2014 simple machines basic concepts

8/12/2019 2014 Simple Machines Basic Concepts

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What do Machines do?

Do they allow one to do more work?

Not really, at best they make completing a task easier.

So then what do Machines do?

Multiply the force.

Multiply the distance.Change the direction of the force.


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Work = Force x Distance an object moves

while the force is applied.

W = F x d

In SI Units:

Force is measured in newtons (N)

distance is measured in meters (m)

Work in N.m which is a joule (J).

Named after James Prescott Joule


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What does work do?

Work causes a change in Energy. Inother words, it can do any of the

following:

Make something move faster.

Lift something up.

Move something against friction.A combination of the above.


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Examples of Work:

A cart is pushed to the right as illustrated.

How much work is done on the cart?

50.0 N

distance cart is moved 4.00 m.

W = F x d = (50.0 N)(4.00 m) = 200. J


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Examples of Work:

A box is lifted as illustrated.

How much work is done on

the box?

80.0

N

distance

boxismov

ed20.

m.

W = F x d

W = (80.0 N)(20. m)

W = 1600 J


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Class 1 Lever: load on one side of the

fulcrum and the effort on the other side.

Load

Effort

Fulcrum

The Fulcrumis the pivot point.

The Loadis what we are trying to

lift or the output of the machine.

The Effortis the force that is applied to

lift the load or the input of the machine.


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Class 1 Lever: Load on one side of the

fulcrum and the Effort on the other side.

LoadEffort

Fulcrum


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Class 1 Lever: More terminology

Load

Effort

Fulcrum

Note that in lifting the load the Effortmoved

much farther than theLoad.

With a smaller Effortwe could lift a

Loadthat is heavier.


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Load=90

0N

Effort = 300 N

Fulcrum

With an Effort of 300 N we were able to lift a Loadof

900 N. We multiplied the input force by 3.

Effortdistanc

e,

dE=60c

m

Lo

addistance

,d

L=20cm

The Effort moved 60 cm

while the Load moved only

20 cm. We moved 3 times

farther than the LOAD.


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Load=90

0N

Effort = 300 N

Fulcrum

Work, W = F x d

Effortdistanc

e,

dE=60c

m

Lo

addistance

,d

L=20cm

WIN= E x dE= (300 N)(0.60 m) = 180 J

WOUT= L x dL= (900 N)(0.20 m) = 180 J

Note: WorkIN= WorkOUT

We didnt do more work,

we just did it with less

Effortthan if I tried to lift

it without the lever.


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Load=90

0N

Effort = 300 N

FulcrumEffortdistanc

e,

dE=60c

m

Lo

addistance

,d

L=20cm

We say that we have a Mechanical Advantage, MA.

We can lift 3 times more than our input Effort.

MA = Load/Effort

MA = (900 N)/(300 N)

MA = 3

MA = dE/dLMA = (60 cm)/(20 cm)

MA = 3


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Load=90

0N

Effort = 300 N

FulcrumEffortdistanc

e,

dE=60c

m

Lo

addistance

,d

L=20cm

We triple our Effort(input force) at the expense of

moving the Load as much.

MA = Load/Effort

MA = (900 N)/(300 N)

MA = 3

MA = dE/dLMA = (60 cm)/(20 cm)

MA = 3


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Fulcrum

We can also analyze the lever by measuring the

distance from the Effortto the fulcrum (pivot point)and the distance from the Loadto the fulcrum.

This is called the lever arm or just armand is often

given the variable name x.

Effort arm, xE= 3.0 m Load arm, xL= 1.0 m


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Fulcrum

This can also be used to calculate theMechanical Advantage.

MA = xE/xL= (3.00 m)/(1.00 m) = 3

Looks familiar doesnt it.

Effort arm, xE= 3.0 m Load arm, xL= 1.0 m


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More terminology:

Often we use the terms, Ideal MechanicalAdvantage, IMA andActual Mechanical

Advantage, AMA

IMA = xE/xLor dE/dL

AMA = L/E with the load being just what you

ultimately wanted to move, excluding anythingelse that may have to be moved with it.This will become clearer when we look at a 2ndclass lever.


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More terminology:

Often we use the terms, Ideal MechanicalAdvantage, IMA andActual Mechanical

Advantage, AMA

We want to find out how well the particular

machine does its work. This is called

Efficiency, Eff.

