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Using Physics to Make things
Work
P3 Physics
Mr D Powell
Connection
•
•
•
Connect your learning to the
content of the lesson
Share the process by which the
learning will actually take place
Explore the outcomes of the
learning, emphasising why this will
be beneficial for the learner
Demonstration
• Use formative feedback – Assessment for
Learning
• Vary the groupings within the classroom
for the purpose of learning – individual;
pair; group/team; friendship; teacher
selected; single sex; mixed sex
• Offer different ways for the students to
demonstrate their understanding
• Allow the students to “show off” their
learning
Activation
Consolidation
• Construct problem-solving
challenges for the students
• Use a multi-sensory approach – VAK
• Promote a language of learning to
enable the students to talk about
their progress or obstacles to it
• Learning as an active process, so the
students aren’t passive receptors
• Structure active reflection on the lesson
content and the process of learning
• Seek transfer between “subjects”
• Review the learning from this lesson and
preview the learning for the next
• Promote ways in which the students will
remember
• A “news broadcast” approach to learning
Mr Powell 2012
Index
P3 2.1 Moments p226 / P2.3 Moments in Balance p230/
P3 2.3 Stability p232
a) Know that the turning effect of a force is called the moment.
b) Know that the size of the moment is given by the equation: M = F x d
M is the moment of the force in newton-metres, Nm
F is the force in newtons, N
d is the perpendicular distance from the line of action of the force to the pivot in
metres, m
c) Know and understand that if an object is not turning, the total clockwise moment
must be exactly balanced by the total anticlockwise moment about any pivot.
d) HT only: Be able to calculation of the size of a force, or its distance from pivot, acting
on an object that is balanced.
e) Appreciate the idea of simple levers limited to levers as force multipliers.
f) HT only: Know and understand that if the line of action of the weight of an object lies
outside the base of the object there will be a resultant moment and the body will tend to
topple. Your knowledge of applications should include vehicles and simple balancing
toys.
Mr Powell 2012
Index
Moments
The turning effect of a force is called a moment.
A moment is NOT a period of time. Nor is it the same as momentum.
Moment is the product (two numbers multiplied together) between
the force and the perpendicular distance from the force and the pivot. To show
what this means, look at the picture:
moment (newton metre, Nm) = force (newton, N)× perpendicular distance (metre, m)
 = Fd or moment = Fd
(gamma = )
Mr Powell 2012
Index
Example...
Worked Example
A spanner 0.30 m long has a force of 20 N applied
to it. What is the moment?
Answer
Moment = F × d = 20 N × 0.30 m = 6.0 Nm
Question...
1) A wheel nut is tightened to a
moment of 100 Nm. A motorist has
to undo the nut with a wheel
wrench which is 0.40 m long. What
force must he apply?
F = /d = 100 Nm ÷ 0.40 m = 250 N
Increase the length of the wheel wrench.
2) How can the force applied be
reduced?
Mr Powell 2012
Index
Balancing Moments...
• Moments have two possible directions, clockwise or anti-clockwise. This
means that the object will turn in the direction of the bigger moment.
•
If the clockwise moment is equal to the clockwise moment, then the object
will stay where it is.
This leads to an important rule in Physics, the Principle of Moments:
If the clockwise moment = anticlockwise moment, the system is in equilibrium
This means that the system is balanced:
Mr Powell 2012
Index
Questions.....
1) Show
mathematically that
the first picture is a
balanced situation.
2) If the second seesaw
is balanced. What is
the force F?
Clockwise moment = F × 2d = 2Fd
Anticlockwise moment = 2F × d = 2Fd
Since the anticlockwise moment =
clockwise moment, seesaw is balanced.
Clockwise moment = 40 N × 1.8 m = 72
Nm
Since the seesaw is balanced,
anticlockwise moment = clockwise
moment.
Anticlockwise moment = 72 Nm = F
×0.8 m
F = 72 Nm ÷ 0.8 m = 90 N
Mr Powell 2012
Index
Moments and Levers
One of the simplest machines is a lever. Here are two
forces:
•
•
the effort FE, which is the force we apply to the
lever;
the load FL, which is the force that acts on the
object we want to move.
The distance from the effort to the pivot is d, and the
distance from the pivot to the load is x.
When we apply the effort, there is a clockwise moment What this tells us is that the
effort is smaller than the
about the pivot: M = FEd
load. The ratio d/x is
sometimes called the
At the same time there is an opposing anticlockwise
mechanical advantage. As
moment which is equal in value: M = FLx
the load force is much
bigger than the effort force,
Hence... Fed = FLx
the lever is called a force
multiplier.
OR..... d/x = FL/Fe
Index
Mr Powell 2012
Worked Example...
