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M1  Chapter 3  Dynamics of a Particle moving in a straight line
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by
TweetPeter Scott
on 7 November 2016Transcript of M1  Chapter 3  Dynamics of a Particle moving in a straight line
me creating art
What is it?
before we start....
the day i learnt about
combustion!
meeting the big cheese!
buy milk
pick up kids
Mechanics is the branch of mathematics which deals with the action of forces on objects.
M1:
Kinematics of a particle moving in a straight line
Mechanics is a very complicated topic and in M1 we want to start with the basics. To do this we will simplify our problems by using a mathematical model. This involves making some modelling assumptions.
Ignore air resistance
there are more but we will meet these as we progress through the module!
with
Mr Scott
Mechanics
To pass this module you will need to be confident in your ability to manipulate algebra.
Can you rearrange this equation to make each letter the subject?
=
(b+2)
c
a
2
7
CHAPTER 1+2:
1
2
Constant force due to gravity (9.8m/s)
3
Many objects modelled as a particle.
We will learn quite a few formulae before the exams. Here are our first 2
v = u + at
and
s =
u + v
2
( )
t
Where . . . .
s
u
v
a
t
= displacement (distance)
= initial velocity
= final velocity
= acceleration
= time
when decelerating this value will be negative!
Dynamics of a particle moving in a straight line
CHAPTER 3:
A
FORCE
is what causes a particle to accelerate.
This guy said an object will remain at rest or continue to move in a straight line at a constant speed unless it is acted upon by a resultant force.
Gravity acting on the laptop is causing a downward force.
but the laptop is not moving so the force must be canceled out somewhere
the table is exerting a Normal Reaction Force. This is denoted by
R
Newtons second law states ...
F = ma
acceleration (ms )
force (N)
mass (kg)
2
Some different forces you need to be aware of are . . .
FRICTION

a force which opposes the motion between 2 rough surfaces
TENSION

a force which acts on an object being pulled by a
string
.
THRUST
COMPRESSION

a force acting on an object that is being pushed.
in M1 we use a modelling assumption when using string. Will will model all string as inextensible. It cannot stretch.
or
RESISTANCE

a force which acts on an object as it moves through air or a liquid.
WEIGHT

the force due to gravity acting on an object. Using F = ma , WEIGHT can be calculated using . . . .
W = mg
When there is more than one force acting on an object you can RESOLVE the forces in a certain direction to find the resultant force in that direction.
You usually resolve in the direction of acceleration and
perpendicular
to the direction of acceleration.
In your answers you should use the letter R (resolving) and an arrow to indicate the direction in which you are resolving the forces
R( )
for example if you were resolving vertically up.
EXAMPLE
4 kg
Find the magnitudes of the unknown forces X and Y
X
Y
20 N
80 N
4g N
2 ms
2
Issac Newton 1643  1727
Component Forces
Lesson 2:
If you are being pushed by two people from 2 different angles you will only move in one direction. So these 2 people could be replaced by one single person pushing you in that direction.
When we resolve the forces acting on a object we resolve in the direction of motion and perpendicular to motion.
but what if a force acts at an angle that is in a different direction?
4 kg
X
20 N
80 N
4g N
well then we can split the force into two separate components.
4 kg
? N
2 ms
2
2 ms
2
4g N
5
N
4
N
3
N
we do this with our good friend
Trigonometry!
x
xcosØ
Ø
Example
4 kg
20 N
? ms
2
A block is pulled across a frictionless surface. The block does not lift off the surface. At what rate does it accelerate?
30
o
Component in the xdirection
= F
cos
Ø
= 20
cos
30
Using
F = ma
20
cos
30 = 4a
20
cos
30 = a
4
a = 4.33 ms (2dp)
Or vice versa
2
Friction
Lesson 2:
The Coefficient of Friction
5 kg
F
R
P
5g N
No friction jokes allowed. I wont let them slide.
Friction is a force that opposes motion between 2 rough surfaces. It occurs when the 2 surfaces move relative to one another (or there is a
tendency
for them to move).
There is a maximum or limiting value for friction. An object will only move once the force applied overcomes this value.
The limiting value of friction depends on 2 things:
the normal reaction
R
between the 2 surfaces in contact
the roughness of the 2 surfaces in contact
F = µR
max
where
µ is the coefficient of friction
R is the normal reaction force
Example
A block of mass 5kg lies at rest on a rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A horizontal force P is applied to the block. Find the magnitude of the frictional force and the acceleration of the block when P is a) 10N b) 19.6N c) 30N
R( ) R = 5g = 49N
F = µR = 0.4 x 49
max
= 19.6 N
a) When P = 10N the friction will only need to be 10N to prevent the block from sliding and the block will remain at rest in
equilibrium
b) When P = 19.6N the friction will be at its maximum to prevent the block from sliding and the block will remain at rest in
limiting
equilibrium
c) When P = 30N the friction will be at maximum and unable to prevent the block from sliding and the block will
accelerate
.
5 kg
19.6N
R
30N
5g N
R( ) F = ma
a ms
2
30  19.6 = 5a
10.4
= a
5
a = 2.08ms
= 2.1 ms (2s.f.)
why?
2
2
Full transcriptWhat is it?
before we start....
the day i learnt about
combustion!
meeting the big cheese!
buy milk
pick up kids
Mechanics is the branch of mathematics which deals with the action of forces on objects.
M1:
Kinematics of a particle moving in a straight line
Mechanics is a very complicated topic and in M1 we want to start with the basics. To do this we will simplify our problems by using a mathematical model. This involves making some modelling assumptions.
Ignore air resistance
there are more but we will meet these as we progress through the module!
with
Mr Scott
Mechanics
To pass this module you will need to be confident in your ability to manipulate algebra.
Can you rearrange this equation to make each letter the subject?
=
(b+2)
c
a
2
7
CHAPTER 1+2:
1
2
Constant force due to gravity (9.8m/s)
3
Many objects modelled as a particle.
We will learn quite a few formulae before the exams. Here are our first 2
v = u + at
and
s =
u + v
2
( )
t
Where . . . .
s
u
v
a
t
= displacement (distance)
= initial velocity
= final velocity
= acceleration
= time
when decelerating this value will be negative!
Dynamics of a particle moving in a straight line
CHAPTER 3:
A
FORCE
is what causes a particle to accelerate.
This guy said an object will remain at rest or continue to move in a straight line at a constant speed unless it is acted upon by a resultant force.
Gravity acting on the laptop is causing a downward force.
but the laptop is not moving so the force must be canceled out somewhere
the table is exerting a Normal Reaction Force. This is denoted by
R
Newtons second law states ...
F = ma
acceleration (ms )
force (N)
mass (kg)
2
Some different forces you need to be aware of are . . .
FRICTION