Efficiency, Eff = (AMA/IMA) x 100%


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2ndclass lever:

Notice that the Effortand the Loadare on the same side of the Fulcrum

and the Loadis between the Effort

and the Fulcrum.

Fulcrum

LoadEffort


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2ndclass lever:

Again the Effort moves much farther than the Load. We

are getting more force out than what we put in, but theload only moves a short distance. We are also lifting

the lever along with the load.

Fulcrum


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2ndclass lever:

IMA = xE/xL= (2.8 m)/(0.35 m) = 8

Fulcrum

LoadEffort

Effort arm, xE= 2.8 m

Load arm, xL= 0.35 m

The load is 900 N and since the Effort has

to lift the Load and the lever, lets say that

the Effort is 150 N.


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2ndclass lever:

IMA = xE/xL= (2.8 m)/(0.35 m) = 8

Fulcrum

L = 900 NE = 150N

Effort arm, xE= 2.8 m

Load arm, xL= 0.35 m

AMA = L/E = (900 N)/(150 N) = 6

Eff = AMA/IMA = 6/8 x 100% = 75%


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3rdclass lever:

Fulcrum

LoadEffort

Notice that the Effortand the Loadare on the same side of the Fulcrum

and the Effortis between the Load

and the Fulcrum.


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3rdclass lever:

Fulcrum

AMA = L/E = (200 N)/(1000 N) == 0.2

Eff = AMA/IMA = (0.2/0.25) x 100% = 80%

L = 200 NE = 1000 N

xE= 50. cm

xL= 2.0 m

IMA = xE/xL= (0.50 m)/(2.0 m) = = 0.25


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There is a lab part of the competition.

Lets look at some of the basic concepts

for a lever that is in static equilibrium.


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The easiest lever to analyze is the first class

lever (seesaw), that is balanced by itself.

The center of gravity of the lever is on the

fulcrum.

c.g.


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If a lever is not moving (rotating) then it is said to

be at static equilibrium. When an object is at

static equilibrium the following is true:

F = 0, that is netF = 0, no unbalanced forces.

= 0, that is there are no unbalanced torques.

If you place a seesaw so that its center of gravityis on the fulcrum, it will balance. That is, the left

side balances the right side.


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Torque is the tendency of a force to

cause an object to rotate around an

axis. In the case of a lever, the axis isthe fulcrum.

Force In this case the force would make the leftside of the lever go down or rotate the

lever counterclockwise, ccw.


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Torque is the tendency of a force to

cause an object to rotate around an

axis. In the case of a lever, the axis isthe fulcrum.

Force

In this case the force would make the leftside of the lever go down or rotate the

lever counterclockwise, CCW.


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What if the force is 24 N, what torque is applied?

x = 1.2 m

F = 24 N

Earlier we talked about the lever arm or arm being

the distance from the fulcrum (axis) to the force. Wewill use the letter x as the symbol for lever arm.

The symbol for torque is the Greek letter tau,

= (F)(x) = (24 N)(1.2 m)

= 28.8 N.m = 29 N.m

A torque of 29 N.m will rotate the lever CCW.


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The weight of the seesaw on the left creates a torque

that tries to rotate the seesaw counter-clockwise, CCW,

so that the left side would go down.

F

The weight of the seesaw on the right creates a torquethat tries to make it rotate clockwise,CW, so that the right

side would go down.

F

The two balance each other

and it does not rotate.


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Another way to look at this is that we can place all

the weight of the seesaw ( FL ) at its center of

gravity.

FL

c.g.

The center of gravity of the seesaw is at the axis of

rotation (fulcrum) so the value of the lever arm is

zero and the force creates no torque.

Note: The center ofgravity may not be at

the geometric center.

Especially whenusing wooden

meter sticks!


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Sample problem: Two identical 40.0 kg twin girls

are sitting on opposite ends of a seesaw that is 4.0

m long and weighs 700 N, so that the center of

gravity of the seesaw is on the fulcrum.

c.g.

How do we analyze this situation?


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F d F b h h h


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FN= 1500 N

x1= 2.0 m x2= 2.0 m

FL

= 700 N

c.g..

