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What are we learning today?
You will be a successful learner today if you
can:
1. Describe the effects of balanced forces.
2. State Newton’s first Law.
Success Criteria: Complete LO’s.
Isaac Newton
Isaac Newton was a great mathematician and
scientist. His discoveries made major advances
to our knowledge of the physical laws that
govern the Universe.
Isaac Newton
1643 - 1727
He invented a new branch of mathematics
called calculus, that enabled mathematicians
to solve many problems.
He researched into gravity, light, rotating bodies, planetary orbits,
and tidal motion. He had amazing powers of concentration,
sometimes studying and solving problems continuously for 19
hours a day.
In 1687, Newton published his most important discoveries in a
famous book called “Principia Mathematica”.
• Youtube dog on skate board
http://www.youtube.com/watch?
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Newton’s First Law
When a vehicle travels at a constant speed, the frictional forces
exactly balance the driving force, so the resultant force is zero.
Newton’s First Law
When a vehicle travels at a constant speed, the frictional forces
exactly balance the driving force, so the resultant force is zero.
Applying Newton’s first law
In everyday life, most things don’t move unless a force is
applied. Why is this?
What forces act on this car as it travels at a steady speed?
Air resistance and
friction acts against the
direction of motion.
Engine force from the car
pushes the car forward.
The car is travelling at a steady speed, so these forces
(air resistance + friction and engine force) must be equal.
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An object in space will continue at the same speed forever,
unless it is acted upon by the gravity of a celestial body.
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• Colorado Phet “Motion and Force”
• Colorado Phet “ Force”
Newton’s Laws of Motion
Newton stated three laws concerning force and motion.
Newton’s First Law
If the forces acting on an object are
balanced then the object will continue
travelling at constant speed in a straight
line or remain at rest.
Examples:
 If the resultant force acting on a stationary body is zero, the
body will remain stationary.
 If the resultant force acting on a moving body is zero, the body
will continue to move at the same speed and in the same
direction (i.e. with the same velocity).
Balanced forces
Force is a vector. The vector sum of the forces on an object
gives a net or resultant force.
Look at the forces on this
object:
To see the vector sum of the
forces, add the horizontal forces
together.
3N
3N
3N
3N
There is no ‘net’ or resultant
force on the mass: the forces
are balanced.
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Balanced forces
Force is a vector. The vector sum of all the forces on an
object gives a net or resultant force.
Look at the forces on this
object:
4N
To see the vector sum of the
forces, add the horizontal forces
together first then add the
vertical forces.
3N
4N
3N
4N
3N
4N
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3N
Again there is no ‘net’ or
resultant force on the mass:
the forces are balanced.
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Relative motion
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Newton’s first law: true or false?
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What are we learning today?
You will be a successful learner today if you
can:
1. Describe the effects of unbalanced forces.
2. State Newton’s second Law.
3. Analyse skydiving.
4. Calculate tension in haulage systems.
5. Calculate forces in rocket motion.
6. Analyse forces in lifts.
Success Criteria: Complete LO’s.
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There are 2 parts to Newton’s Second Law
Part 1:The acceleration of a body is
proportional to the resultant force and
in the same direction as the force.
Newton’s second law is represented by the equation:
F =
m x a
Unbalanced Force mass
acceleration
 If the resultant or unbalanced force acting on a stationary body
is not zero, the body will accelerate in the direction of the
resultant force.
Demonstrating Newton’s Second Law
Using F = ma
F = mxa
m = F/a
a = F/m
Unbalanced forces
The forces on this 1kg toy car are unbalanced:
3N
18 N
What is the resultant force?
18 N
3N
15 N
F = 18 - 3 = 15 N
θ = 180°
Force is a vector, resultant force is 15 N to the right
There is a resultant force of 15 N at a direction of 180° from
the horizontal.
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Unbalanced forces
The forces on this mass are unbalanced:
What is the resultant force? Add the
vectors together again, nose-to-tail.
8N
3N
F
θ
3N
4N
4N
3N
F = √32 + 42 = 5 N
4N
θ = tan-1(4/3) = 53°
There is a resultant force of 5 N at a direction of 53° from
the horizontal.
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Using F = ma
F = ma connects the resultant force to the mass of the object and
its acceleration:
Force = mass x acceleration
N
kg
m/s2
Force is measured in Newtons (N). One Newton is the force
required to accelerate a mass of 1kg by 1m/s2.
The formula, F = ma can be rearranged using a formula triangle...
