week 1-mechanics introduction

8/6/2019 Week 1-Mechanics Introduction

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Mechanics Introduction Dr. Stuart McLachlan

1

Mechanics

Gravity, Weight and Moments

Gravity is the force of attraction between all masses.

Gravity attracts all masses, noticed when one of the masses is really big, i.e. a planet.

Anything near a planet or a star is attracted to it very strongly.

Three Important Effects

1. Gravity makes all things accelerate towards the ground, all with the same acceleration,

g, which is 9.81 m/s2 on Earth (or approximately 10 m/s2).

- Leaning tower of Pisa:

Made during the time whencountries were making the best

out of cannons and wanted toknow how to drop them precisely

onto their neighbours.

To know this they carried out

experiments by dropping cannonballs of different masses from the

tower, to show that they landed

onto the ground at the same time

and therefore with the same

acceleration.

What about the feather?

The air resistance would slow it down.

- If we put a cannon ball and a feather into a

vacuum jar, both would fall and land at the

same time.

?


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2

- Thekitchen-weighing machine There are two types:

Balance Scales Spring Scales

Which measures mass and which measures weight?

Clue: Imagine moving these scales to the moon.

2. Gravity gives everything a weight.

3. Gravity keeps planets, moons and satellites in their orbits.

Orbit is a balance between forward motion of the object and the force of gravity

pulling it inwards.

1kg massSugar

1kg

1kg mass

Sugar

1kg massreference

Thrust

Forward Velocity

Force of

Gravity

Resultant Orbit


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3

Weight and Mass

1. Mass is the amount of matter in an object.

i.e. Mass of an object is the bulk of ALL particle masses (atoms: electron / proton /

neutron / etc). Therefore an object will have the SAME MASS anywhere in the

Universe.

2. Weight is the effect of mass pulled by gravity.

Balance Scales Spring Scales

The Balance Scales compares Mass

The Spring Scales measures weight by measuring the force on a spring exerted by themass of the flour pulled by gravity.

3. An object has the same mass whether it is on Earth or on the Moon but its weight

will be different.

1 kg mass weighs less on the moon of 1.6 Newtons compared with that on the Earth of

10 Newtons.

4. Weight is a force measured in Newtons (N) measured using a spring balance or a

Newton meter.

Weight = Mass x Gravitational Acceleration

W = m g

e.g. 5 kg on Earth w = 5 x 9.81 = 49.05 N

Moon w = 5 x 1.6 = 8.00 N

1kg massSugar

1kg

1kg mass

Sugar

1kg mass

reference

Force due to

gravityForce due to

gravity

Force due to gravity pressing

against a spring

Forces are in balance on the pivot

pivot

Force

Spring displacement

Hookes Law - springs

Force ~ distance


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4

Moments

If we move the pivot 50 mm towards the 1 kg reference weight, how much sugar do I need to

balance the scales?

For the balance scales to be in equilibrium (balanced) then:

Left Hand Side = Right Hand Side

Force x Distance LHS = Force x Distance RHS

( ) msmMmsmkg s 200.0/81.9050.0150.0/81.9122=

Thus kgMs 5.02.081.9

01.081.91=

=

Try:

Findx

mmmx

x

350350.02

6.01.0

3.081.921.081.9181.92

==+=

+=

When a force acts on something, which has a pivot, it creates a

turning force called a moment.

Moments are calculated by:

moment = force x perpendicular distance

1kg

150 mm

50 mm

150 mm

1 kg sugar Ms1 kg reference

weight

AntiClockWise

Turning force

ClockWiseTurning

force

For a system to be in equilibrium:

TOTAL CLOCKWISE MOMENT = TOTAL ANTICLOCKWISE MOMENT

2kg

x mm100 mm

300 mm

2 kg Flour2 kg reference

weight

1 kg sugar


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5

Force Diagrams

Force is a push or a pull.