F1

= 400 N F2= 400 N

= 0 or CCW = CW

F1x1= F2x2

FLand FNboth act through

the axis of rotation, so their

lever arm is zero, making

their torque 0.

(400 N)(2.0 m) = (400 N)(2.0 m)

800 Nm = 800 Nm

The torques balance so the

seesaw can be in static

equilibrium.


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It is important that the seesaw be

level, so that the force applied by eachof the girls is acting downward and is

perpendicular to the lever arm. If the

Force and the Lever Arm are not

perpendicular, then the equation for

the torque becomes complex. It isbetter that we avoid that situation.


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One 400 N girl sits on one end of a seesaw that is

centered on the fulcrum, is 4.0 m long, and weighs 700 N.

Where must her 650 N brother sit in order for the seesaw

to be in static equilibrium?

c.g.

?

So, what do you do to balance the seesaw if the two

people are not the same weight (mass)?

Option #1, move the

heavier person closer to

the fulcrum.

F d F b th t th h


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FN= 1750 N

xG= 2.0 m xB= ?? m

FL

= 700 N

c.g..

FG

= 400 N FB= 650 N

= 0 or CCW = CW

FGxG= FBxB

FLand FNboth act through

the axis of rotation, so their

lever arms are zero.

(400 N)(2.0 m) = (650 N)xB

800 = 650xB

xB= (800 Nm)/(650 N)

xB= 1.23 m

Option #2 move the center of gravity of the seesaw so that


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One 400 N girl sits on one end of a 4.0 m long seesaw

weighing 700 N That has moved the center of gravity ofthe lever 0.2 meters towards her. Where must her 650 N

brother sit in order for the seesaw to be in static

equilibrium?

c.g.

?

Option #2, move the center of gravity of the seesaw so that

more of the seesaw is on the side of the lighter person,

Now the weight of the

seesaw creates a torque

helping the girl.

0F acts through the axis of


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FN= 1750 N

xG= 2.2 m xB= ?? m

FL

= 700 N

c.g..

FG

= 400 N FB= 650 N

= 0 or CCW = CW

FGxG+ FLxL= FBxB

FNacts through the axis of

rotation, so its lever arm is zero.

(400)(2.2) + (700)(0.2) = (650 N)xB

880 + 140 = 650xB

xB= (1020 Nm)/(650 N)

xB= 1.57 m

xL= 0.2 m


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For the Middle School(Division B) Competition,

you will need to build a

simple first class leversystem. The lever may not

be longer than 1.00 meter.

Si l M hi (Si l )


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Simple Machines. (Simple case.)

Given small mass placed on one side.

Given unknown large mass on the other.

Unless the values are too extreme, you may be able to

move the large mass close enough to the fulcrum.

c.g.

If this is the setup, you

dont have to worry about

the weight of the lever.

Si l M hi (Si l )


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c.g.

c.g.

FS FBFL

FN

Small MassBig Mass

.

xBxS

Si l M hi (Si l )


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c.g.

FS FBFL

FN

.

xBxS

In this case the torque equation is: CCW= CW

(FS)(xS) = (FB)(xB) and you can solve for any value.


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So far we have been dealing with the force applied by the

hanging mass. This force is known as the weight of the

object or the force of gravity (Fg) acting on the object.

The force of gravity acting on an object is the product of

the mass of the object multiplied by the gravity constant

on the planet Earth (9.8 N/kg).

Fg= mg = m(9.8 N/kg) so, mass, m = Fg/(9.8 N/kg)

This gets quite confusing because weight is measured in

newtons and mass is measured in grams or kilograms (kg).

You may have been told to weigh something but you

actually measured its mass in grams.

Si l M hi (Si l )


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c.g.

FS= mSg FB= mBgFL

FN

.

xBxS

Knowing that Fg= mg

(FS)(xS) = (FB)(xB) This equation can be written:

(msg)(xS) = (mBg)(xB) Dividing by g we get:

(msg)(xS)/g = (mBg)(xB)/g (ms)(xS) = (mB)(xB)

Si l M hi (Si l )


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c.g.

mS mBFL

FN

.

xBxS

We can now solve for a mass using this equation

and modify our torque diagram as shown:

(ms)(xS) = (mB)(xB)

Simple Machines (Simple case )


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c.g.

mS= 125 g mB= ??FL

FN

.

xB= 10.0 cmxS= 47.6 cm

Suppose that you were given a small mass of 125

grams and an unknown large mass. You set upyour lever so it balances as shown:

Simple Machines (Simple case )


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c.g.