Using F = ma
If you know any two of these values, you can use the
formula triangle to find the remaining value...
m = F
a = F
F = mxa
a
m
Using F = ma
Example 1 (General / NAT 4 level)
A lorry with a mass of 3500kg decelerates steadily at 3m/s2.
What is the braking force applied?
F = m x a
F = 3500 x 3
F = 10500N
The braking force applied is 10500N
Using F = ma
Example 2
A force of 350N acts on an object of mass 25kg.
If the object accelerates at 10 ms-2. What is the force of friction?
F = mx a
F = 25 x 10
F = 250 N
But this is the unbalanced force
100 N
350 N
So the force of friction is 100N
Summarising Newton’s second law
If there is a resultant force on an object, it accelerates in the
direction of that force. Its acceleration is directly proportional
to the force, and inversely proportional to the object’s mass.
This can be stated in a vector equation:
F = ma
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Newton’s second law example 1 Haulage System
A car pulls a caravan with an engine force of 4000N.
1. Find the acceleration of the car and caravan.
4000N
3000 kg
5000 kg
The engine must accelerate both the car and the caravan:
a = F/m = 4000 / (3000 kg + 5000 kg) = 0.5 ms-2
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Newton’s second law example 1 Haulage System
Both the car and caravan are accelerating at 0.5 ms-2.
2. Find the tension (force) in the tow-bar.
?
4000 N
3000 kg
5000 kg
To answer this type of question consider putting your
hand over the car and ask the question what force
would cause the caravan to accelerate at 0.5 ms-2.?
F (or tension)= ma = 5000 kg × 0.5 ms-2 = 2500 N
This is the tension in the tow-bar.
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Newton’s second law example 2 Skydiving
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Newton’s second law example 3 Rockets and Missiles
A rocket of mass 2 x 106 kg is fired
vertically into the air. The rocket
motors provide a thrust of 50 x 106 N,
and there is a drag force of 2 x 106 N.
Find the acceleration of the rocket.
F1
a = F/m = unbalanced force / mass
unbalanced force = F1 – F2
F2
F1 = 50 x 106 N
F2 = 2 x 106 N + mg
F2 = 2 x 106 N + 20 x 106 N = 22 x 106 N
a = (50 x 106 – 22 x 106 )
2 x 106
Acc =14 ms-2
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Why would the acceleration of the rocket change?
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Newton’s second law example 4 Lift experiment
Stand in a lift with a mass on
a spring balance and record
the reading.
Note if the reading increases
or decreases as the lift:
accelerates upwards
accelerates downwards
decelerates upwards
decelerates downwards
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Newton’s second law example 4 Lifts
What is the weight
of the woman in
this lift which is
staionary?
W = mg
W = 60 x 10 (9.8)
W = 600 N
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Newton’s second law example 4 Lifts
What is the weight
of the woman in
this lift which is
moving at a
constant speed up
or down?
W = mg
W = 60 x 10 (9.8)
W = 600 N
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Newton’s second law example 4 Lifts
The lift can accelerate
and decelerate upwards.
It can also accelerate
and decelerate
downwards.
We now require to
calculate the apparent
weight of the women in
each of these 4
situations.
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Newton’s second law example 4 Lifts
What is the apparent weight of the woman when the lift
accelerates upwards at 1ms-2? (speeding up on the way up)
To answer this question we
need to consider the 2 forces:
weight and apparent weight.
apparent As the lift accelerates it is the
apparent weight which the lady
weight
feels and is displayed on the scales.
Her weight, the force acting
downwards, is always W = mg
W = 60 x 10 (9.8)
W = 600 N
weight
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Newton’s second law example 4 Lifts
If the lift accelerates upwards then there must be a greater
upward force acting on the lady than downward force?
So…the unbalanced force is?
F1– F2
F1– F2 = ma
apparent
weight F1 Apparent weight – weight = ma
Apparent weight – 600 = 60 x 1
Apparent weight = 660 N
She feels heavier!
weight F2
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Newton’s second law example 4 Lifts
What is the apparent weight of the woman when the lift
deccelerates upwards at 1ms-2? (slowing down on the way up)
To answer this question we
need to consider the 2 forces:
weight and apparent weight.
apparent As the lift decelerates upwards it is
the apparent weight which the lady
weight
feels and is displayed on the scales.
Her weight, the force acting
downwards, is always W = mg
W = 60 x 10 (9.8)
W = 600 N
weight
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Newton’s second law example 4 Lifts
If the lift decelerates upwards then there must be a greater
downward force acting on the lady than upward force?
