Post on 07-Apr-2018
8/6/2019 Week 1-Mechanics Introduction
1/22
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.
?
8/6/2019 Week 1-Mechanics Introduction
2/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
3/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
4/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
5/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
6/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
7/22
Mechanics Introduction Dr. Stuart McLachlan
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.
8/6/2019 Week 1-Mechanics Introduction
8/22
Mechanics Introduction Dr. Stuart McLachlan
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)
8/6/2019 Week 1-Mechanics Introduction
9/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
10/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
11/22
Mechanics Introduction Dr. Stuart McLachlan
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 =
=
=
=
8/6/2019 Week 1-Mechanics Introduction
12/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
13/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
14/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
15/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
16/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
17/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
18/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
19/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
20/22
Mechanics Introduction Dr. Stuart McLachlan
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
8/6/2019 Week 1-Mechanics Introduction
21/22
Mechanics Introduction Dr. Stuart McLachlan
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.
8/6/2019 Week 1-Mechanics Introduction
22/22
Mechanics Introduction Dr. Stuart McLachlan
22
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