FL

FN

.

xS= 47.6 cm

(ms)(xS) = (mB)(xB)

mS= 125 g mB= ??

xB= 10.0 cm

(125 g)(47.6 cm) = (mB)(10.0 cm)

mB= (125 g)(47.6 cm)/(10.0 cm) = 595 grams

Simple Machines (More realistic case)


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Simple Machines (More realistic case)

Then place the small mass on the long side.

Place the unknown large mass on the short side.

If the difference between the unknown mass and

the known mass is large, move the fulcrum nearone end of the lever.

c.g.

Now the lever helps

balance the large weight.

FNacts through the axis of = 0 or =


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N g


FN

xS= cm

xB= cm

FL= N

c.g..

FS

= N FB

= ?? N

xL = cm

= 0 or CCW = CW

FSxS+ FLxL= FBxB

FNacts through the axis of = 0 or =


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N g


FN

xS= cm

xB= cm

FL= mLg

c.g..

FS= mSg FB

= mB

g

xL = cm

= 0 or CCW = CW

FSxS+ FLxL= FBxB

If you divide through by g you get:

mS

xS

+ mL

xL

= mB

xB

(mSg)xS+ (mLg)xL= (mBg)xB

Sample: Suppose that you have set up


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FN

xS= cm

xB= 10.0 cm

FL= mLg

c.g..

FS= mSg FB

= mB

g

xL= 20.0 cm

p pp y p

your 1.00 meter long lever of mass 83.4

grams so that the center of gravity is

20.0 cm from the fulcrum. You have alsodetermined that the big unknown mass

will be placed 10.0 cm from the fulcrum.


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mSxS + mLxL = mBxB


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FN

xB= 10.0 cm

mL= 83.4 g

c.g..

mS= 76.2 g mB

= ??

xL= 20.0 cm

mSxS+ mLxL mBxB

xS= 62.8 cm

(76.2 g)(62.8 cm) + (83.4 g)(20.0 cm) = mB(10.0 cm)

mB= [(76.2 g)(62.8 cm) + (83.4 g)(20.0 cm)]/(10.0 cm)

mB= 645 grams


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As long as the levers are horizontal

and in static equilibrium, you canuse the equations with mass

instead of force or weight. Your

Physics teacher probably will not

be too happy, but the equation is

mathematically correct for thesituation.


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c.g.

B

c.g.

First Class Lever Second Class Lever

For the High School Competition you will need to

build a compound lever system made up of a first

class lever connected to a second class lever. Each

lever may not be longer than 50.0 cm.


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c.g.

xS= cmxE1= cm

mL1= g

c.g..

mS= g E = ?? gxL1 = cm

mSxS+ mL1xL1= ExE1

First part of lever system


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xE2= cm

xB= cmmL2= g

c.g..

E = g

mB= ?? g

xL2 = cm

ExE2= mL2xL2+ mBxB

Second part of lever system

B

c.g.

5 0Sample:


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c.g.

B

c.g.

Suppose that you built your lever system so thatthe fulcrum in the class 1 lever was 5.0 cm from

the string connecting the levers.

xE1= 5.0 cmSample:

Also suppose that in the second class lever thestring connecting the levers is 35.0 cm from its

fulcrum and you set up the lever so that the

unknown big mass is 10.0 cm from the fulcrum.

xE2= 35.0 cm

xB= 10.0 cm

x = 5 0 cmSample 3 0


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c.g.

B

c.g.

You are given a small mass of 86.0 grams and anunknown large mass that you place at the 10.0

cm mark.

xE1= 5.0 cmSample

You slide the small mass along the class 1 leverand manage to get the levers to balance when

the small mass is 32.7 cm from its fulcrum.

xE2= 35.0 cm

xB= 10.0 cmxS= 32.7 cm

First part of lever systemxE1= 5.0 cm


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c.g.

xS= 32.7 cmxE1= 5.0 cm

mL1= 27.4 g

c.g..

mS= 86.0 g E = ?? gxL1= 16.0 cm

mSxS+ mL1xL1= ExE1

E1

xS= 32.7 cm

You also measured the mass of the

lever as 27.4 g and arranged the

lever so that its center of gravity is16.0 cm from the fulcrum.

m + m EFirst part of lever system


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xS= 32.7 cmxE1= 5.0 cm

mL1= 27.4 g

c.g..

mS= 86.0 g E = ?? gxL1= 16.0 cm

mSxS+ mL1xL1= ExE1

(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm) = E(5.0 cm)

E = [(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm)]/(5.0 cm)

E = 650 g

Second part of lever systemxE2= 35.0 cm


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xE2= 35.0 cm

xB= 10.0 cm

mL2= 31.4 g

c.g..