So…the unbalanced force is?
F2– F1
F2– F1 = ma
apparent
weight F1 Weight – apparent weight = ma
600 – apparent weight = 60 x 1
Apparent weight = 540 N
She feels lighter!
weight F2
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Newton’s second law example 4 Lifts
What is the apparent weight of the woman when the lift
accelerates downwards at 1ms-2? (speeding up on the way down)
To answer this question we
need to consider the 2 forces:
weight and apparent weight.
As the lift accelerates downwards it
apparent is the apparent weight which the
weight
lady feels and is displayed on the
scales.
Her weight, the force acting
downwards, is always W = mg
W = 60 x 10 (9.8)
W = 600 N
weight
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Newton’s second law example 4 Lifts
If the lift accelerates downwards then there must be a greater
downward force acting on the lady than upward force?
So…the unbalanced force is?
F2– F1
F2– F1 = ma
apparent
weight F1 Weight – apparent weight = ma
600 – apparent weight = 60 x 1
Apparent weight = 540 N
She feels lighter!
weight F2
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Newton’s second law example 4 Lifts
What is the apparent weight of the woman when the lift
decelerates downwards at 1ms-2? (slowing down on the way down)
To answer this question we
need to consider the 2 forces:
weight and apparent weight.
As the lift decelerates downwards it
apparent is the apparent weight which the
weight
lady feels and is displayed on the
scales.
Her weight, the force acting
downwards, is always W = mg
W = 60 x 10 (9.8)
W = 600 N
weight
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Newton’s second law example 4 Lifts
If the lift decelerates downwards then there must be a greater
upward force acting on the lady than downward force?
So…the unbalanced force is?
F1– F2
F1– F2 = ma
apparent
weight F1 Apparent weight – weight = ma
Apparent weight - 600 = 60 x 1
Apparent weight = 660 N
She feels heavier!
weight F2
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Newton’s second law example 5 Components
How would you calculate the acceleration of the vehicle shown below? What
information do you need?
500 grams
F1– F2
acceleration = force
mass
50 grams
Weight = mg = 50 x 10-3 x 10 (9.8) = 0.5 N
weight
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acceleration = force = 0.5 = 1 ms-2
mass 0.5
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Newton’s second law example 5 Components
When this experiment was performed these results were
recorded on the QED: 0.69 ms-2, 0.68 ms-2, 0.72 ms-2,
0.71 ms-2 and 0.71 ms-2. Can you think why?
Before we consider why lets calculate the average acceleration
and the random uncertainty.
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Newton’s second law example 5 Components
Average acceleration = 0.69 + 0.68 + 0.72 + 0.71 + 0.71
Average acceleration = 0.70 ms-2
Random uncertainty = largest value – smallest value
number of values
Random uncertainty = 0.72– 0.68 = 0.04 = 0.008 = 0.01
5
5
Random uncertainty = + 0.01
-
Average acceleration = 0.70 ms-2
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+- 0.01 ms-2
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Newton’s second law example 5 Components
Now go ahead and try this experiment. Calculate the theoretical value then
obtain the experimental value with your mass and vehicles mass.
? grams
F1– F2
acceleration = force
mass
? grams
Theoretical value
Weight = mg = ? x 10 (9.8) = ? N
weight
acceleration = force = ? = ? ms-2
mass ?
The experimental value was ? ms-2.
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Newton’s second law example 5 Components
The value obtained from the QED is less than the calculated
value. Where have we gone wrong?
F1– F2
This equation is OK
acceleration = force
mass
This mass is OK
But the force we use is too big.
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Newton’s second law example 5 Components
To understand why the experimental value is less than the
calculated value we must now consider components of a force.
A force can be split horizontal and vertical components.
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Newton’s second law example 5 Components
We must use the force which acts in the same direction to the
one in which the vehicle accelerates.
We must therefore find the horizontal component of force.
Cos 45 = horizontal component of force
0.5
Horizontal component of force = 0.5 x cos45 = 0.35 N
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Newton’s second law example 5 Components
horizontal component of force:
Cos 45 = horizontal component of force
0.5
Horizontal component of force = 0.5 x cos45 = 0.35 N
acceleration = force = 0.35 = 0.7 ms-2
mass 0.5
The value obtained experimentally
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Newton’s second law example 6 Components
A man pulls a garden roller of mass 100 kg with a force of 200 N
acting at 300 to the horizontal.
If there is a frictional force of 100 N between the roller and the
ground, what is the acceleration of the roller?
