1. Vectors and Scalars
Define Vector and Scalar quantities and give examples of each.

2. Scalars
A scalar quantity is a quantity that has magnitude only and has no direction in space
Examples of Scalar Quantities:
Length
Area
Volume
Time
Mass

3. Vectors
A vector quantity is a quantity that has both magnitude and a direction in space
Examples of Vector Quantities:
Displacement
Velocity
Acceleration
Force

4. Vector Diagrams Vector diagrams are shown using an arrow
The length of the arrow represents its magnitude
The direction of the arrow shows its direction

5. The Parallelogram Law The Triangle Law
When two vectors are joined tail to tail
Complete the parallelogram
The resultant is found by drawing the diagonal
The Triangle Law
When two vectors are joined head to tail
Draw the resultant vector by completing the triangle

6. Problem: Resultant of 2 Vectors
Two forces are applied to a body, as shown. What is the magnitude and direction of the resultant force acting on the body?
Solution:
Complete the parallelogram (rectangle)
The diagonal of the parallelogram ac represents the resultant force
The magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc
12 N a d θ
5 N
13 N 5 b c 12
Resultant displacement is 13 N 67º with the 5 N force

7. Resolving a Vector Into Perpendicular Components
When resolving a vector into components we are doing the opposite to finding the resultant
We usually resolve a vector into components that are perpendicular to each other
Here a vector v is resolved into an x component and a y component v y x

8. Practical Applications
Here we see a table being pulled by a force of 50 N at a 30º angle to the horizontal
When resolved we see that this is the same as pulling the table up with a force of 25 N and pulling it horizontally with a force of 43.3 N
50 N
y=25 N
30º
x=43.3 N
We can see that it would be more efficient to pull the table with a horizontal force of 50 N

9. Problem: Calculating the magnitude of perpendicular components
A force of 15 N acts on a box as shown. What are the horizontal and vertical components of the force?
Solution:
12.99 N
15 N
Component
Vertical
60º
Horizontal
Component
7.5 N

10. Moments: turning forces
What is the equation for calculating moment of a force?

11. Moment of a force = force x perpendicular distance
Keywords:
Force
Moment
Couple
Torque
COM
Weight
Turning
All: calculate the moment of a force about a point (torque) (C)
Most: apply principle of moments to simple balanced situations (B)
Some: employ principle of centre of mass in calculations (A)
Key Equation:
Moment of a force = force x perpendicular distance
moment = fd 11

12. = The Principle of moments: When an object is in equilibrium:
Keywords:
Force
Moment
Couple
Torque
COM
Weight
Turning
All: calculate the moment of a force about a point (torque) (C)
Most: apply principle of moments to simple balanced situations (B)
Some: employ principle of centre of mass in calculations (A)
Note definition
The Principle of moments:
When an object is in equilibrium:
Sum of clockwise moments about any point
Sum of anti-clockwise moments about any point =
clockwise moments – anticlockwise moments = zero (resultant force) 12

13. Centre of gravity is important for balancing

14. You have to keep the centre of gravity above the base if you want to balance.

15. My head is big for my body compared to an adult’s head
My head is big for my body compared to an adult’s head. That’s why I find it harder to balance.

16. Poor balance produces an overall turning force.
This requires a turning force in the opposite direction to counteract it.

17. Calculate the turning effect of the weight about point P.
Keywords:
Force
Moment
Couple
Torque
COM
Weight
Turning
All: calculate the moment in single support situations (C)
Most: calculate the moment in twin support situations (B)
Some: determine the moment of a couple (torque) (A)
A uniform horizontal shelf of width 0.38 m is attached to a wall a shown in the diagram. The total weight of the shelf and books is 70 N. This weight acts from the middle of the shelf (0.19 m from the wall).
Calculate the turning effect of the weight about point P.
Calculate the tension in the support wire. T
40o
70 N P

18. Keywords:
Force
Moment
Couple
Torque
COM
Weight
Turning
All: calculate the moment in single support situations (C)
Most: calculate the moment in twin support situations (B)
Some: determine the moment of a couple (torque) (A)
1. Moment = 70 N  0.19 m = 13.3 N m
2. The component of T at 90 to the shelf must provide the moment to balance the moment of the weight.
T sin 40°  0.38 m = 13.3 N m
T sin 40° = 13.3 / 0.38 = 35 N
T = 35 / sin 40°
T = 54.5 N
Note that we take moments about point P. This is because there is a third force which acts on the shelf; this is the contact force (or ‘reaction’) of the wall on the shelf. We do not know its magnitude or direction but, since it acts through point P, it has no turning effect about P. T
40o
70 N P

19. 1) The plank is set up as shown and the balance zeroed
1) The plank is set up as shown and the balance zeroed. When the student lies on the plank the reading is 600 N. The balance is 2 m from the student’s feet and the centre of gravity of the student is 1.5 m from their feet. What is the student’s weight?

