1. Unit 9: Forces and Their Effects: Dynamics- Resolving Vectors

2. Objectives: Additional skills gained: Diagram Vs Calculation
9.2 Explain the difference between vector and scalar quantities using examples
9.3 Use vector diagrams to illustrate resolution of forces, a net force, and equilibrium situations (scale drawings only)
9.4 Draw and use free body force diagrams 4a, 5a, 5b
9.5 Explain examples of the forces acting on an isolated solid object or a system where several forces lead to a resultant force on an object and the special case of balanced forces when the resultant force is zero
Additional skills gained:
Diagram Vs Calculation
Vector Notations

3. Lesson Link Acceleration Height Momentum Displacement Length Energy
Scalar quantities have only a magnitude whereas vector quantities have both a magnitude AND direction
Draw a table with ‘Scalar’ and ‘Vector’ headings then sort the following into each column!
Acceleration
Height
Momentum
Displacement
Length
Energy
Speed
Velocity
Temperature
Distance
Weight
Magnetic Field
Electric Field
Force
Density
Time
Width
Volume
Mass

4. Scalar Vector Scalar Vs Vector Density Acceleration Distance Direction
Scalar quantities have only a magnitude whereas vector quantities have both a magnitude AND direction
Scalar Vector
Density Acceleration
Distance Direction
Energy Displacement
Height Electric Fields
Length Force
Mass Magnetic Fields
Speed Momentum
Temperature Velocity
Time Weight
Volume
Width

5. Resolving a Vector A D B C 6m 9m
This is the first brand new topic. This was not discussed in GCSE before!
Demo: Resolving a vector from its components!
Needed: 2 volunteers, tape measure
Distance: The total number of meters covered.
6m + 9m = 15m A D
Displacement: The distance between start point and end point (connect them)
??????????????? 6m B C 9m

6. Resolving a Vector Pythagoras! A D a2 + b2 = c2 B C 6m 9m
How would you discover the value for displacement?
Pythagoras!
Psssst….I know…
That’s me btw! a b c A D
a2 + b2 = c2 6m
Calculate the displacement for this problem…. B C 9m

7. Distance-Time Graphs 62 + 92 = s2 62 + 92 = 117 𝟏𝟏𝟕 s 6m 9m = 10.8m
Note: This can be done by drawing a ‘to scale’ diagram. This is often allowed in the exam…..though risky…

8. Seeing it in real-life…
Horizontal Component
Vertical Component
Needed: two new volunteers!

9. by wires. Each wire is 8 metres long and is
Practice
1. A telegraph pole is connected to the ground
by wires. Each wire is 8 metres long and is
fixed to the ground 4.5 metres from the pole. Use this information to calculate the height of the telegraph pole.
2. ABCD is a rhombus whose diagonals are
24 cm and 40 cm.
Calculate the perimeter of this rhombus.
3) The diagram opposite shows the side view of a shed. Calculate the width of the shed.

10. Important
I have used the example of displacement so far, but this works with ANY vector. I.e. if you did it with Force, it’d work in the same way!
Sometimes the question is only worded and no diagram provided…In this case you need to draw the diagram in steps!
Steps for drawing your diagram:
1) Draw the first component mentioned (only to scale if you’re using the drawing method).
2) Starting from the end of the first component, draw the second
3) Connect the beginning of the first component with the end of the second with an arrow (ending at the end of the second) 1)
E.g.
A man runs 6ms-1 east whilst on a boat sailing north at 5ms-1. Calculate the resultant 2) 3)

11. Isolating Components from Vectors
This almost always comes up in exams!
Just as we can resolve a vector from its components, we can also isolate the components from a vector (i.e. identify its horizontal and vertical parts).
This time though we need the angle!
Steps for answering these questions:
Draw a diagram of what you are dealing with
Sketch along the horizontal and vertical components
(Hint: when drawing the triangle, it’s easier to make sure the angle given is within the triangle)
3) Use trigonometry to work out each component
15m
20°
15m
20° sv sH
SOHCAHTOA