Worked example
An force of 120 N is applied to a crowbar that is 2
metres long. The distance from the effort to the pivot
is 1.8 m, while the distance from the pivot to the load
is 0.2 m
1.
What is the load force?
2.
What is the mechanical advantage?
Answer
Use:
FL = d = 1.80 = 9.0
FE x 0.20
FL = 9.0 × 120 = 1080 N
The mechanical advantage is 9.0, which means that
the force is multiplied by 9.0.
Mr Powell 2012
Index
Centre of Mass and Stability
A stable object does not tip over. Objects that have a
high stability do not tip over easily. Their centre of
mass is low down, near the base. This candlestick has a
low centre of mass so that it does not tip over so easily.
1. The candlestick is in stable equilibrium. If you push
the top, it will drop back to where it was.
2. Now suppose we put it upside down. This time the
centre of mass is high up. It is in unstable
equilibrium and if you pushed the candlestick, it
would tip over easily.
3. Now we put it on its side:
4.
If you push it, it will roll, but will not tip over. It is
in neutral equilibrium.
Mr Powell 2012
Index
Centre of Mass and Stability
Let us look at how we can explain stability in vehicles. This
bus has a low centre of mass and a wide track (distance
between the wheels)
You can see that there is a line of action of the weight that
acts vertically downwards from the centre of mass.
If the bus tilts, the line of action will pull the bus upright.
Although they are tall, double-decker buses are very
stable. They test buses by putting lots of sandbags on the
seats upstairs (with nothing downstairs) and tilt them over
on a tilting platform. The centre of mass is low enough to
ensure that they are tilted to more than 60o off the vertical
before they tip over.
Lorries have a higher centre of mass on their trailers, due
to the load. If you live in the country and get stuck behind
a hay-lorry, you may see it swaying alarmingly. This kind of
accident tends to happen with lorries when they drive
through strong cross-winds on exposed roads.
Mr Powell 2012
Index
Tilting or Toppling?
You can see the line of action from
the centre of mass. It is in the middle of
the track (distance between the wheels)
of the bus.
Q) Now suppose the bus goes
even faster round a sharp bend.
Which way is the moment
now? What will happen to the
bus?
Now suppose the bus goes fast round a
corner and tilts over. Which way is the
moment now? What will happen to the
bus?
Mr Powell 2012
Index
Plenary
Higher Tier...
Look at the picture above. – sketch it out in outline then draw on....
(a) Where is the centre of mass in the bottle?
(b) Where is the pivot?
(c) Where does the line of action from the centre of mass of the bottle act?
(d) Use the principle of moments to explain how this system balances.
Mr Powell 2012
Index
Plenary
Higher Tier...
Answers....
(a) Where is the centre of mass in the bottle?
(b) Where is the pivot?
(c) Where does the line of action from the centre of mass of the bottle act?
(d) Use the principle of moments to explain how this system balances.
Mr Powell 2012
Index
P3 2.1 Moments p226 / P2.3 Moments in Balance p230/ P3 2.3 Stability p232
a-c)
I already know....
a) Know that the turning effect of a force is called the moment.
b) Know that the size of the moment is given by the equation: M = F x d
M is the moment of the force in newton-metres, Nm
F is the force in newtons, N
d is the perpendicular distance from the line of action of the force to the
pivot in metres, m
c) Know and understand that if an object is not turning, the total clockwise
moment must be exactly balanced by the total anticlockwise moment about any
pivot.
(Formulae - Basic)
d)
d) HT only: Be able to calculation of the size of a force, or its distance from pivot,
acting on an object that is balanced.
e) Appreciate the idea of simple levers limited to levers as force multipliers.
f) HT only: Know and understand that if the line of action of the weight of an
object lies outside the base of the object there will be a resultant moment and
the body will tend to topple. Your knowledge of applications should include
vehicles and simple balancing toys.
(Harder – example calculation)
e)
f)
(Diagram - Harder)
(Maths)
P3 2.2 Centre of Mass p 228 / P2.7 The Pendulum
a) Know and understand that the centre of mass of an object is that point at which the
mass of the object may be thought to be concentrated. You will be expected to be able
to describe how to find the centre of mass of a thin, irregular sheet of a material.
b) Know and understand that if freely suspended, an object will come to rest with its
centre of mass directly below the point of suspension.
c) Know the centre of mass of a symmetrical object is along the axis of symmetry.
d) Know that for a simple pendulum: and be able to use the formula: T = 1/ f
T is periodic time in seconds, s
f is frequency in hertz, Hz
Know that applications of the pendulum should include simple fairground and
playground rides. The equation T = 2√(l/g) is not required.
e) Know that the time period depends on the length of a pendulum
Mr Powell 2012
Index
P3 2.2 Centre of Mass p 228 / P2.7 The Pendulum
a) & b)
I already know....
a) Know and understand that the centre of mass of an object is that point
at which the mass of the object may be thought to be concentrated. You
will be expected to be able to describe how to find the centre of mass of a
thin, irregular sheet of a material.
b) Know and understand that if freely suspended, an object will come to
rest with its centre of mass directly below the point of suspension.