There are mainly 6 different forces:

1. Gravity or weight always acting straight down

2. Reaction force from a surface, usually acting straight up

3. Thrust or push or pull due to an engine or rocket speeding something up

4. Drag or air resistance or friction which is slowing the thing down

5. Lift due to an aeroplane wing

6. Tension in a rope or cable

These result in 5 different force diagrams

1. Stationary object in balance

Static weight

2. Steady horizontal velocity all forces in balance

Moving car

3. Steady vertical velocity all forces in balance

Sky diver

- If there is an unbalanced force then you get an acceleration, NOT steady speed

4. Horizontal acceleration unbalanced forces

Moving car throttle down

5. Vertical acceleration unbalanced forces

Sky diver crouched

- Only get acceleration with overall resultant unbalanced force

- The bigger the unbalanced force the greater the acceleration

reaction

sideside

weight

1 kg

reaction

dragthrust

weight

drag

sideside

weight

reaction

dragthrust

weight

acceleration

drag

sideside

weight

acceleration


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6

Friction

1. Friction is always there to slow things down

a. Neutral friction slows car down

b. Steady 30 mph friction

Friction always occurs:

a. Between solid surfaces which are gripping. e.g. between the tyres and the road

when cornering.

There is a limit on how the two surfaces can grip each other. i.e. brake too hard

and youll skid.

b. Between solid surfaces which are sliding past each other. e.g. Between brake pads

and brake discs which use sliding friction.

c. Resistance or Drag from fluids (air or liquids). i.e. Cars are streamlined to reduceair resistance (drag).

The opposite of this is a parachute, which requires high amounts of drag.

2. Friction always increases as speed increases

The car has much more friction to work

against when travelling at 60 mph compared

to 30 mph.

- worse fuel economy

- worse engine life engine works harder

Note: Compare speed and fuel economy:

Drag car 300+ mph 1 gal / second

Sports car 100+ mph < 20 mpg

Family car around 100 mph around 50 mpg

Record vehicles 10 20 mph > 10,000 mpg

3. Need friction to move and stop

e.g. walk / run, race off traffic lights, hold fixings together (nuts and bolts) . . .

4. Friction causes wear and heating

a. Always acts between surfaces that are sliding over each other.

b. Produces heat and wear of surfaces.

c. Lubricants used to keep friction as low as possible.

d. Lubricants enable machines to run more freely using less power with reduced

wear.

e. Heating effect of friction can be large.

e.g. Brakes glowing red hot causing brake fade, engine without oil seizes.

forcein balance

30 mph drag

60 mph more drag


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7

Laws of Motion

Sir Isaac Newton published Philosophioe Naturalis Principia Mathematica 1st

edition 1687

1

st

LawEvery body remains stationary or in uniform motion in astraight line unless it is made to change that state by external

forces.

2nd Law

Change of momentum (acceleration) is proportional to the

impressed force, and acts along the same straight line.

3rd Law

Action is always equal and in the opposite direction toreaction.


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8

Velocity and Acceleration

Example

If I walk such that my steps are 1 m apart and my pace is 10 steps in 9 seconds. What is myspeed?

( )

velocityspeed

hkm

hkm

smonds

metersspeed

==

=

=

==

time

distance

/4

60601000

1

9

10

/11.1sec9

10

Example

An aircraft flies for 50 seconds in a straight line.

Time (s) 0 10 20 30 40 50

Distance

(km)

0 6 12 18 24 30

skmvelocity /6.050

30==

Distance (km)30

24

18

12

6

0

0 10 20 30 40 50 Time (s)


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9

Examples

a. A cars speedometer registers 72 km/h.

Express this velocity in m/s

smsm /206060

1

100072 =

b. A motorist is driving along a motorway, which runs due N and S from a junction

known as Romans Cross. He starts 20 km N of the junction at 9:00am, drives for

43 h N at 80 km/h, waits for

43 h at a service station and then drives S at a

distance of 120 km at 60 km/h. After another2

1 h wait he returns to Romans

Cross, arriving there at 2:00pm.

Represent his journey on a graph.

QP h PL 60 km

hkmhkm /804

360since =

This fraction is called SLOP or GRADIENT of graph

UT (displacement at U) (displacement at T)

( ) kmkm 40400 +==

SR 2 h RM

(displacement at S) (displacement at R)

( ) ( ) kmkmkm 1208040 =+= South

Test

Find:

a. Average velocity over journey

b. Represent this on graph

c. Average speed over journey

S km

80

60

40

20

0

-20

-409:00 9:45 10:30 11:15 12:00 12:30 1:00 2:00 Time

P

Q

L

R

S T

U


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10

Solution

Theory:

periodtimetotal

ntdisplacemetotalperiodtheovervelocityaverage =

periodtimetotal

travelleddistancetotalperiodtheoverspeedaverage =

Therefore total displacement of the car between 9:00am and 2:00pm

= (displacement at U) (displacement at P)

( ) kmkm 20200 =+=

The time interval is 5 h therefore the velocity is:

hkmh

km/4

5

20=

=

i.e. 4 km/h South represented by the slope of the line UP

Average speed the car travels km2204012060 =++=

hkmspeedAverage /445

220==

start

finish


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11

Acceleration

Is how quickly you are speeding up

Acceleration is how quickly the velocity is changing

takentimevelocityinchangeonaccelerati =

e.g.

