3 eso technology mechanism To print

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Unit 2. Mechanism and machines

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 engine

Unit 2. Machines and mechanisms

Which one of these objects is a mechanism and which one is a structure?

2.1 Introduction

Structures and mechanisms resists forces and transmit them, but mechanism can transform these forces and movement in our benefit.

A machine is a group of elements that help us do a job. Inside we can find, mechanism, engines and structures

Unit 2. Machines and mechanisms 2.2 Rectilinear into an equivalent

Rectilinear Rectilinear

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2.2 Rectilinear into an equivalent

In this group we will find machines that transform a rectilinear movement into another rectilinear movement. The simplest one is the lever

Lever: It is a mechanism made up of a rigid bar and a point of support which is also called a fulcrum.



Archimedes said once: Give me a place to stand on, and I will move the Earth

Resistance (R) is a force (normally the weight of an object) that has to be overcome by the use of the applied Force (F).

2.2 Rectilinear into an equivalent Lever elements

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.

Fulcrum

Force Resistance

dRarm dFarm

2.2 Rectilinear into an equivalent Lever elements

Fulcrum

Force Resistance

dRarm dFarm

The lever's Law RdR=FdF

2.2 Rectilinear into an equivalent In physics we define mechanical work as the amount of energy transferred by a force acting through a distance

W= F•d d= distance between A and B F= Force applied to move the object

F d

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Levers behave according to a law of physics, called the LAW OF THE LEVER, that is derived from the Newton’s second Law. Equilibrium means that all forces applied to an object are neutralized ∑ F=o

2.2 Rectilinear into an equivalent Therefore, if we apply the Newton's law, we get that, when there is an equilibrium, all forces and works applied to an object are equal to cero Equilibrium ∑ Fd=∑W=0 ∑W= Wr+Wf=0

Wr= Wf


Wr= Wf

The lever's Law RdR=FdF


Units: R,F= [N] D=[m] W=[Nm]=[j]

Exercise: Calculate the weight of the man to be able to raise the old lady. Data: Man’s distance to fulcrum= 1 m Lady’s distance to fulcrum= 2 m Lady’s weight= 90 Kg 1Kg= 9,8N

1º Ex 2.2 Rectilinear into an equivalent

Solution

1. We read the text

2. We identify the mechanism, and write all the related formulas

3. We draw the diagram of this mechanism

How to do an exercise

Lever RdR=FdF

Fulcrum

Force Resistance

dRarm dFarm

4. We write all the data that we need to solve any exercise. Change all units to IUPAC: Meters, Kg, Newtons, etc..

5. We read the text again and write the value of the magnitudes needed.

F=? R= 882N DR=2m DF=1m 4. We calculate the magnitude


Distance Mass Force Time Meters Kilograms Newtons Seconds

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Exercise: 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

Resistance Force


Solution

Exercise: 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 • Ant height= 1m


Solution

There are three classes of levers and each class has a fulcrum, load and effort which together can move a heavy weight.

2.2 Rectilinear into an equivalent First Class lever: Fulcrum is situated between the Force and Resistance


Force

Resistance

Arm Arm

Second Class lever: the Resistance is situated between the Force and the Fulcrum


Force Resistance

Arm Arm

Resistance Force

Third Class lever: the Force is situated between the Resistance and the Fulcrum


Force

Arm Arm Resistance

Force

Resistance

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Pulleys: A pulley is a wheel with a slot. It makes easy to overcome a resistance offered from an object

2.2 Rectilinear into an equivalent 2.2 Rectilinear into an equivalent

A pulley is a group of mechanisms forming a machine. And as a machine a lever is able to do work

But what is work?