100 N
acceleration = unbalanced force = 200 - 100 = 1 ms-2
mass
100
Is this a good answer?
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Newton’s second law example 6 Components
No, it is not a good answer.
When calculating the acceleration you must use only the
bit of the force which acts in the direction the roller moves.
You must calculate the……..
100 N
Horizontal component of force
Cos 30 = horizontal component of force
200
Horizontal component of force = 200 x cos30 = 173 N
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Newton’s second law example 6 Components
100 N
friction
173 N
acceleration = unbalanced force = 173 - 100 = 0.73 ms-2
mass
100
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History lesson: Galileo (1564 – 1642)
Galileo was a hugely influential Italian astronomer, physicist and
philosopher.
Galileo Galileo was born on 15 February 1564 near Pisa, the son of a musician. He began to
study medicine at the University of Pisa but changed to philosophy and mathematics. In 1589,
he became professor of mathematics at Pisa. In 1592, he moved to become mathematics
professor at the University of Padua, a position he held until 1610. During this time he worked
on a variety of experiments, including the speed at which different objects fall, mechanics and
pendulums.
In 1609, Galileo heard about the invention of the telescope in Holland. Without having seen an
example, he constructed a superior version and made many astronomical discoveries. These
included mountains and valleys on the surface of the moon, sunspots, the four largest moons of
the planet Jupiter and the phases of the planet Venus. His work on astronomy made him
famous and he was appointed court mathematician in Florence.
In 1614, Galileo was accused of heresy for his support of the Copernican theory that the sun
was at the centre of the solar system. This was revolutionary at a time when most people
believed the Earth was in this central position. In 1616, he was forbidden by the church from
teaching or advocating these theories.
In 1632, he was again condemned for heresy after his book 'Dialogue Concerning the Two Chief
World Systems' was published. This set out the arguments for and against the Copernican
theory in the form of a discussion between two men. Galileo was summoned to appear before
the Inquisition in Rome. He was convicted and sentenced to life imprisonment, later reduced to
permanent house arrest at his villa in Arcetri, south of Florence. He was also forced to publicly
withdraw his support for Copernican theory.
Although he was now going blind he continued to write. In 1638, his 'Discourses Concerning
Two New Sciences' was published with Galileo's ideas on the laws of motion and the principles
of mechanics. Galileo died in Arcetri on 8 January 1642.
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Galileo's Inclined Plane Experiment
Galileo's use of the inclined plane to study the motion of objects is one of
his most important contributions to science. As this video segment from
NOVA illustrates, the inclined plane allowed Galileo to accurately measure
the effect of gravity on falling objects and develop a universal law
describing this effect. Galileo hypothesized that rolling balls down an
inclined plane would cause them to accelerate uniformly.
Dropping things vertically happened too quickly so he studied the motion
on inclined planes. The speeds were very slow and this allowed him to
analyse the motion
Based on his book Two new Sciences Galileo. He measured the speed of a
ball using water clocks and a swinging pendulum. He found that in the first
unit of time it travelled 1 unit of time, in the second unit of time it travelled 3,
in the next it travelled 5 and then 7
Galileo found that the velocity = g t and distance travelled was always
proportional to 1/2 gt2.
He had found laws which apply to all falling bodies and this was a key used
by Newton who established mechanics and gravity as a science later in the
century.
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Galileo
In physics Galileo is remembered for discovering the
laws of failing bodies and motions of projectiles; in
astronomy he is known for pioneering telescopic
observation and, through this, discovering sunspots,
the irregular surface of the Moon, the satellites of
Jupiter, and the phases of Venus. Most of all, he is
remembered for championing Copernicus' Suncentred universe and prompting the split that freed
scientific knowledge from the restrictions of spiritual
belief. Imprisoned in old-age for repeatedly publishing
his views, his name has become immortalised as an
emblem of resistance against the oppression of Truth.
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Simple machine
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Roads on hills
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Wheel chair access
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Inclined planes
An inclined plane, commonly referred to as a ramp, is an
even surface that is tilted at an angle. It helps reduce the
force necessary to move an object by increasing the
distance it must be moved. Picture a vertical wall two
metres (m) tall. You would have to apply a lot of force to lift
a 10 kilogram (kg) object over the wall. Now picture a 5 m
ramp leading up to the top of the wall. It would be far easier
to move the 10 kg object up the ramp than it would be to lift
the object straight up. Ramps are used this way in many
applications. They allow people with physical disabilities to
move up to another floor in a building, for example, and
they help to move objects into a truck. Some common
applications of the use of the inclined plane are wheel chair
access ramps, the ramps used in multilevel parking garages
and road building: roads tend to wind up mountains instead
of going straight up.
