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Main topics in M1: using the particle model
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An overview of M1 and M2
Objectives of M1 and M2
• They are about modelling and basic concepts
• Students formulate models, appreciate assumptions and make
simple deductions
• No great emphasis on algebraic, calculus (or trigonometric)
skills but interpretation of the results is required.
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•
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vectors
kinematics
force
static equilibrium
Newton’s laws of motion
projectile motion.
the moment of a force
the centre of mass of an object
forces in pin-jointed light frameworks
friction
work, energy and power
impulse, momentum and impacts
the conservation of linear momentum and the conservation of
mechanical energy.
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Main topics in M2: using a rigid body model
where appropriate
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Basic modelling assumptions
• the particle model
• dealing with bodies
• Approximations to ideal behaviour
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light
smooth
rigid
inextensible
value of g
Hooke’s law
Coulomb’s law
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Mathematical models
• Our aim is to develop insights into and be able to predict the
behaviour of a real system
• The extent to which a model is suitable depends on
– the accuracy required
– an ability to solve the equations
• Models can only be justified by comparing their predictions
with the results of experiments (measurements)
• Models are often refined using an iterative process
Real-world
problem
Simplifying
assumptions
Mathematical
model
(equations
etc)
Experiment
Prediction
Analysis and
solution
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Why do experiments in class?
• Students who have prior practical experience of a situation are
– more likely to recognise it
– be more aware of approximations
– less likely to accept inappropriate answers
• Experiments
– provide new experiences
– help students interpret experiences they already have
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The particle and rigid body models
• Reduction to particles in M1
• Need for a rigid body model when equilibrium of a body
involves lines of action and so moments (to be dealt with in
more detail later by Stephen)
• The associated need for being able to calculate the centre of
mass of a body
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Kinematics
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Increasing sophistication using graphical representations
Constant acceleration results
The use of calculus in 1 dimension
Motion in 2 and 3 dimensions
Applications to projectile motion (dealt with in more detail by
Stephen in another session)
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Why use simulations in class?
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•
Experiments
– may be influenced by large ‘hidden’ factors
– may not be easy to use over a wide range of values
– can be very time-consuming
Simulations allow students to familiarise themselves with the predictions of
a mathematical model
– quickly and over a wide range of controlled parameter values
– interactively and individually or in groups
– augment the experience gained from working through (graded)
examples
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The nature and effects of forces
• Newton’s laws of motion - one of the intellectual peaks of all
time (to be dealt with in more detail later by Stephen).
• The use of resistance in M1 and a more detailed look at how it
can be modelled in M2
• The interesting simplifications to calculations that are found in
pin-jointed light frameworks
• When do bodies topple?
• The changes caused by forces over given periods of time and
over given distances moved, derived from N2L (to be dealt
with in more detail by David later and in another session)
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Two fundamental conservation laws
• Linear momentum
– When is linear momentum is conserved for a system?
• Mechanical energy
– When is mechanical energy conserved for a system?
(To be dealt with in more detail later by David; more on
mechanical energy is also the subject of another session)
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For this topic…
• Consider idea of Newton’s laws
Examples of:
• Particles (on horizontal, on an incline)
• Connected particles
• Lifts
Newton’s laws
Newton’s second law
Newton’s first law (N1L)
Every particle continues in a state of rest or uniform motion
in a straight line unless acted on by a resultant external force.
N2L
Newton’s second law (N2L)
When a force acts on a particle, the change in momentum is
proportional to the force. For constant mass, F = ma.
In a given direction F = ma.
Example
A toy train of mass 0.5 kg moves on a horizontal straight
track. There is a driving force of 16 N and a resistance to
motion of 12 N. Find the acceleration.
Newton’s third law (N3L)
When one object exerts a force on another there is always a
reaction that is equal in magnitude and opposite in direction
to the applied force.
Newton’s second law
N2L
a m s -2
0.5 kg
12 N
16 N
N2L in the direction of motion gives 16  12  0.5a
so a = 8 and the acceleration is 8 m s -2 in the direction of the
16 N force.
Newton’s second law
Some points to note
In a given direction F = ma.
Example
A particle of mass 4 kg slides down a slope inclined at
to the horizontal where sin  = 0.2. The resistance to the
motion is 1.84 N. Find the acceleration.
s -2
RN
1.84 N
am
Applying N2L down the slope
4g N
4 g sin   1.84  4a

so 4a = 7.84 – 1.84 = 6
Hence a = 1.5 and the acceleration is 1.5 m s -2 down the slope
Diagrams establish positive sense as well as help identify all
the relevant forces.
It is often useful to consider separately the directions parallel
to and perpendicular to the motion.
A change to one external force may lead to a change in
another.
Students sometimes learn wrong forms, e.g. F  mg  ma
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Connected particles
Connected particles
The motion of connected particles or bodies.
R3
m3
N2L may be applied to any part of the system.
F3
Whether a force is internal or external depends on
what part of the system is being considered.