20. The plank is set up as shown and the balance zeroed
The plank is set up as shown and the balance zeroed. When the student lies on the plank the reading is 600 N. The balance is 2 m from the student’s feet and the centre of gravity of the student is 1.5 m from their feet. What is the student’s weight?
Taking moments about the feet.
600 x 2 = W x 1.5 so W = 800 N

21. Keywords:
Force
Moment
Couple
Torque
COM
Weight
Turning
All: calculate the moment in single support situations (C)
Most: calculate the moment in twin support situations (B)
Some: determine the moment of a couple (torque) (A)
Couples
A couple is a pair of equal and opposite forces NOT acting along the same line

22. Couples Show that the total moment = Fd Keywords: Force Moment Couple
Torque
COM
Weight
Turning
All: calculate the moment in single support situations (C)
Most: calculate the moment in twin support situations (B)
Some: determine the moment of a couple (torque) (A)
Couples
Show that the total moment = Fd F x d F

23. Couples Show that the total moment = Fd Keywords: Force Moment Couple
Torque
COM
Weight
Turning
All: calculate the moment in single support situations (C)
Most: calculate the moment in twin support situations (B)
Some: determine the moment of a couple (torque) (A)
Couples
Show that the total moment = Fd F x d
Clockwise moment = Fx
Clockwise moment = F(d-x)
Total = Fx + F(d-x)
Total = Fx + Fd – Fx
Total = Fd F

24. The equations of motion
Keywords:
Velocity
Acceleration
Displacement
Initial
Final
Time
Constant
All: make calculations using the equations of motion (C)
Most: link algebraic and graphical methodologies (B)
Some: derive the equations of motion from 1st principles (A)
The equations of motion
Only work when the acceleration is constant! 24

25. (Take acceleration due to gravity as = 9
(Take acceleration due to gravity as = 9.8 ms-2 unless otherwise stated) 1. A car accelerates from rest at 3 ms-2 for 4s. How far has it travelled? 2. A stone is dropped from a cliff: (a) How far will it have fallen in 4s? (b) What will its velocity be at that point? (c) What is the average velocity during the 4s? 3. Starting from rest a car travels for 2 minutes with a uniform acceleration of 0.3 ms-2 after which its speed is kept constant until the car is brought to rest with a uniform decceleration of 0.6ms-2 if the total distance travelled is 4500m how long did the journey take? 4. A stone is thrown upward with a velocity of 12 ms-1 (a) How far will it have risen in 1s? (b) What will its velocity be at that paint? (c) What is the maximum height that it will reach before coming down again?
Lesson 2

26. Distance travelled = ½ at2 = ½x3x42 = 24 m
1. A car accelerates from rest at 3 ms-2 for 4s. How far has it travelled?
Distance travelled = ½ at2 = ½x3x42 = 24 m
2. A stone is dropped from a cliff:
How far will it have fallen in 4s?
(a) s = ½ gt2 = ½ x 9.8 x 42 = 78.4 m
What will its velocity be at that point?
(b) v = at = 9.8x4 = 39.2 ms-1
What is the average velocity during the 4s?
(c) average velocity = 39.2/2 = 19.6 ms-1
Lesson 2

27. Distance travelled in that time = ½ 0.3x1202 = 2160 m
3. Starting from rest a car travels for 2 minutes with a uniform acceleration of 0.3 ms-2 after which its speed is kept constant until the car is brought to rest with a uniform retardation of 0.6ms-2 if the total distance travelled is 4500m how long did the journey take?
Initial acceleration time = 2 minutes = 120s (note conversion to seconds)
Distance travelled in that time = ½ 0.3x1202 = 2160 m
Speed after this initial acceleration = 0.3x120 = 36 ms-1
Distance travelled during final deceleration = 362/2x0.6 = 1080 m
Time of final deceleration = 1080/18 = 60 s
Distance of constant velocity section = 4500 – 2160 – 1080 = 1260
Time of constant velocity section = 1260/36 = 35 s
Therefore total time for the journey = = 215 s = 3 min 35 s
4. A stone is thrown upward with a velocity of 12 ms-1
(a) How far will it have risen in 1 s?
(b) What will its velocity be at that paint?
(c) What is the maximum height that it will reach before coming down again?
(a) s = ut – ½ gt2 = 12 – ½ x9.8x1 = 7.1 m
(b) v = u –gt = 12 – 9.8x1 = 2.2 ms-1
(c) maximum height (h) when vertical velocity = 0
v2 = 0 = 122 – 2x9.8xh h = 7.5m
Lesson 2

28. Projectile : an object acted upon only by the force of gravity
Projectile Motion
Projectile : an object acted upon only by the force of gravity
Putting our SUVAT equations to good use

29. Starting points....
We need to consider the vertical and horizontal components of motion separately.
The acceleration involved is always g (downwards) & only affects the vertical component
Any horizontal velocity is constant and unaffected by g
We’ll consider up as positive and down as negative

30. Note : Horizontally there is no change in velocity, (1 square per photo all the way across)
Note : Vertically, the distance travelled between each photo increase (acceleration)
Note : Vertically, the motion of the dropped ball and the thrown ball is identical

31. A worked example....
An object is projected at a horizontal velocity of 15 ms-1 from the top of a 35m tower. Calculate the following
How long it takes to reach the ground
How far it travels horizontally
It’s speed at impact

32. A worked example....
An object is projected at a horizontal velocity of 15 ms-1 from the top of a 35m tower. Calculate the following
How long it takes to reach the ground
(Using S=ut + ½ at2 = 2.67s)
How far it travels horizontally
(Using S = ½ (u + v) t or S = velocity x time = 40m)
It’s vertical speed at impact
(Using V2 = U2 + 2aS = 26.2 ms-1)

33. An example....
A ball is projected horizontally at 0.52 ms-1 across the top of an inclined board which is 60cm wide (horizontally) and 120cm high (vertically in the direction of the incline). The ball reaches the bottom of the board in 0.9s. Calculate the following:
The distance travelled across the board [0.468m]
Its acceleration on the board [2.96ms-2]
Its speed at the bottom of the board [2.8ms-1]