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Newton’s second law example 7 Inclined planes
It is easier to climb stairs to get to a second floor than to
climb straight up a rope. It is easier to walk up a long,
gentle hill than to climb up a short, steep hill. Here we
examine
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why
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Forces on Inclined Planes
How difficult is it to keep this car stationary?
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Forces on Inclined Planes
How difficult is it to keep this car stationary?
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Forces on Inclined Planes
How difficult is it to keep this car stationary?
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Forces on Inclined Planes
Comment on the weight
(force) acting down the
incline ……….
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Forces on Inclined planes
Comment on the force acting into the surface ………
As 0, the angle of the slope increases; the force acting into
the slope decreases.
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Force on an inclined plane
Perpendicular
component
Weight
(mg)
Parallel
component
When a mass sits on a slope we need to consider the
components of its weight when studying its motion and to
do this we use trigonometry
Sin
Cos
= parallel component
mg
= perpendicular component
mg
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parallel component = mg sin
perpendicular component = mgcos
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Forces on Inclined planes
mgcos90
mgcos0
0
mgcos45
As 0, the angle of the slope increases; the bit of weight
(component of weight) acting into the slope (mgcos0)
decreases.
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Forces on inclined planes
mgsin0
mgsin0
mgsin0
0
mgsin0
mgsin0
mg
sin0
As the angle of the slope decreases, sin0 decreases
and the component acting down the slope
decreases.
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mgsin0 & mgcos0
• In this course we will concentrate only on
the component acting down the slope:
mgsin0
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Force on an inclined plane
Components parallel and perpendicular to slope
Sin
= parallel component
mg
Cos
= perpendicular component
mg
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parallel component = mg sin
perpendicular component = mg cos
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Worked examples on inclined planes
A mass of 10kg is placed on a plane inclined at 300 to the
horizontal.
If there is a constant frictional force between the mass and
the plane of 10N, what is the acceleration of the mass when
released? (4ms- )
10 N
2
mgsin0
10 Kg
300
acceleration = unbalanced force = mgsin30 - 10 = 4 ms-2
mass
10
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Worked examples on inclined planes
A mass of 10kg is launched up a plane inclined at 300 to the
horizontal with a force of 300N.
Calculate the frictional force between the mass and the
plane when the acceleration of the mass is 10ms-2. (150N)
10 Kg
300
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Worked examples on inclined planes
A mass of 10kg is launched up a plane inclined at 300 to the
horizontal with a force of 300N.
Calculate the frictional force between the mass and the
plane when the acceleration of the mass is 10ms-2. (150N)
Friction (now acts
Component of weight acting down slope
downwards)
10 Kg
mgsin0
300N
300
Force acting up the slope
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Worked examples on inclined planes
Friction (now acts
Component of weight acting down slope
downwards)
10 Kg
mgsin0
300N
300
Force acting up the slope
acceleration = unbalanced force = 300 –(mgsin30 +friction)
m
10
10 ms-2 = unbalanced force = 300 –(50 +friction)
10
10
10 = 300 –(50 +friction)
10
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friction = 150N
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Action and reaction
A football rests on flat ground. There are four key forces
acting between the ball and the Earth. What are they?
 The ball presses down
on the ground.
 The ground reacts to
the ball.
 Earth’s gravity acts on
the ball.
 The ball’s gravity acts
on the Earth.
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Newton’s third law
For every action there is an equal and opposite reaction.
Forces always come in pairs.
List all the horizontal forces acting here, between the car,
the caravan and the road:




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The car pushes the road backwards.
The road pushes the car forwards.
The car pulls the caravan forwards.
The caravan pulls the car backwards.
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Action and reaction
“If every action has an equal and opposite reaction,
how do unbalanced forces ever occur?”
Newton’s third law applies only to forces between objects.
Pairs of action and reaction forces are of the same kind.
This diagram shows a pair
of frictional forces between
a box and the ground:
friction of the box on the
ground, and friction of the
ground on the box.
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Balanced forces
When considering the forces that cause acceleration, they
must all act on the same object. These forces can be
considered individually and can be of different kinds.
For instance the pulling
force on this box is not
equal to the frictional
force acting against it,
so the box will start to
move to the right.
Free body diagrams only include the forces that act on one
object. This makes it easy to work out whether the forces on it
are balanced or not, and whether the object will accelerate.
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Balanced forces: true or false?
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