(Diagram/practical/video)
d)
c) Know the centre of mass of a symmetrical object is along the axis of
symmetry.
d) Know that for a simple pendulum: and be able to use the formula: T =
1/ f
T is periodic time in seconds, s
f is frequency in hertz, Hz
Know that applications of the pendulum should include simple fairground
and playground rides. The equation T = 2√(l/g) is not required.
(Maths/ practical/ examples)
e) Know that the time period depends on the length of a pendulum
c)
e)
(practical - harder)
(basic)
Centre of Mass
In Physics we find it a lot easier to think of
objects as point masses. All objects have
a point at which they balance, called
the centre of mass. We think of all the
mass as being concentrated at the centre
of mass.
The centre of mass is the point at which
the weight of the object is said to act. The
green arrow is the line of action of the
force from the centre of mass. Force due
to gravity on a mass is the weight.
Note that it is called the centre of
mass not centre of weight. This is because
if the object were in space, it would still
have a centre of mass, even though it
were weightless. Sometimes the centre of
mass is called the centre of gravity.
Q) What happens if the line of action
of the force is to the right of the
pivot?
A) It would tip over to the right, as
there would be a clockwise moment.
Mr Powell 2012
Index
Centre of Mass
One important principle in Physics is that any object, however
sophisticated, large, or complex can be treated as a point mass. When
we work out the effect of forces, we show the forces all acting at a single
point. This point is called the centre of mass.
We treat objects as point masses referring to a single point called
the centre of mass.
In regular objects like a cube or a sphere, the centre of mass is in the
middle. In some objects the centre of mass is outside the object.
Mr Powell 2012
Index
Finding the centre of mass for irregular objects (video)
We can find the centre of mass of an
irregular object quite easily. If we let it
hang freely, the centre of mass is directly
below where we hang it from. In old text
books, an object like this is called
an irregular lamina. (The word lamina is a
Latin word for leaf.)
We draw a line vertically downwards.
If we then hang the object from a couple
of other points and draw the lines that go
vertically downwards, the centre of mass
is where the lines meet.
When the object is hanging freely, the
centre of mass is vertically below the
hanging point. The vertical arrow is called
the line of action of the weight.
Mr Powell 2012
Index
Centre of Mass
For any regular object, e.g. box, a cylinder,
etc., the centre of mass is in the very
centre of the object.
If we allow an object to dangle freely from
a single point, we find that the centre of
mass is on a line vertically underneath the
point from which the object is hung.
We can trace the line by hanging a plumb
line (heavy object on a string) which
always hangs vertically.
Now if we turn the rectangle so that it
hangs off one of the holes in the corner,
we can use the plumb line to trace a
second line like this:
We could do the same hanging the
rectangle from the opposite corner.
Mr Powell 2012
Index
Pendulums...
A pendulum is a simple system consisting of
a mass (often called a bob) that hangs freely below a
fixed point on a thread.
Normally the mass will hang vertically below the fixed
point.
That is the idea you used when working out the centre
of mass of an irregular object.
It doesn't matter what shape the bob is.
We treat it as a single point mass and the thread is light
and does not stretch. If we push the bob to one side,
we find that the pendulum will swing from side to side.
It oscillates. Each oscillation (complete to-and-fro
movement) has a time period.
Mr Powell 2012
Index
The Maths....
Period depends on the length of the thread.
It does NOT depend on the mass, or the material, of
the bob.
As long as the angle of swing is small, the angle does
not affect the periodicity of the swing either.
However, if we swing through a large angle, the
period is different (by a small amount) from the
period of a small swing.
The periodicity is important, as it enables pendulum
clocks to measure time with good accuracy.
1
T
f
l
T  2
g
NB: for GCSE you will need
the top formula but
the bottom one will
help you work out the
explanation using
grown up maths!
Many clocks in churches and other public buildings
have their time-keeping governed by the swing of a
pendulum.