A skulking cat accelerates from 2 m/s to 6 m/s in 5.6 s. Find its acceleration?

Acceleration = a

Velocity = v

Time = t

( ) 2/71.06.5

26sm

t

va =

=

=

When the velocity of a body is not uniform, weintroduce a quantity, which measures the rate at

which velocity is changing. This is called the

acceleration of a body.

Example

If a train is moving at 18 km/h at one instant, and at 90 km/h two minutes later, its velocity

has increased by 72 km/h in 2 minutes.

If we suppose that it is gaining speed at a steady rate, we say that it has an acceleration of 36

km/h per minute.

We therefore need to convert the kilometres to metres, hours to seconds, and minutes to

seconds:

2/6

1

603600

36000

606060

100036

60

1

6060

1100036 sm

ssma =

=

=

=


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12

If the velocity of a body is decreasing, the body is said to have negative acceleration, usually

called retardation or deceleration.

i.e. + m/s2

acceleration

- m/s2

deceleration or retardation

It is frequently useful to draw a graph of the velocity of a body plotted against time. When the

velocity is uniform, the graph is a straight line parallel to the time axis.

Velocity Time Graphs:

If the velocity increases by equal amounts in equal times, so that the graph is a straight line

inclined at an angle to the time axis, the body is said to have UNIFORM ACCELERATION.

The value of this acceleration is represented by the slope of the line.

Whether the acceleration is uniform or not, the average acceleration over a period of time is

measured by dividing the increase in velocity during that time period, by the time taken.

Velocity

in m/s

60

50

40

30

20

10

0

0 10 20 30 40 50 60 70 Time in secs

acceleration

Steady speed

deceleration

Steady speed

Increasingacceleration


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13

Example

A train has a motion shown above in the velocity-time graph, has a velocity of 40 km/h at

10:00 am and one of 120 km/h at 10:20 am.

The average acceleration between 10:00 and 10:20 is:

( ) 22403

1/40120

hkmhhkm =

This is represented on the graph by NU / TN

i.e. The slope of the line TU

Try:

Find acceleration if 9:10 am velocity is 20 km/h and at 9:35 am is 120 km/h.

Summary Definitions

The VELOCITY of a body is the rate at which its displacement is increasing with

respect to the time.

The ACCELERATION of a body is the rate at which its velocity is increasing with

respect to the time.

dt

dsv ==

timeinchange

distanceinchange

2

2

timeinchangeofrate

distanceinchangeofrate

dt

sda ==

v km/h140

120

100

80

60

40

20

0

9:00 9:30 10:00 10:30 Time

T

N

U


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14

Example

In catching a train a man runs for 80 s at 2.5 m/s, then walks for 100 s at 2.0 m/s and finally

runs for 20 s at 3.0 m/s. Sketch the velocity-time graph and find the total distance that he

covers.

The velocity-time graph consists of 3 straight lines parallel to the time axis.

Since the velocity over each leg is uniform, the total distance the man goes is:

mm 4602031002805.2 =++

NOTE: distance = velocity time = area of graph

0 50 100 150 200 t s

v m/s

3

2

1

0

80100

20


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15

Example

An electric train starts from a station and maintains an acceleration of 0.9 m/s2

for 20 seconds.

It then travels for 80 seconds with uniform velocity, and finally has a uniform retardation

which brings it to rest in a further 10 seconds. Find the retardation and the total distance gone.

The velocity-time graph consists of 3 lines OA, AB, and BC. The slope of the line OA

represents the original acceleration (0.9 m/s2) since MA/OM = 0.9, and OM = 20, it follows

that MA = 18 m/s.

This means that the maximum velocity attained is 18 m/s

MN = 80, the time in seconds for which the train has uniform velocity

NC = 10, the time in seconds for which the train has uniform retardation

The slope of the line BC is BN/NC

Representing an acceleration 2/8.110

/18sm

s

sm=

=

Negative indicates the train is slowing down with a retardation of 1.8 m/ss.

