= F•d F= Force applied to move the object F d
13. 13. 2.2 Rectilinear into an equivalentLevers behave according to a law of physics,called the LAW OF THE LEVER, that isderived from the Newton’s second Law.Equilibrium means that all forces applied toan object are neutralized ∑ F=o
14. 14. 2.2 Rectilinear into an equivalentTherefore, if we apply the Newtons law, weget that, when there is an equilibrium, allforces and works applied to an object areequal to ceroEquilibrium ∑ Fd=∑W=0 ∑W= Wr+Wf=0 Wr= Wf
15. 15. 2.2 Rectilinear into an equivalent Wr= Wf Units: R,F= [N] D=[m] W=[Nm]=[j] The levers Law RdR=FdF
16. 16. 1º Ex 2.2 Rectilinear into an equivalentExercise: Calculate the weight of the man tobe able to raise the old lady.Data:Man’s distance to fulcrum= 1 mLady’s distance to fulcrum= 2 mLady’s weight= 90 Kg1Kg= 9,8N Solution
17. 17. How to do an exercise1.  We read the text2.  We identify the mechanism, and write all the related formulas Lever RdR=FdF3.  We draw the diagram of this Force Resistance mechanism dRarm dFarm Fulcrum
18. 18. How to do an exercise4.  We write all the data that we need to solve any exercise. Change all units to IUPAC: Meters, Kg, Newtons, etc.. Distance Mass Force Time Meters Kilograms Newtons Seconds5.  We read the text again and write the value of the magnitudes needed.F=? R= 882N DR=2m DF=1m4.  We calculate the magnitude
19. 19. 2º Ex 2.2 Rectilinear into an equivalentExercise: Calculate the force that has to be applied to break this nut.Extra data:•  Dstance between the nut and fulcrum =2cm•  Nut weight= 15gr•  Nut Break limit Resistance= 1 N•  Force distance to fulcrum= 15cm•  Resistance distance to fulcrum= 5cm Force Resistance Solution
20. 20. 3º Ex 2.2 Rectilinear into an equivalentExercise: What must the distance be between the ant and the fulcrum in order to rise an elephant that weights 1 ton.Extra data:•  Distance between elephant and fulcrum =1cm•  Ant weight= 1gr•  Fulcrum weight= 30kg Solution•  Ant height= 1m
21. 21. 2.2 Rectilinear into an equivalentThere are three classes of levers and eachclass has a fulcrum, load and effort whichtogether can move a heavy weight.
22. 22. 2.2 Rectilinear into an equivalentFirst Class lever: Fulcrum is situatedbetween the Force and Resistance Force Resistance Arm Arm
23. 23. 2.2 Rectilinear into an equivalentSecond Class lever: the Resistance issituated between the Force and the Fulcrum Resistance Force Arm Arm Force Resistance
24. 24. 2.2 Rectilinear into an equivalentThird Class lever: the Force is situatedbetween the Resistance and the Fulcrum Force Resistance Arm Arm Resistance Force
25. 25. 2.2 Rectilinear into an equivalentPulleys: A pulley is awheel with a slot. Itmakes easy toovercome aresistance offeredfrom an object
26. 26. 2.2 Rectilinear into an equivalentA pulley is a group of mechanisms forming amachine. And as a machine a lever is able todo work But what is work?
27. 27. 2.2 Rectilinear into an equivalentPulleys: A pulley is a Axle: it holds the wheelwheel with a slot. WheelThere is a rope,chain or strap thatgoes around it’s axle Slot: gap where the rope goes around Resistance Force
28. 28. 2.2 Rectilinear into an equivalent Fixed Pulleys: they have only one wheel therefore they only change the direction of the ForceIt is used to raise and lower weight easily. Forexample in wells
29. 29. 2.2 Rectilinear into an equivalentIf we analyze the Fixed pulley we see that is alever with equal distance to the fulcrum, so wecan apply the Levers lawRRdR=FFdFSince dR=dF R=F balance
30. 30. 2.2 Rectilinear into an equivalent Mobile Pulleys: It is group of two pulleys, one of them is fixed and the other one can move linearly.In this case we only have to apply half of theresistance to get the balance
31. 31. 2.2 Rectilinear into an equivalentMultiple MobilePulleys: If we canhave severalcombinations ofthis mechanism.In this case, this is the formula used to definethe equilibrium (where n is the number ofmobile wheels)
32. 32. 2.2 Rectilinear into an equivalent Hoist: It has multiple mobile wheels that decrease exponentially the Force needed to achieve the balanceWhere n is the numberof mobile wheels
33. 33. 2.2 Rectilinear into an equivalentExercise:We want to rise a fixed pulley that has awater bucket hanging from the hook. Whatis the force that we have to apply to getbalance? Data: Water volume: 5l Wheel diameter: 30cm Well depth: 15m 1L=1kg 1kg=9,8N
34. 34. 2.2 Rectilinear into an equivalent Data: Water mas=5L x 1kg/L=5Kg R=5kg x 9,8N/kg= 49N F=? R=F 49N=F
35. 35. 2.2 Rectilinear into an equivalent Exercise: We have this hoist and we want to raise a heater. What is the force needed to get at least balance? Data: Heater weight: 50kg Heater volume: 39L Heater Brand: Fagor
36. 36. 2.2 Rectilinear into an equivalentSloping flat: It’s a flat that forms an angle thathelps to raise an object. The smaller the angle is, less force will be needed to raise the object and the distance will be longer
37. 37. 2.2 Rectilinear into an equivalentThe formula is obtained using the trigonometrylaws b α a Fα b
38. 38. 2.2 Rectilinear into an equivalentWedge: It’s a double Sloping flat. Theforce applied is proportional to the faceslength.