Pulleys: A pulley is a wheel with a slot. There is a rope, chain or strap that goes around it’s axle


Wheel

Slot: gap where the rope goes around

Axle: it holds the wheel

Force Resistance

Fixed Pulleys: they have only one wheel therefore they only change the direction of the Force


It is used to raise and lower weight easily. For example in wells

If we analyze the Fixed pulley we see that is a lever with equal distance to the fulcrum, so we can apply the Levers law


RRdR=FFdF

Since dR=dF R=F balance

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 the resistance to get the balance

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Multiple Mobile Pulleys: If we can have several combinations of this mechanism.


In this case, this is the formula used to define the equilibrium (where n is the number of mobile wheels)

Hoist: It has multiple mobile wheels that decrease exponentially the Force needed to achieve the balance


Where n is the number of mobile wheels

Exercise: We want to rise a fixed pulley that has a water bucket hanging from the hook. What is the force that we have to apply to get balance?


Data: Water volume: 5l Wheel diameter: 30cm Well depth: 15m 1L=1kg 1kg=9,8N


Data: Water mas=5L x 1kg/L=5Kg R=5kg x 9,8N/kg= 49N F=?

R=F 49N=F

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

Sloping flat: It’s a flat that forms an angle that helps to raise an object.


The smaller the angle is, less force will be needed to raise the object and the distance will be longer

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2.2 Rectilinear into an equivalent The formula is obtained using the trigonometry laws

α b

a

α F

b

2.2 Rectilinear into an equivalent Wedge: It’s a double Sloping flat. The force applied is proportional to the faces length.

2.2 Rectilinear into an equivalent Screw: It’s a multiple Sloping flat rolled up. The force applied is proportional to the number of teeth.

2.3i Circular into Rectilinear

Circular Rectilinear

Handle-winch: A handle is a bar joined to the axle that makes it turn. A winch is a cylinder with a rope around it that is used to raise an object

2.3i Circular into Rectilinear This mechanism is equal to a lever, so we can apply the same lever’s law:


F

R

DF DR

RDR=FDF

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Calculate the force needed to raise a water bucket that has 10L of fresh water. Name the mechanism, draw its diagram and the formulas applied

Extra data: Handle size Df =30cm Bar radius Dr= 15 cm Water density 1kg/L 1Kg= 9,8N Bucket material: iron Bucket color: Black

solution

2.3i 1º Ex Circular into Rectilinear Rack and Pinion: This mechanism is used to transmit high efforts like a car transmission or a lift:


2.3ii Circular into an Equivalent

Circular Circular

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…


Both wheels turn in the opposite direction.

All the teeth must have the same shape and size.


driven gear driver gear

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.


The most common use is in bicycles and motorbikes.

Both gears turn simultaneously in the same direction

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The gear that provides the energy is called driver gear and the one that receives driven gear


Force is applied in this gear


ω

2.3ii Circular into an Equivalent Friction wheels: System with two or more wheels that are in direct contact.

These wheels can't transmit high forces but they can resist vibration and movements

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 them

These wheels can't either transmit high forces but they can resist vibration and movements

2.3ii Circular into an Equivalent Pulleys and strap system are used also to change movement direction in many mechanism like motor engines, industrial mechanism, etc

2.3ii Circular into an Equivalent Pulleys and strap system shown in this picture has driven pulley A and five driven pulleys. Indicate each wheel movement direction.

2.3ii Circular into an Equivalent The speed of the wheel is measured in rpm (revolutions per minute) that describe the angular speed ω

v =rw

ω = angular speed r= radio v= linear speed

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Gears are used to increase or decrease the angular speed. To describe the equilibrium we have to know the number of teeth and angular speed E= driver S=driven


WS= ZS=


WS= ZS=

2.3ii Circular into an Equivalent In these mechanisms the ratio between the speed of the driven wheel and speed of the driver wheel is called transmission ratio i

DriveN DriveR DriveN

DriveR DriveN DriveR

Exercise: We have a pulley and strap system formed by two wheels as you can see in the picture. Which is the angular speed of the driver wheel?