The distance travelled is the area of the trapezium OABC which is:

(110 + 80) x 18 = 1710

Therefore the distance between the two stops is 1710 m or 1.71 km.

v m/s

18

0

0 M N C t s

20 80 10

A B


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16

Equations of Motion

For uniform acceleration in a straight line

u = velocity at the start of the period

v = velocity at the end of the perioda = uniform acceleration throughout the period

s = displacement from initial to final position

t= time taken

2/smt

uv

time

velocityinchangea

==

Thus,

atuv += .............................................................. 1

The velocity-time graph is a straight line, and the area between the graph and the time axis is:

( ) tvus += 2

1........................................................ 2

substituting v from 1

( ) tatutus ++=2

1

2

2

1atuts += ..................................................... 3

similarly 2

2

1atvts =

substituting tinto 2 from 1

( )a

uvvus

+=

2

1

asuv 222 += ..................................................... 4

velocity

uv

t


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17

Examples

A marble is rolled down a gentle slope. It has a constant acceleration, and its velocity

increases from 10 cm/s to 32 cm/s in a distance of 77 cm. Find the acceleration.

10=u 32=v 77=s required a

222

22

22

/6154

924

772

1032

7721032

2

scma

a

asuv

==

=

+=

+=

A train starts from rest and, moving with constant acceleration, passes through a station 9 kmaway after 5 minutes. Find the acceleration.

0=u 9000=s 300=t required a

2

2

2

/2.090000

90002

3002

109000

2

1

sma

a

atuts

=

=

+=

+=

A car approaching a speed limit applies its brakes. It takes 4 s to cover the next 100 m, and 5 s

to cover the succeeding 100 m.

Find the retardation, and the speed at which it was moving when the brakes were applied.

The average speeds over the two 100 m stretches are 25 m/s and 20 m/s

These are average speeds and represent periods

Therefore 25 m/s at a time 2 seconds after brakes applied

20 6.5

The speed decreases by 5 m/s in 4.5 seconds

Thus retardation is 5 m/s 4.5 = 1.1 m/s2

In the first 2 seconds after the brakes are applied the speed decreases by 2.2 m/s and at

the end of this time the speed is 25 m/s

Therefore the initial speed is 27.2 m/s


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18

Forces and Acceleration

Newtons 3 Laws

1. Body remains stationary or in uniform motion in a straight line unless

changed by external forces.2. aF maF=

3. Action is always equal and opposite to the reaction force.

e.g. A man drags a dinghy of mass 200 kg across the beach with a force of 200 N. The

motion is hindered by a friction force of 170 N. Find the acceleration of the dingy.

2/15.020030

200

30170200

smaa

kgm

NF

=

=

=

==

One person pushing a 1000 kg car can accelerate it at 0.1 m/s2. Two people pushing

equally hard with the same force as before can accelerate it at 0.3 m/s2. How large is

the resistance to motion?

Let the force with each person pushes be x newtons and the resistance to motion R

newtons.

With one person pushing:

RxF = 1000=m 1.0=a

1001.01000 ==Rx

With two persons pushing:

RxF = 2 1000=m 3.0=a

3003.010002 ==Rx

Deduced that R = 100 Newtons

0.1x

R0.3

2x

R


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19

Work and Energy

When a body moves under the action of a force, we measure the usefulness of the force by its

work done:

Work done by the force = force x distance = F s

The amount of work done when a force of 1 N is applied through a distance of 1 m is 1 Nm.

This is called aJoule. i.e. 1 Nm = 1J

(from James Joule (1818-1889),who first developed the modern concept of energy)

e.g. If a luggage-truck is pulled along a platform by applying a force of 20 N to a handle, and

if the truck is moved a distance of 150 m, the work done is 3000J.

Kinetic Energy KE

When a body is in motion:

Kinetic Energy of a body = mass x velocity2

2

2

1mvKE=

Work Done by Force = KE at end KE at start

( )2222

2

1

2

1

2

1uvmmumvFsKE ===

A car of mass 1200 kg starts from rest and is observed to reach a speed of 20 m/s after it has

travelled 250 m. Neglecting resistances, find the driving force if it assumed to be constant.