12. Isolating Components from Vectors
Nb: When using the calculator, press enter after ‘sine (angle)’ before doing anything else. Or keep it in parentheses
Hypotenuse (vector)
S 𝑂 𝐻 C 𝐴 𝐻 T 𝑂 𝐴
15m
20° sv
Opposite
(to the angle) sH
In this instance we need to know the Opposite and the Adjacent and have been given the Hypotenuse and the angle! Which do we use?
Adjacent
(to the angle)
Vertical= 5.13m, Horizontal= m
We use Sine to work out the opposite (vertical component) and Cosine to work out the adjacent (horizontal component)
Task: Rearrange the formulae to calculate the value for the components!
E.g.
Sin (20°) = 𝑂 𝐻

13. Free-Body Force Diagrams
A free body force diagram is the fancy name given to a diagram that shows all the forces acting upon one object.
Take the example on the left, this shows a box that has 4 forces acting upon it:
Weight of 6N downward
Contact force of 6N upward
3N pull to the left
20N pull to the right
You would usually use these diagrams to help you determine the resultant force left on the object.
Task: Calculate the Resultant Force
If there is more than one force acting in the same direction then you just add them together for that direction. If they oppose each other directly (e.g. the 3N and 20N, you take one from the other!)

14. Free-Body Force Diagrams
Task: Draw a free body force diagram for a box falling that has a weight of 600N and an air resistance of 200N. Then calculate the resultant Force. CHALLENGE: Thus calculate the acceleration on the box assuming it’s mass is 40kg
So, as the 200N and the 600N are in the same plane of action they interact.
The resultant is the left-over Force which would be 400N downwards
Challenge: as F=ma, then a= F/m, a=400/40= 10m/s2
If the object is ever at rest (stationary) or moving at a constant velocity don’t forget that it must have no resultant so every force in every plane is equal!

15. Free-Body Force Diagrams (with a twist)
Consider this ball on a rope. If we were to take this snapshot, we can draw a simple free body force diagram for the ball
Task: Draw a free-body force diagram for the ball on the left
Though you may be tempted to suggest that there is a force going to the left, the only forces acting upon the ball are its weight and the tension of the string.
What however is balancing the weight of the ball so that it doesn’t fall?

16. Free-Body Force Diagrams (with a twist)
The Tension is not acting in this instance in only one plane instead it has both a vertical and a horizontal component!
So…the only pare of the Tension that can be balancing out the weight must be the vertical component (as weight acts vertically too!).
Thus by knowing the weight of the object, I can determine the vertical component of tension or vice versa (if I knew the angle anyway)!

17. Resolving Vectors with multiple factors (worked example)
Often, the questions you’ll be given have one resultant force produced by the resolution of multiple factors (i.e. lots of forces, lots of velocities…)
E.g. The following figure shows a uniform steel girder being held stationary in a horizontal position by a crane using two cables. The tension (T) in each of these cables is 850N. Calculate the weight (W) of the girder.
Task: Attempt it first…
1) Draw the free body diagram
42°
42°

18. Resolving Vectors with multiple factors (worked example)
1) Draw the free body diagram
850N
850N
42°
42°
42°
42° W
2) Weight is solely a vertical acting component. We can ignore any horizontal forces here. You can see that each of the cables is applying both a horizontal and vertical force, as it’s a resultant in a diagonal.
3) As the girder is stationary , the value of weight must be equal to the total vertical components in the other direction (i.e. no resultant)
850N
850N
4) So use the same approach to find one vertical component…(Sine 42 x 850 = 569N). Total (each cable)= 569N x 2 = 1138N!
42°
42°

19. Practice Assume negligible friction
Force for one = cos 35 x 6500 = Both = 10648N. A= F/m = 10.6ms-2
Assume negligible friction

20. Dr Henshaw is enjoying the snow and is sledding down a ramp of 35°
Extension/IS
Dr Henshaw is enjoying the snow and is sledding down a ramp of 35°
assuming his mass is 65kg and he is travelling at a constant speed. First draw a free body force diagram for him and then use this to calculate the force of friction. *HINT* consider what would be matching this plane!
Force for one = cos 35 x 6500 = Both = 10648N. A= F/m = 10.6ms-2

21. Objectives: Additional skills gained: Diagram Vs Calculation
Spec point 12: understand scalar and vector quantities and know examples of each type of quantity and recognise vector notation
Spec point 13: be able to resolve a vector into two components at right angles to each other by drawing and by calculation
Spec point 14: be able to find the resultant of two coplanar vectors at any angle to each other by drawing, and at right angles to each other by calculation
Additional skills gained:
Diagram Vs Calculation
Vector Notations