2.3ii Ex 1 Circular into an Equivalent

Sol

Exercise: We have a gear system formed by two gears with 20 and 40 gears teeth (driven and driver wheels respectively). Calculate: • Which is the transmission ratio? • If the driver gear is moving at 300 r.p.m., how fast is the driven gear moving?

2.3ii Ex 2 Circular into an Equivalent

Sol

2.3ii Circular into an Equivalent The transmission ratio I indicates if the gear increase or decrease the driven gear speed

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2.3ii Circular into an Equivalent I>1 indicates that the mechanism increases the driven gear speed, but decreases its power

F

driven driver

2.3ii Circular into an Equivalent I<1 indicates that the mechanism decreases the driven gear speed, but decreases its power

driven driver

2.3ii Ex 3 Circular into an Equivalent Exercise. This pulley and strap system it’s used to modify the speed of a drill, changing the pulleys combination.

2.3ii Ex 3 Circular into an Equivalent a. 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? Si se elige la posición que aparece representada en la

figura ¿A qué velocidad girará la broca? Solution

2.3ii Circular into an Equivalent Gears are also used to raise heavy objects applying 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

2.3ii Circular into an Equivalent Therefore, we can apply the lever’s law

RDR=FDF

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2.3ii Circular into an Equivalent Mechanical associations

We can create a mechanical association connecting several elements. With this association we can decrease or increase the out speed or the force applied


When we analyze this mechanism we study how the energy and the movement is transmitted in each step






So, when we have a mechanical association, the transmission ratio between the first and the last one is:

€

itotal =D1⋅ D3 ⋅ D5 ⋅ ⋅ ⋅D2 ⋅ D4 ⋅ D6 ⋅ ⋅ ⋅

=WS

WE

itotal =D or Z driversD or Z driven

=WS

WE

itotal = i1−2 ⋅ i3−4


Indicate in which direction moves each wheel and where is applied more force

Solution

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Analyze the next mechanism and answer the following questions: 1. What’s the name of the system formed by 1 and 2? And 3 and 4? 2. If 1 spins clockwise, how do the 2,3 and 5? 3. If 1 is spinning at 6 rpm, what’s 2 and 3? speed? 4. If 3 is spinning at 90 rpm and it’s 10cm Ø, what’s 4 speed if it’s 2cm Ø? 5. What is the global transmission ratio between 1 and 4 Data Z1 =4 Z2 =16

Solution

2.3ii Ex 4 Circular into an Equivalent Mechanical associations 2.3ii Circular into an alternative rectilinear

Crank-connecting rod: are also used to raise heavy objects applying a low force at a low speed.

video

2.3ii Circular into an alternative rectilinear We can create also an alternative rectilinear movement also with: Excentric: it’s a regular wheel that has it’s axle off-center Excentric video Cam: it’s a wheel with a oval shape. Video

Name Movement transformation

……..into……………

……..into……………

……..into……………

……..into……………

……..into……………

……..into……………

Final exercises Final 1 Complete these table solution

Object/Use Justification

Toy car engine

MILL

ELEVATOR

CAR ENGINE

Bicycle

Solution Final exercises Final 2 Complete these table Solution

Final exercises Final 2 Complete these table

• Do a Cheating log • 5X5 cm

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Final exercises Pag 72, 73 ex 13, 18, 19, 20, 21, 22, 23, 26

5.5 Thermal machines

New rubrics: • Thermal machines, how do they

work? • Classification • Main applications


The thermal machines transform the thermal energy from combustion into mechanical energy.


We classify these machines according where the combustion takes place:

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 engine

A steam engine transform heats water using fuel combustion in order to obtain high pressure steam that is used to move a mechanism.

5.5 Thermal machines. External combustion Steam engine

Watt used this engine to move trains, ships and the first industrial machines that created the first Industrial Revolution.


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The pressure of the steam moves the piston in both directions thanks to the mechanism that changes the entrance of the high pressured steam.


