1. Using Physics to Make things Work
P3 Physics
KS4 Triple Science
Mr D Powell

2. 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
Consolidation
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
Activation
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

4. 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.

5.  = Fd or moment = Fd (gamma = )
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 = )

6. Example... Question... 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...
F = /d = 100 Nm ÷ 0.40 m = 250 N
Increase the length of the wheel wrench.
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?
2) How can the force applied be reduced?

7. 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:

8. Show mathematically that the first picture is a balanced situation.
Questions.....
Show mathematically that the first picture is a balanced situation.
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

9. 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 about the pivot: M = FEd
At the same time there is an opposing anticlockwise moment which is equal in value: M = FLx
Hence... Fed = FLx
OR d/x = FL/Fe
What this tells us is that the effort is smaller than the load.  The ratio d/x is sometimes called the mechanical advantage.  As the load force is much bigger than the effort force, the lever is called a force multiplier.

10. 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
What is the load force?
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.

11. 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.
 The candlestick is in stable equilibrium.  If you push the top, it will drop back to where it was.
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.
Now we put it on its side:
If you push it, it will roll, but will not tip over.  It is in neutral equilibrium.

12. 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.

13. Tilting or Toppling?
Q) Now suppose the bus goes even faster round a sharp bend. Which way is the moment now?  What will happen to the bus?
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.
Now suppose the bus goes fast round a corner and tilts over. Which way is the moment now?  What will happen to the bus?

14. 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.

15. 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.

16. P3 2. 1 Moments p226 / P2. 3 Moments in Balance p230/ P3 2
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.
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.
(Formulae - Basic) d)
(Harder – example calculation) e) f)
(Diagram - Harder)
(Maths)

17. 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

18. 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.
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
(Diagram/practical/video) d)
(Maths/ practical/ examples) c) e)
(practical - harder)
(basic)

19. 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.

20. 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.

21. 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.

22. 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.

23. 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.

24. 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. 
Many clocks in churches and other public buildings have their time-keeping governed by the swing of a pendulum.
NB: for GCSE you will need the top formula but the bottom one will help you work out the explanation using grown up maths!

25. P Hydraulics p234
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.
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.
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

26. P3 2.5 Hydraulics p234 a) b) e) I already know....
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
(Particle Model + Hydraulic System Diagram) b)
(Examples of Areas?) e)
(Diagram and Maths Examples - Harder)

27. 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.

28. Bottle Jack....

29. Trolley Jack....

30. 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!)

31. P3 2.6 Circular Motion p236 a) b) c) I already know....
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!)
(Link ideas of velocity v speed) b)
(Link real world situations) c)
(Use the formulae to explain factors - Harder)

32. 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.

33. 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

34. 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!

35. ..... a) & b) d) e) I already know.... c) a) Know ..... (.......)