E = 650 gmB= ?? g

xL2= 18.0 cm

ExE2= mL2xL2+ mBxB

p y

B

c.g.

xB= 10.0 cm

Suppose the mass of the lever is

31.4 g and its center of gravity is

18.0 cm from the fulcrum.

Ex = m x + m x Second part of lever system


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xE2= 35.0 cm

xB= 10.0 cm

mL2= 31.4 g

c.g..

E = 650 gmB= ?? g

xL2= 18.0 cm

ExE2= mL2xL2+ mBxBp y

(650 g)(35.0 cm) = (31.4 g)(18.0 cm) + mB(10.0 cm)

mB= [(650 g)(35.0 cm) - (31.4 g)(18.0 cm)]/(10.0 cm)

mB= 2218 g = 2.218 kg


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Things to note:

You must build your own lever system.

You may want to have two set places to have

your fulcrum depending on the given masses.

You may want to have the unknown mass at a

predetermined spot and thus notching the leverat that point.

Make sure that you know the mass of your lever

and have marked the location of its center ofgravity.


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Fi d P ll


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Fixed Pulley

A fixed pulley is basically a First

Class Lever that can rotate. The

mechanical advantage is 1. All a

fixed pulley does is change the

direction of the force.

LoadEffort

Load

Effort

Fulcrum

M bl P ll


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Movable Pulley

A movable pulley is basically a

Second Class Lever that can rotate.The mechanical advantage is 2.

Load

Effort

Load

Effort

Fulcrum

Bl k & T kl


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Block & Tackle

A Block and Tackle is a combination

of a movable pulley connected to afixed pulley. In this case the

mechanical advantage of the

movable pulley is 2 and the MA of

the fixed pulley is 1. Combined themechanical advantage is 2.

In order to calculate the Ideal

Mechanical Advantage, IMA, of a

Block and Tackle, you count the

number of supportingropes.

Load

Effort

Block & Tackle


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Block & Tackle

A Block and Tackle is a combination

of a movable pulley connected to afixed pulley. In this case the

mechanical advantage of the

movable pulley is 2 and the MA of

the fixed pulley is 1. Combined themechanical advantage is 2.

Work is done to lift the Load, but you also

must lift the movable pulley with it. The

AMA will be less, not only because of

friction in the system but because the

weight of the movable pulley also has to

be lifted by the Effort.

Load

Effort

Wheel and Axled


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Wheel and Axle

A Wheel and Axle is two different

diameter cylinders on the same shaft. Thisalso is a first class lever that can rotate.

Load

Effort

dL

dE

FulcrumEffortLoad

The Ideal Mechanical Advantage is the

ratio of the diameters. IMA = dE/d

L.


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All these are examples of a wheel and Axle.

Inclined Plane


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Inclined Plane

An inclined plane is basically a ramp that is stationary.

A load is pushed up the ramp instead of being lifted

straight up. The Ideal Mechanical Advantage of aramp is the ratio of the length of the ramp (x) to the

height of the ramp (h). IMA = x/h

x

h

Wedge


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Wedge

A wedge is like an inclined plane, but instead of being

stationary the wedge is driven into something or

between things.

All these are examples of wedges.


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LEVERS.


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Class 1 Levers.

Class 1 Levers. One of these can also be used

l 2 L Whi h ? H ?


81/85

as a class 2 Lever. Which one? How?


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COMPOUND

MACHINES.

This is actually a compound machine.

Th t th d th t t i t th d


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The teeth are wedges that cut into the wood.

The whole blade is a wheel and axle.

These too are compound machines.

Th t th d th t t th i


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The teeth are wedges that cut the wires.

The handles make a class 1 lever.


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2014 simple machines basic concepts

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