The first explosion engine was a gas engine where the gas was introduced in two entrances. The high pressure gasses of the explosion move the piston that moves the wheel attached.

5.5 Thermal machines. Internal combustion Explosion engines

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.

5.5 Thermal machines. Internal combustion The four stroke engine

1º Intake: The admission valve is opened and at the same time the piston goes down creating vacuum absorbing air and fuel

piston

piston Rod

Crank

Cylinder Combustion chamber

2.Compression: Both valves are closed and the piston goes up compressing the mixture of air and fuel.

Oil and air At high pressure

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This movement is created by an electric engine of the starting mechanism. After the first complete process the inertia energy reboot the process over again.

Inertia wheel

3º Power: An electric discharge from the spark plug explode the mixture creating high pressure gases that makes the piston goes down

spark plug

4. 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

Admision Valve

Escape Valve

Piston movement

Intake OK

Compression

Power

Exhaust OK

5.5 Thermal machines. Internal combustion 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.

5.5 Thermal machines. Internal combustion The two stroke engine

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5.5 Thermal machines. Internal combustion The diesel engine

The diesel engines use gasoil instead of gasoline . The mixture explodes it self when is compressed thus it doesn’t need a spark plug

5.5 Thermal machines. Internal combustion Jet engine

Newton's third law says: For every force acting on a body there is an equal and opposite reaction.

Newton's third law says: For every force acting on a body there is an equal reaction.


1600 years before any flying machine could ever fly and before Newton announced the third mechanical law, Hero of Alexandria invented the Aeolipile.”


Rocket It carries two deposits with oxigen and fuel

that when they are mixtured explode creating a high preassure gases that push up the rocket.

5.5 Thermal machines. Internal combustion 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.


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Turbofan: used by most of the comercial planes because it generate les noise.

The plane moves due to the gases and the helix movement.

It is similar to a turbojet. It consists of a large fan with a smaller turbojet engine mounted behind, so we may may that a turbofan is a Fan + a turbo jet.


Turboprop. In this case, the main shaft is directly coupled to a gearbox at the front of the jet engine to drive a propeller.

Data: Man’s distance to fulcrum= 1 m

Lady’s distance to fulcrum= 2 m

Lady’s weight= 90 Kg 1Kg= 9,8N

F=?

R=90Kgx9,8N/Kg= 882N DR=2m

DF=1m

Fulcrum

Force Resistance

dRarm dFarm

RdR=FdF

882x2=Fx1

1764N=F

F=1764N Weight=180Kg

1º Ex Solution 2.2 Rectilinear into an equivalent

Exercise

Data: Dr= 2cm=0,02m Df= 15cm=0,15m R= 1N F=?

Fulcrum

Force Resistance

dRarm dFarm

RdR=FdF

1x0,02=Fx0,15

0,02=Fx0,15 F=0,02/0,15

F=0,133N

2º Ex Solution 2.2 Rectilinear into an equivalent

Exercise

1. We read the text



1º Grade Lever RdR=FdF

Fulcrum

Force Resistance

dRarm dFarm

3º Ex solution 2.2 Rectilinear into an equivalent


1 ton= 1000kg • Distance between elephant and fulcrum =1cm= 0,01m

• Ant weight= 1gr= 0,001Kg≠0,001x9,8N • Fulcrum weight= 30kg • Ant height= 1m 5. We read the text again and write the value of the magnitudes needed

F= 0,001Kgx9,8N/kg= 0,0098N DF=? R=1000kgx9,8N/Kg=9800N DR=0,01m


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5 We write all the data that we need to solve any exercise. We read the text again and write the value of the magnitudes needed

F= 0,001Kgx9,8N/kg= 0,0098N DF=? R=1000kgx9,8N/Kg DR=0,01m 6. Calculate the magnitude

