2. Vectors and scalars Scalar has magnitude only
e.g. 10 mph
Vector has magnitude and direction
e.g. 10 mph in a NE direction
Can be represented by a line with and arrow on it drawn to scale
10 mph

3. Vectors and scalars
Vectors
Scalars

4. Distance and displacement
Displacement is distance travelled in a definite direction
Nottingham (N)
London (L)

5. Speed and velocity Units distance (metres) time (sec) speed (m/s) x s

6. Average velocity

7. Distance-time graphs Steady speed gradient = a/b
Sign of gradient indicates direction of motion
time x a b

8. Velocity time graphs Uniform velocity acceleration
a= change in velocity
time
uniform acceleration
gradient = a/b
Sign of gradient indicates acceleration or deceleration
time
vel a b

9. Area under a velocity – time graph
Area under graph indicates the displacement (distance travelled)
time t
vel u v

10. Acceleration-time graphs
Gradient at any point gives rate of change of acceleration at that time
Area under graph represents change in velocity experienced
accel
time t a

11. Equations of motion Where u = initial velocity (m/s)
v = final velocity (m/s)
x = displacement (m)
t = time (s)
a = acceleration (m/s2)

12. Acceleration under gravity
Sign convention
Acceleration is always downwards
Accel (a) is +ve when moving down and a is –ve when moving up
Accel (a) = g (the accel due to gravity, in magnitude)

13. Projectile motion Horizontal motion is independent of vertical motion
Constant velocity for horizontal component of motion
Constant acceleration for vertical component of motion

15. Equations of projectile motion
An object in freefall:
moves at a constant horizontal (x) velocity
ax = 0
moves at a constant vertical (y) acceleration.
ay = g
The following equations can therefore be applied. Can you see how they have been derived?
x = vxt
constant x velocity
vy = uy + gt
vy2 = uy gy
suvat equations for uy and vy with a = g
uy + vy
y = uyt + ½gt2
y = t 2

16. Height of a projectile
A tennis player hits a volley just above ground level, in a direction perpendicular to the net.
The ball leaves her racquet at 8.2 ms-1 at an angle of 34° to the horizontal.
Will the ball clear the net if it is 2.3 m away and 95 cm high at this point?
What assumptions should you make to solve this problem?
Photo credit: © Shutterstock 2009, Nicholas Rjabow
no air resistance
no spin
initial height is zero.

17. Height of a projectile
We need to calculate the value of y at x = 2.3 m and determine whether or not it is greater than 0.95 m.
What are the relevant equations of motion?
8.2 ms-1
x = vXt
0.95 m
34°
y = uyt + ½gt2
2.3 m
First, use the x equation to calculate t when x is 2.3.
2.3 = 8.2 × cos34° × t
t = 0.34 s

18. Height of a projectile So the ball reaches x = 2.3 m when t = 0.34 s.
Now substitute this value of t into the y equation to find y, and determine whether or not it is greater than 0.95 m.
8.2 ms-1
0.95 m
35°
2.3 m
y = uyt + ½gt2
y = ((8.2 × sin34°) × 0.34) + (½ × × 0.342)
y = 0.99 m
So y is greater than 0.95 and the ball clears the net!

23. Vector addition and resolution
Vectors have
Magnitude and direction
Can be represented by lines drown to scale with arrows on them
Can be added to find the resultant
Resultant can be found by drawing or calculation

24. Resultant of 2 vectors
Resultant = 4N 1N 3N + =
Resultant = -2N + =

25. Vector addition and resolution
Resultant can be found by drawing or calculation or

26. Pythagoras’ Theorem

27. Resolving vectors
Resolving vectors

28. Resolving vectors
Resolving vectors
The vector F can be resolved into 2 components at right angles to each other
X component = F cos 
Y component = F sin 
Tan  = F sin 
F cos 
F sin 
F cos  F

29. Types of Force Forces acting at a distance Contact forces
Gravitational
Electrostatic
Electromagnetic
Nuclear
Contact forces
Frictional forces
Reaction forces

30. Forces
Thrust
Weight

31. Lift, Thrust, Drag

32. Reaction and weight

33. Action and Reaction
Equal and opposite

34. Introduction to turning forces
Forces can make things accelerate. They can also make things rotate.
What’s wrong with these pictures?
too short!
too short!
Teacher notes
Although some doors do have handles in the middle as shown, less force is needed to open doors with handles at the opposite side to the hinges as usual.
wrong place!
We know instinctively that we need to apply a force at a large distance from the pivot for it to be effective.