Special care is needed with diagrams and labelling.
W3
R1
R2
TB
TB
TA
m2
F2
W2
TA
P
m1
F1
W1
a
Applying N2L to the whole system
P  ( F1  F2  F3 )  (m1  m2  m3 )a
Applying N2L to the LH truck
TB  F3  m3a
Forces in couplings
Particles connected round a pulley
The diagram shows the forces acting on a trailer in its line of
motion. Its mass is 800 kg and its deceleration is 2 m s -2.
The force in the tow-bar is T N and the resistance to the
motion of the trailer is R N.
A box of mass 4 kg and a sphere of mass 8 kg are connected
by a light inextensible string that passes over a smooth pulley.
The sphere hangs freely; there is a resistance of 10 N on the
box. Find the tension in the string and the acceleration.
For the box, N2L →
a m s -2
RN
T  10  4a
10 N
TN
4
kg
For the sphere, N2L ↓
TN
8 g  T  8a
4g N
Adding gives
8 kg
a m s -2
8 g  10  12a
8g N
so a = 5.7 and T = 32.8
RN
2 m s -2
TN
+ ve
N2L in the + ve direction gives
T  R  800  2 so T  R  1600
If R < 1600 then T < 0 and the tow-bar is in compression.
If R > 1600 then T > 0 and the tow-bar is in tension even
though the minibus and trailer are decelerating.
Newton’s third law (N3L)
N3L: motion in a lift
When one object exerts a force on another there is always a
reaction that is equal in magnitude and opposite in direction
to the applied force.
Forces ‘at a distance’
Example
A boy of mass 50 kg is standing in a
lift of mass 200 kg. The lift is raised
by a vertical cable. Find the reaction
of the floor on the boy and the tension
in the cable when the acceleration
is ¼g upwards.
Lifts and quicksands
For the lift, N2L ↑, T  R  200 g  200  14 g so T  312.5g
‘Pairs’ of forces
TN
¼ g m s -2
200g N
RN
50g N
RN
For the boy, N2L ↑, R  50 g  50  14 g so R  62.5g
Or, for the lift and boy, N2L ↑, T  250 g  250  14 g so T  312.5 g
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For this topic…
• Consider the idea of equilibrium
– For a particle
– For a body
Examples of:
• Particles (on horizontal, on an incline)
• Bodies on an incline (topple/slide)
• Moments
Equilibrium
• For a particle to be in equilibrium, it is necessary and sufficient that the
resultant of all the external forces acting on it is zero.
• For a body to be in equilibrium it is necessary and sufficient that the
resultant of all the external forces acting on it is zero and that the total
moment of these forces is zero about any axis.
• Two important results that may be used when considering equilibrium
In many situations a body may be modelled as a single particle
The resolved parts of forces in any chosen direction must have zero
sum
Three identical boxes each of weight W are stacked one on top of the other and
stand in equilibrium on a smooth horizontal floor. Calculate the normal
reaction at each interface.
R2
W
top box
R1 = W
R1
R1
R1
R2
W
W
R1
R1
W
middle box
R2 = W + R1
R2 = 2W
R3
W
R2
W
bottom box
R3 = R2 + W
R3 = 3W
The diagram shows a box on a rough horizontal floor. The box does not tip and is
on the point of sliding.
R
P
60°
box is in limiting
equilibrium
F
W
Since the system is equilibrium,
resolving horizontally
←
F – P cos 60° = 0
so F = ½ P
Resolving vertically
↑
R + P sin 60° – W = 0 so R = W – Psin60
Since the box is in limiting equilibrium, F = R,
eliminating F and R gives ½ P =  (W - P sin 60° )
Given either P or , you can find the other.
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A box is just being stopped from sliding down a rough plane by a light string.
T
Parallel to the plane
+
+
N
60°
A question could ask if a box, of given dimensions, topples or slides first as the
angle of the incline is increased.
Parallel to the plane
+
F + Tcos60 – Wsin30 = 0
F = W sin a
+
R
F
Perpendicular to plane
F
R = W cos a
Perpendicular to plane
When on the point of
sliding F =  R
N – W cos30 + Tsin60 = 0
30°
W sin a =  W cos a
a°
W
tan a = 
W
To see at what angle it topples, check the
dimensions of the box, and use
trigonometry to calculate the angle.
A three force problem
Moment
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A small block of weight 20N is at rest on a smooth plane inclined at 30° to
the horizontal. The block is held in position by a light inextensible string.
The angle between the string and the plane is 15°. Find the magnitude of
the tension in the string and the normal reaction of the plane on the box.
TN TN
RN
45
15°
30°
20 N
RN
20 N
30
So for a given value of 
you can find a.
The moment of a force F about an axis through O perpendicular to the
plane containing O and the line of action of F is Fd, where d is the
perpendicular distance from O to the line of action of F.
O
20
R
T


sin 105 sin 45 sin 30
d
F
T = 10.35 and R = 14.64
Calculating moments
Calculating moments
• Finding the perpendicular distance onto the line of
action of a force is not always easy. The following
approach is often efficient and effective.