€

F ∗DF = R ∗DR

0,0098∗DF = 9800∗0,01

DF =9800∗0,010,0098

=10000m

DF =10000m exercise

3º Ex 2.2 solution Rectilinear into an equivalent

1. We read the text



1º Fixed pulley R=F

Fulcrum

Force Resistance



1 ton= 1000kg • Distance between elephant and fulcrum =1cm= 0,01m

• Ant weight= 1gr= 0,001Kg≠0,001x9,8N • Fulcrum weight= 30kg • Ant height= 1m 5. We read the text again and write the value of the magnitudes needed

F= 0,001Kgx9,8N/kg= 0,0098N DF=? R=1000kgx9,8N/Kg=9800N DR=0,01m

How to do an exercise 5 We write all the data that we need to solve any exercise. We read

the text again and write the value of the magnitudes needed F= 0,001Kgx9,8N/kg= 0,0098N DF=? R=1000kgx9,8N/Kg DR=0,01m 6. Calculate the magnitude

€

F ∗DF = R ∗DR

0,0098∗DF = 9800∗0,01

DF =9800∗0,010,0098

=10000m

DF =10000m exercise


Calculate the force needed to raise a water bucket that has 10L of fresh water. Name the mechanism, draw its diagram and the formulas applied

2.3i 1º Ex Circular into Rectilinear

Exercise Extra data: Handle size Df =30cm Bar radius Dr= 15 cm Water density 1kg/L 1Kg= 9,8N Bucket material: iron Bucket color: Black

1. We read the text 2. We identify the mechanism, and write all the related formulas

Handle-winch that has the same structure as a 1º grade lever.


RdR=FdF

Fulcrum

Force Resistance


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Handle size DF= 30cm=0,3m Bar radius DR= 15 cm=0,15m Water= 10L=10kg 5. We read the text again and write the value of the magnitudes needed

F= ? DF=0,3m Resistance=10kgx9,8N/Kg= 98N DR=0,15m

How to do an exercise 5 We write all the data that we need to solve any exercise. We read

the text again and write the value of the magnitudes needed F= ? DF=0,3m Resistance=10kgx9,8N/Kg= 98N DR=0,15m 6. Calculate the magnitude

€

F ∗DF = R ∗DR

F ∗0,3 = 98∗0,15

F =98∗0,150,3

=14,70,3

= 49N

F = 49Nexercise


Exercise: We have a pulley and strap system formed by two wheels as you can see in the picture. Which is the angular speed of the driver wheel?

2.3ii Ex 1 Sol Circular into an Equivalent


Pulley and Strap



WE

DE

WS

DS


Driver wheel DE= 20cm= 0,2m DriveN wheel DS= 60 cm=0,6m DriveN speed WS= 250 rpm 5. We read the text again and write the value of the magnitudes needed

DE= 20cm= 0,2m DS= 60 cm=0,6m WS =250 rpm WE =? rpm



DE= 20cm= 0,2m DS= 60 cm=0,6m WS =250 rpm

WE =? rpm 6. Calculate the magnitude

€

WS

WE=

DE

DS= i

250WE

=0,20,6

⇒ WE =250 • 0,6

0,2WE = 750rpm

exercise


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Exercise: We have a gear system formed by two gears with 20 and 40 gears teeth (driver and driven). Calculate: • Which is the transmission ratio? • If the driver gear is moving at 300 r.p.m., how fast is the driven gear moving?


Z=20 Z=40


Gears





Driver wheel ZE= 20 DriveN wheel ZS= 40 DriveN speed WE= 300 rpm 5. We read the text again and write the value of the magnitudes needed