Mr Powell 2012
Index
P3 2.5 Hydraulics p234
a) Know and understand that liquids are virtually incompressible, and the
pressure in a liquid is transmitted equally in all directions. You should
understand that this means that a force exerted at one point on a liquid will
be transmitted to other points in the liquid.
b) Know and understand that the use of different cross-sectional areas on the
effort and load side of a hydraulic system enables the system to be used as a
force multiplier.
c) Know that the pressure in different parts of a hydraulic system is given by:
P=F/A
P is the pressure in pascals, Pa
F is the force in newtons, N
A is the cross-sectional area in metres squared, m2
Mr Powell 2012
Index
P3 2.5 Hydraulics p234
I already know....
a)
a)
Know and understand that liquids are virtually incompressible,
and the pressure in a liquid is transmitted equally in all
directions. You should understand that this means that a force
exerted at one point on a liquid will be transmitted to other
points in the liquid.
b) Know and understand that the use of different cross-sectional
areas on the effort and load side of a hydraulic system enables the
system to be used as a force multiplier.
(Particle Model + Hydraulic System Diagram)
b)
c) Know that the pressure in different parts of a hydraulic system is
given by:
P=F/A
P is the pressure in pascals, Pa
F is the force in newtons, N
A is the cross-sectional area in metres squared, m2
(Examples of Areas?)
e)
(Diagram and Maths Examples - Harder)
Example....
The force applied to the pump is 5 N
and the area is 10 cm2. The ram that
acts as the motor has an area of 250
cm2. Calculate:
(a) The pressure in Pascals in the
system;
(b) The force from the ram;
(c) The factor by which the force is
multiplied.
Answers...
(a) Pressure = force ÷ area = 5 N ÷ (10 × 10-4) = 5000 Pa
(b) Force = pressure × area = 5000 × (250 × 10-4) = 125 N
(c) Factor by which force is multiplied = 125 ÷ 5 = 25 times.
Notice that the area of the ram is 25 times bigger than the area of the pump.
Mr Powell 2012
Index
Bottle Jack....
Mr Powell 2012
Index
Trolley Jack....
Mr Powell 2012
Index
P3 2.6 Circular Motion p236
a) Know and understand that when an object moves in a circle it continuously
accelerates towards the centre of the circle.
Know that this acceleration changes the direction of motion of the body, not its
speed.
b) Know and understand that the resultant force causing this acceleration is called the
centripetal force and is always directed towards the centre of the circle.
You should be able to identify which force(s) provide(s) the centripetal force in a
given situation.
c) Know and understand that the centripetal force needed to make an object perform
circular motion increases as:
•
the mass of the object increases,
•
the speed of the object increases,
•
the radius of the circle decreases,
•
The equation F = mv2 / r is not required (but can be used!)
Mr Powell 2012
Index
P3 2.6 Circular Motion p236
I already know....
a)
a) Know and understand that when an object moves in a circle it
continuously accelerates towards the centre of the circle.
Know that this acceleration changes the direction of motion of
the body, not its speed.
b) Know and understand that the resultant force causing this acceleration
is called the centripetal force and is always directed towards the centre of
the circle. You should be able to identify which force(s) provide(s) the
centripetal force in a given situation.
(Link ideas of velocity v speed)
b)
c) Know and understand that the centripetal force needed to make an
object perform circular motion increases as:
•
the mass of the object increases,
•
the speed of the object increases,
•
the radius of the circle decreases,
•
The equation F = mv2 / r is not required (but can be used!)
(Link real world situations)
c)
(Use the formulae to explain factors - Harder)
Circular Motion...
We have looked at motion in a straight line. In
this topic we are going to explore some of the
basic ideas of circular motion.
Think about a toy aeroplane tied to a pole:
The aeroplane is flying at a constant
speed. However: Its velocity is changing all the
time.
If there is a change in velocity the Laws of
Physics state that there must also be
an acceleration. And if there is acceleration,
there must a force causing the acceleration.
The force is called centripetal force and its
direction is towards the centre of the circle.
Mr Powell 2012
Index
Centripetal Forces...
The centripetal force is always towards
the centre of the circle.
The force always acts at 90 degrees to the
direction of the movement.
If the string were to snap, the aeroplane
would immediately fly in a straight
line because there would be no sideways
force to keep it in a circular path.
It would fly off at a tangent as shown in
the picture below:
Centripetal force is increased when:
•
•
•
the linear speed is increased;
the mass is increased;
the radius is decreased
mv
F
r
2
Mr Powell 2012
Index
Centrifugal Force
Centrifugal force is a very common bear
trap:
There is no such thing as centrifugal force.
This runs counter to what you might think,
because you feel a force pushing you away
from the centre of the circle.
In reality you are trying to fly off at a
tangent. If you do come unstuck, that is
exactly what will happen.
Any reference to centrifugal force is bad
physics and you will get no marks for it in
the exam.
So don't use it!
Mr Powell 2012
Index
.....
a) & b)
I already know....
(.......)
d)
a) Know .....
(.......)
c)
e)
(.......)
(.......)