a force which opposes the motion between 2 rough surfaces
TENSION

a force which acts on an object being pulled by a
string
.
THRUST
COMPRESSION

a force acting on an object that is being pushed.
in M1 we use a modelling assumption when using string. Will will model all string as inextensible. It cannot stretch.
or
RESISTANCE

a force which acts on an object as it moves through air or a liquid.
WEIGHT

the force due to gravity acting on an object. Using F = ma , WEIGHT can be calculated using . . . .
W = mg
When there is more than one force acting on an object you can RESOLVE the forces in a certain direction to find the resultant force in that direction.
You usually resolve in the direction of acceleration and
perpendicular
to the direction of acceleration.
In your answers you should use the letter R (resolving) and an arrow to indicate the direction in which you are resolving the forces
R( )
for example if you were resolving vertically up.
EXAMPLE
4 kg
Find the magnitudes of the unknown forces X and Y
X
Y
20 N
80 N
4g N
2 ms
2
Issac Newton 1643  1727
Component Forces
Lesson 2:
If you are being pushed by two people from 2 different angles you will only move in one direction. So these 2 people could be replaced by one single person pushing you in that direction.
When we resolve the forces acting on a object we resolve in the direction of motion and perpendicular to motion.
but what if a force acts at an angle that is in a different direction?
4 kg
X
20 N
80 N
4g N
well then we can split the force into two separate components.
4 kg
? N
2 ms
2
2 ms
2
4g N
5
N
4
N
3
N
we do this with our good friend
Trigonometry!
x
xcosØ
Ø
Example
4 kg
20 N
? ms
2
A block is pulled across a frictionless surface. The block does not lift off the surface. At what rate does it accelerate?
30
o
Component in the xdirection
= F
cos
Ø
= 20
cos
30
Using
F = ma
20
cos
30 = 4a
20
cos
30 = a
4
a = 4.33 ms (2dp)
Or vice versa
2
Friction
Lesson 2:
The Coefficient of Friction
5 kg
F
R
P
5g N
No friction jokes allowed. I wont let them slide.
Friction is a force that opposes motion between 2 rough surfaces. It occurs when the 2 surfaces move relative to one another (or there is a
tendency
for them to move).
There is a maximum or limiting value for friction. An object will only move once the force applied overcomes this value.
The limiting value of friction depends on 2 things:
the normal reaction
R
between the 2 surfaces in contact
the roughness of the 2 surfaces in contact
F = µR
max
where
µ is the coefficient of friction
R is the normal reaction force
Example
A block of mass 5kg lies at rest on a rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A horizontal force P is applied to the block. Find the magnitude of the frictional force and the acceleration of the block when P is a) 10N b) 19.6N c) 30N
R( ) R = 5g = 49N
F = µR = 0.4 x 49
max
= 19.6 N
a) When P = 10N the friction will only need to be 10N to prevent the block from sliding and the block will remain at rest in
equilibrium
b) When P = 19.6N the friction will be at its maximum to prevent the block from sliding and the block will remain at rest in
limiting
equilibrium
c) When P = 30N the friction will be at maximum and unable to prevent the block from sliding and the block will
accelerate
.
5 kg
19.6N
R
30N
5g N
R( ) F = ma
a ms
2
30  19.6 = 5a
10.4
= a
5
a = 2.08ms
= 2.1 ms (2s.f.)
why?
2
2