ZE= 20 ZS= 40 WS =? rpm WE =300 rpm


€

WS

WE=ZEZS

= i


ZE= 20 ZS= 40 WS =? rpm

WE =300 rpm 6. Calculate the magnitude

€

WS

WE=

ZE

ZS= i

WS

300=

2040

⇒ WE =20 • 300

40WE =150rpm



ZE= 20 ZS= 40 WS =? rpm

WE =300 rpm 6. Calculate the magnitude

€

WS

WE=

ZE

ZS= i

i =2040

⇒ i = 0,5

i = 0,5exercise

2.3ii Ex 1 Sol Circular into an Equivalent 2.3ii Circular into an Equivalent 1º Read 2º Identify the mechanism It is a pulley and strap with three

different positionsl

exe

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2.3ii Circular into an Equivalent 1º Read 2º Identify the mechanism It is a pulley and strap with three

different positionsl 3º Draw the mechanism diagram

Engine

Driver Driven

X 3

Drill

2.3ii Circular into an Equivalent 4º Formulas and data related

1º Position 2º Position 3º Position DE 80cm 70cm 60cm

DS 100cm 120cm 140cm

WE 1400rpm 1400rpm 1400rpm

WS ? ? ?

i ? ? ?

1º Position 2º Position 3º Position

2.3ii Circular into an Equivalent 5º Calculate the missing magnitudes

1º Position DE 80cm

DS 100cm

WE 1400rpm

WS 0,8

i 1120rpm



DS 120cm

WE 1400rpm

WS 0,583

i 816,7rpm



DS 140cm

WE 1400rpm

WS 0,583

i 816,7rpm

2.3ii Circular into an Equivalent 4º Formulas and data related

1º Position 2º Position 3º Position DE 80cm 70cm 60cm

DS 100cm 120cm 140cm

WE 1400rpm 1400rpm 1400rpm

WS 1120rpm 816rpm 600rpm

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

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Indicate in which direction moves each wheel and where is applied more force

Back

Analyze the next mechanism and answer the following questions: 1. What’s the name of the system formed by 1 and 2? And 3 and 4? 2. If 1 spins clockwise, how do the 2,3 and 5? 3. If 1 is spinning at 6 rpm, what’s 2 and 3? speed? 4. If 3 is spinning at 90 rpm and it’s 10cm Ø, what’s 4 speed if it’s 2cm Ø? 5. What is the global transmission ratio between 1 and 4 Data Z1 =4 Z2 =16

2.3ii Ex 4 sol Circular into an Equivalent Mechanical associations

1. What’s the name of the system formed by 1 and 2? And 3 and 4? Gears with chain and Pulley and strap

2. If 1 spins clockwise, how do the 2,3 and 5? Data Z1 =4 Z2 =16


Analyze the next mechanism and answer the following questions: 3. If 1 is spinning at 6 rpm, what’s 2 and 3 speed? Data Z1 =4 Z2 =16 W1=6rpm W2=? W3=?

Driver Driven

W2=W3=1,5rpm


4. If 3 is spinning at 90 rpm and it’s 10cm Ø, what’s 4 speed if it’s 2cm Ø? Data D3 =10cm D4 =2cm W3=90rpm W4=?

W4=450rpm


5. What is the global transmission ratio between 1 and 4 Data D3 =10cm D4 =2cm

Back

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Name Movement transformation

Friction wheels Circular into equivalent

Pulley and strap Circular into equivalent

Mobile pulley Circular into equivalent Cranck-connecting

rod Circular into linear

Lever Linear into equivalent

Rack and pinion Circular into linear

Final exercises Final 1 solution Complete these table exercise

Object/Use Justification

Toy car engine Thanks to the strap we can change the movement direction and adapt the mechanism to the car shape. Not high efforts are applied

MILL The vibrations from the mill are absorbed by the wheels while they transmit high efforts at slow spped

ELEVATOR The high effort from the engine is transformed into a high speed linear movement to rise the elevator.

CAR ENGINE The high effort from the explosion is transformed into circular movement transmitting all the energy created

Bicycle The chain transmit high effort to the wheel separated and allows to absorb vibrations

Final exercises Final 2 Complete these table. Solution Exercise

A machine is a group of elements that help us do a job. Inside we can find, mechanism, engines and structures

Unit 2. Machines and mechanisms

Technology

3 eso technology mechanism To print