Replace the 10 N force with components parallel to and
perpendicular to OP.
Example
Suppose that the force of 10 N acts through P which is 4 m
from O and that the force is at an angle of 60° with OP. We
want the moment of the force about an axis through O
perpendicular to the plane containing the force and OP
10 sin 60 N
O
4m
10 N
60°
P 10 cos 60 N
10 N
O
60°
4m
P
The anti clockwise moment about an axis perpendicular
to the plane through O is now seen to be
(10  sin 60)  4  (10 cos 60)  0 so 20 3 N m
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Effects of forces
• An unbalanced force acting on a particle with fixed mass
causes
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– acceleration
– which leads to a change in velocity and position over a period of time
• Examples
Linear momentum, mechanical energy and
conservation
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What do we mean by linear momentum?
• Linear momentum is an attribute of a moving body.
It is a vector quantity defined as mv, where m is the mass and
v is the velocity.
In elementary work we usually refer to ‘linear momentum’ as
‘momentum’
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Comparing linear momentum and kinetic energy
• Linear momentum
 mv   M 
L ML

T
T T2
• Kinetic energy
L
ML
1 2
 2 mv   M  T 2  T 2  L
mv
Force × time
SI unit: N s
½ mv ²
2
Force × distance
SI unit: J
a car accelerating away from traffic lights in a straight line
a cyclist halving her speed in a straight line
catching a ball
a ball falling without air resistance
a ball falling with air resistance
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What do we mean by energy?
• Types of energy
• Mechanical energy of bodies
Kinetic Energy (KE)
due to motion
Potential Energy (PE)
determined by position
• KE of a body is defined as ½ mv², where m is mass and v is
speed. KE is a scalar quantity.
• The Gravitational PE (GPE) of a body of mass m increases by
mgh when displaced vertically upwards a height h. (GPE is
also a scalar quantity)
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What is work?
• Work is done on a body when a force displaces it.
• The amount of work done is Fs, where s is the displacement in
the direction of the force.
• When work is done on a body, its mechanical energy changes
• When conservative forces displace a body, they change only
its potential energy (i.e. energy determined only by position)
• When dissipative forces displace a body they cause energy to
be lost from it
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A car accelerating away from traffic lights in a
straight line
Suppose the car has a mass of 900 kg and accelerates from
rest to 16 m s -1.
In the direction of motion
the momentum changes from 0 N s to 900 × 16 = 14 400 N s.
A positive change of 14 400 N s,
the kinetic energy increases from
0 to ½ × 900 × 16² , an increase of 115 200 J
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Conservation of linear momentum
• If a system of particles (or bodies) has no external force acting
on it, then the total linear momentum of the system is constant.
• This is the Principle of Conservation of Linear Momentum
(PCLM)
• Since linear momentum is a vector quantity, this principle may
be applied to the components in any direction
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A cyclist halving her speed in a straight line
Suppose the cyclist and bicycle have a combined mass of
60 kg and she reduces her speed from 12 m s -1 to 6 m s -1
In the direction of motion
the momentum changes from 60 × 12 = 720 N s
to 60 × 6 N s. = 360 N s.
A change of 360 – 720 = – 360 N s.
The kinetic energy reduces from ½ × 60 × 12² = 4320 J
to ½ × 60 × 6² = 1080 J
a decrease of 4320 – 1080 = 3240 J
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Model trucks on a level straight track
• A truck of mass 0.2 kg moves at 2 m s -1 towards a stationary
truck of mass 0.3 kg. The trucks become entangled in the
collision and move off together. Assuming that there is
negligible resistance in the direction of motion, how fast are
they travelling after the collision and in which direction?
2 m s -1
v m s -1
0
0.3 kg
0.2 kg
0.5 kg
before
after
PCLM  2  0.2  0  0.3  0.5v so v  0.8
The speed is 0.8 m s -1 in the direction of motion shown.
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Conservation of mechanical energy
• If the only forces acting on a body are conservative then its
total mechanical energy stays constant
In M2 terms, this means that KE + GPE is constant
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A ball of mass 2.5 kg falls 2 m from rest to the
floor
Suppose the air resistance is negligible and g is 9.8 m s -2.
Measure the GPE relative to the floor
initially PE is 2.5 × 9.8 × 2 = 49 J: KE is 0
total is 49 J
total is 49 J: PE is 1.5 × 9.8 × 2 = 29.4 J
KE is 49 – 29.4 = 19.6 J
2m
1.5 m
total is 49 J: PE is 0 J KE is 49 – 0 = 49 J
½ × 2.5 × v² = 49 so v = 6.26…
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Things to emphasize
• Compare and contrast the changes in the linear momentum and
the energy of a body
– the different conditions under which linear momentum and
mechanical energy are conserved
• The need for clear diagrams showing all the relevant forces,
the direction of motion and the direction taken as positive
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