39. 39. 2.2 Rectilinear into an equivalentScrew: It’s a multiple Sloping flat rolled up.The force applied is proportional to thenumber of teeth.
40. 40. 2.3i Circular into RectilinearCircular Rectilinear
41. 41. 2.3i Circular into RectilinearHandle-winch: A handle is a bar joined tothe axle that makes it turn. A winch is acylinder with a rope around it that is used toraise an object
42. 42. 2.3i Circular into RectilinearThis mechanism is equal to a lever, sowe can apply the same lever’s law: DR DF F R RDR=FDF
43. 43. 2.3i 1º Ex Circular into Rectilinear Calculate the force needed to raise a water bucket that has 10L offresh water. Name the mechanism, draw its diagram and theformulas applied 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
44. 44. 2.3i Circular into RectilinearRack and Pinion: This mechanism isused to transmit high efforts like a cartransmission or a lift:
45. 45. 2.3ii Circular into an EquivalentCircular Circular
46. 46. 2.3ii Circular into an EquivalentGEARS: Wheels with “teeth” that fit intoeach other, so that, each wheel moves theother one.Used in cars, toys, drills, mixers, industrialmachines, etc…
47. 47. 2.3ii Circular into an EquivalentBoth wheels turn in the opposite direction. driver gear driven gearAll the teeth must have the same shape andsize.
48. 48. 2.3ii Circular into an Equivalent Gears with chain system: It consists of two gears placed at a certain distance that turn simultaneously in the same direction thanks to a chain that joins them. Both gears turn simultaneously in the same directionThe most common use is inbicycles and motorbikes.
49. 49. 2.3ii Circular into an Equivalent The gear that provides the energy is called driver gear and the one that receives driven gear driver gear driven gear ωForce is applied in this gear
50. 50. 2.3ii Circular into an Equivalent Friction wheels: System with two or more wheels that are in direct contact.These wheels cant transmit high forces but theycan resist vibration and movements
51. 51. 2.3ii Circular into an Equivalent Pulleys and strap system: Group of pulleys placed at a certain distance that turn simultaneously thanks to a strap that joins themThese wheels cant either transmit high forcesbut they can resist vibration and movements
52. 52. 2.3ii Circular into an EquivalentPulleys and strap system are used also to changemovement direction in many mechanism like motor engines,industrial mechanism, etc
53. 53. 2.3ii Circular into an EquivalentPulleys and strap system shown in thispicture has driven pulley A and five drivenpulleys. Indicate each wheel movementdirection.
54. 54. 2.3ii Circular into an EquivalentThe speed of the wheel is measured in rpm(revolutions per minute) that describe theangular speed ω ω = angular speed r= radio v =rw v= linear speed
55. 55. 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 speedE= driver S=driven WS= Z S=
56. 56. 2.3ii Circular into an Equivalent WS= Z S=
57. 57. 2.3ii Circular into an EquivalentIn these mechanisms the ratio between thespeed of the driven wheel and speed of thedriver wheel is called transmission ratio i DriveN DriveR DriveR DriveR DriveN DriveN
58. 58. 2.3ii Ex 1 Circular into an EquivalentExercise:We have a pulley and strap system formedby two wheels as you can see in thepicture. Which is the angular speed of thedriver wheel? Sol
59. 59. 2.3ii Ex 2 Circular into an EquivalentExercise:We have a gear system formed by two gearswith 20 and 40 gears teeth (driven anddriver 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
60. 60. 2.3ii Circular into an EquivalentThe transmission ratio I indicates if thegear increase or decrease the driven gearspeed
61. 61. 2.3ii Circular into an EquivalentI>1 indicates that the mechanism increasesthe driven gear speed, but decreases itspower F driven driver
62. 62. 2.3ii Circular into an EquivalentI<1 indicates that the mechanismdecreases the driven gear speed, butdecreases its power driven driver
63. 63. 2.3ii Ex 3 Circular into an EquivalentExercise. This pulley and strap system it’s used to modifythe speed of a drill, changing the pulleys combination.
64. 64. 2.3ii Ex 3 Circular into an Equivalenta. Which positions allows us to get the 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?SolutionSi se elige la posición que aparece representada en la figura ¿A qué velocidad girará la broca?
65. 65. 2.3ii Circular into an EquivalentGears are also used to raise heavy objectsapplying a low force at a low speed. This mechanism is also a lever, if we want to raise something heavy we need a small driver gear and a big driven gear
66. 66. 2.3ii Circular into an Equivalent Therefore, we can apply the lever’s law RDR=FDF
67. 67. 2.3ii Circular into an Equivalent Mechanical associationsWe can create a mechanical association connecting several elements. With this association we can decrease or increase the out speed or the force applied
68. 68. 2.3ii Circular into an EquivalentMechanical associationsWhen we analyze this mechanism we study how the energy and the movement is transmitted in each step
69. 69. 2.3ii Circular into an EquivalentMechanical associationsWhen we analyze this mechanism we study how the energy and the movement is transmitted in each step
70. 70. 2.3ii Circular into an EquivalentMechanical associationsWhen we analyze this mechanism we study how the energy and the movement is transmitted in each step
71. 71. 2.3ii Circular into an Equivalent Mechanical associationsSo, when we have a mechanical association, the transmission ratio between the first and the last one is: D1 ⋅ D3 ⋅ D5 ⋅⋅⋅ WS itotal = = D2 ⋅ D4 ⋅ D6 ⋅⋅⋅ WE D or Z drivers WS itotal = = D or Z driven WE itotal = i1−2 ⋅ i3−4