35. Moments and torque
A moment is the turning effect of a force. It can also be called a torque. Torque is given the symbol t (the Greek letter tau). Its units are newton metres (Nm).
pivot F d
F – the force applied in newtons (N).
d – the perpendicular distance (in m) between the pivot and the line of action of the force.
t = F × d

36. Moments for non-perpendicular distance

37. The torque of a couple is the rotation force or moment produced.
Couples and torques
A couple is a pair of forces acting on a body that are of equal magnitude and opposite direction, acting parallel to one another, but not along the same line.
Forces acting in this way produce a turning force or moment.
The torque of a couple is the rotation force or moment produced.
The forces on this beam are a couple, producing a moment or torque, which will cause the beam to rotate. F d F

38. perpendicular distance between lines of action of the forces
The torque of a couple
There is a formula specifically for finding the torque of a couple.
A point P is chosen arbitrarily. Take moments about P. F P x
d – x
total moment = Fx + F(d – x) F
= Fx + Fd – Fx d
= Fd
Teacher notes
It could be emphasized that P stands for any point along the beam. The fact that x does not feature in the final equation for moment shows that it does not matter where P is placed.
perpendicular distance between lines of action of the forces
torque of a couple = force ×

40. Centres of mass and gravity
The centre of gravity of an object is a point where the entire weight of the object seems to act.
The centre of mass of an object is a point where the entire mass of the object seems to be concentrated.
In a uniform gravitational field the centre of mass is in the same place as the centre of gravity.
Teacher notes
It could be highlighted that the difference between the definitions is related to the difference between the definitions of mass and weight, which is an issue students should be familiar with.
An alternative definition is that the centre of mass or centre of gravity of an object is the point through which a single force has no turning effect on the body.

41. Teacher notes
Strictly speaking, this method is for finding the centre of gravity of two dimensional (2D) shapes. It can be carried out in the classroom using a cardboard cut out shape suspended from a clamp stand. A plumb line is suspended from the same stand to allow vertical lines to be drawn downwards from two different pivot points.

43. Equilibrium
A body persists in equilibrium if no net force or moment acts on it. Forces and moments are balanced.
Newton’s first law states that a body persists in its state of rest or of uniform motion unless acted upon by an external unbalanced force.
Bodies in equilibrium are therefore bodies that are at rest or moving at constant velocity (uniform motion).
Teacher notes
The diagram shows a uniform body. It is in equilibrium because both forces and moments acting on it are balanced.
See the ‘Dynamics’ presentation for more information about Newton’s laws of motion. F1 F2 F2 F1
equilibrium

44. Balanced moments
If the total clockwise moment on an object is balanced by the total anticlockwise moment, then the object will not rotate.
Provided that there are no other unbalanced forces on it, the object will be in equilibrium, like the beam below: 4m 2m
3 N
6 N
total anticlockwise moments = total clockwise moments
3 × 4 = 6 × 2
12 Nm = 12 Nm

45. The principle of moments
The principle of moments states that (for a body in equilibrium):
total clockwise moments =
total anticlockwise moments
This principle can be used in calculations:
What is d?
5 m d
4 × 5 = 6d
4 N
6 N
20 = 6d
d = 20 / 6
d = 3.3 m

46. Can you make the beam balance?
Teacher notes
Negative forces are like positive forces, but in the opposite direction. Students could be asked for another way of representing negative forces: as positive forces with the arrows reversed. Care should be taken with the signs in the equation and with noting which forces are clockwise and which anticlockwise.

48. Human forearm
The principle of moments can be used to find out the force, F, that the biceps need to apply to the forearm in order to carry a certain weight. When the weight is held static, the system is in equilibrium.
weight of arm = 20 N
60 N
Taking moments about the elbow joint:
schematic diagram
4 cm F
Teacher notes
Students could be asked to calculate F from the two diagrams before being shown the calculation and answer.
4F = (16 × 20) + (35 × 60)
16 cm
20 N
4F = 2420
60 N
35 cm
F = 605 N

49. Centre of gravity and equilibrium
Teacher notes
Consider the moment of the block’s weight about its bottom left corner. Experiment with the shape of the block and the angle of the slope. If the centre of gravity lies outside the base, the moment of the weight will make the block topple. If the centre of gravity lies inside the base nothing will happen, indicating that the block is in equilibrium.
Students could be asked to predict whether the block will topple, and whether it will topple only once, or keep toppling. Pressing play will show what happens for a particular set of variables. After a short pause, the block will reset with the variables set to the same values. Students could be asked what could be changed to place the block in equilibrium. Changing the values and pressing play again will reveal if they are right.