3 eso mechanism unit

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

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.

2.2 Rectilinear into an equivalent

2.2 Rectilinear into an equivalent

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

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

2.2 Rectilinear into an equivalent

Wr= Wf

The lever's Law RdR=FdF

2.2 Rectilinear into an equivalent

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

How to do an exercise

Distance Mass Force Time Meters Kilograms Newtons Seconds

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

2º Ex 2.2 Rectilinear into an equivalent

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

3º Ex 2.2 Rectilinear into an equivalent

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

2.2 Rectilinear into an equivalent

Force

Resistance

Arm Arm

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

2.2 Rectilinear into an equivalent

Force Resistance

Arm Arm

Resistance Force

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

2.2 Rectilinear into an equivalent

Force

Arm Arm Resistance

Force

Resistance

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

2.2 Rectilinear into an equivalent

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

2.2 Rectilinear into an equivalent

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

2.2 Rectilinear into an equivalent

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.

2.2 Rectilinear into an equivalent

In this case we only have to apply half of the resistance to get the balance

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

2.2 Rectilinear into an equivalent

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

2.2 Rectilinear into an equivalent

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?

2.2 Rectilinear into an equivalent

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

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

Exercise: We have this hoist and we want to raise a heater. What is the force needed to get at least balance?

2.2 Rectilinear into an equivalent

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.

2.2 Rectilinear into an equivalent

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

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:

2.3i Circular into Rectilinear

F

R

DF DR

RDR=FDF

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:

2.3i Circular into Rectilinear

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…

2.3ii Circular into an Equivalent

Both wheels turn in the opposite direction.

All the teeth must have the same shape and size.

2.3ii Circular into an Equivalent

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.

2.3ii Circular into an Equivalent

The most common use is in bicycles and motorbikes.

Both gears turn simultaneously in the same direction

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

2.3ii Circular into an Equivalent

Force is applied in this gear

driven gear driver 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

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

2.3ii Circular into an Equivalent

WS= ZS=

2.3ii Circular into an Equivalent

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

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

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

2.3ii Circular into an Equivalent Mechanical associations

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

2.3ii Circular into an Equivalent Mechanical associations

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

2.3ii Circular into an Equivalent Mechanical associations

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

2.3ii Circular into an Equivalent Mechanical associations

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