There is no KE at the start, and the KE at the end is:

joulesmv 0002402012002

1

2

1 22==

If the driving force is F newtons, the work done by this force is:

joulesF 250

NF

F

960that,so

240000250

=

=

If there were a resistance to motion ofR, the term Fin the above equation would be replacedby FR, and the final equation would become:

22

2

1

2

1mumvRsFsKE ==

work done by accelerating force - work done against resistance = increase in KE


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20

Example

The driver of a car of mass 800 kg travelling at 36 km/h sees an obstruction in the road. He

applies his brakes immediately, but does not use the clutch to disengage the engine until he

has travelled a further 10m. He finally comes to rest at a distance of 30m after the brakes were

applied. If the engine was exerting a forward thrust of 140 N, find the retarding force due to

the brakes.

One advantage of using the energy method to solve this problem is that the whole

motion can be considered at once rather than two stages. Because;

During the first stage before the clutch is disengaged:

Work done by driving force - work done against resistance = gain in KE

Or since there is a loss of KE because the car is slowing down:

work done against resistance - work done by driving force = loss of KE

In the second stage the engine makes no contribution, so that

Total work done against resistance = loss of KE

Adding these two equations, we have for the complete motion:

Total work done against resistance - work done by driving force in first stage = total loss of KE

Initially smhkm /106060

100036/36 =

=

The KE at the start 400001080021

21 22 === mu

Since there is no KE at the end, the loss in KE is 40 000 J

The work done by the driving force over the 10m for which it acts is

J140010140 =

If the resistance is R newtons, the work done against this resistance over the distanceof 30m is:

( )

NRR

JR

1380that,so40000140030

30

=

=

The force due to the brakes is thus 1380 N


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21

Potential Energy PE

The (GRAVITATIONAL) POTENTIAL ENERGY of a body in a given position is the work,

which would be done by its weight if it were to move from that position to a point at some

fixed level.

PE = weight of body x height above fixed level

PE = mgh

Combined PE and KE

Loss of PE = Gain in KE

Therefore,

(KE + PE) at start = (KE + PE) at end

That is, as the stone falls the sum of its kinetic energy and its potential energy remainsconstant.

Conservation of Energy

Two quantities of energy have been discussed which we have given the name energy,

kinetic energy and gravitational potential energy.

In general terms a body, which is moving is, by virtue of its movement, capable of doing

work as it loses its speed. For example, a railway engine which runs into a buffer will

compress the buffer; it is thereby doing work, since the force compressing the buffer is

moving its point of application.

A body, which is capable of doing work, is said to posses energy. Broadly speaking, energyis something, which can be fed into a machine or an engine to produce useful results. For

example, in a windmill the kinetic energy of the air is used to drive the sails; in a

hydroelectric plant the potential energy of a mass of water at a height is used to drive a

dynamo.

Energy appears in many different forms. Heat and electricity are forms of energy. A mass of

gas under compression, possess energy, which can be released when they undergo chemical

change; these substances are called fuels. Within the individual atoms of matter is stored a

vast quantity of nuclear energy.

Energy is continually changing from one form into another. For example, the chemical energy

of coal may be used to heat the water in a boiler. Some of this water is turned into steam, partof whose energy is due to its state of compression. This may be used to drive a piston in a

cylinder, thereby creating kinetic energy. This kinetic energy can in turn be used to drive a

dynamo and create electrical energy.

The quantity of energy involved in these changes is governed by a very important principle,

first stated by Joule, known as the CONSERVATION OF ENERGY.

This states that, despite the changes, which its form undergoes, energy is neither created nordestroyed.

In any system out of which and into which no energy passes the total amount of energy

remains unaltered.


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Example

A cyclist reaches the top of a hill with a speed of 3 m/s. He descends 30 m and then rises 20

m on the other side, reaching the top of the next rise with a speed of 2.5 m/s. The total mass of

the cyclist and his machine is 120 kg, the road and air resistances amount to 20 N and the road

distance between the two points is 800 m. Find the work done by the cyclist during this

stretch.

At the start the KE is J54031202

1 2=

And at the end it is J3755.21202

1 2=

In calculating the PE, take as the fixed level that of the point at the end of his ride. Hestarts at a height of 30m 20m = 10m above this; his weight is 120 x 9.81 = 1176 N,

so that

PE at start J11760101176 ==

And the PE at the end is zero.

The work done against resistance is J1600080020 =

The work energy principle takes the form:

(KE+PE) at start + work done by cyclist work done against resistance = (KE+PE) at end

Therefore,

037516000cyclistbydonework11760540 +=++

so that the work done by the cyclist is 4075 J

week 1-mechanics introduction

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