1. Module 1.1 – Displacement and Velocity Vectors
Displacement and velocity vectors were studied in physics 11 for one dimensional motion. This module extends the study of kinematics to two dimensions, allowing us to study a much wider range of situations.

2. 2D Vectors
Vector Magnitude and Direction
Components

3. Trigonometry Review
Pythagorean Theorem

4. Example
Assume that the angle in the diagram below is 30.0° and the magnitude of the position vector is 7.2 cm. Calculate the components.

5. Solution

6. Vector Parts

7. Check Your Learning
Calculate the components for each of the following vectors:

8. Check Your Learning
Taking the signs into consideration (to the right and down), we get

9. Check Your Learning
Taking the signs into consideration (to the left and up), we get

10. Expressing Direction Assume that θ=30.0°
The angle is measured from east
Must go north to get to the vector
30.0° N of E
(Could also be expressed as 60.0° E of N)

11. Check Your Learning
State the direction for each of the following vectors:
20o S of E
80o W of S
40o W of N
50o N of W S
75o E of N

12. Vector Addition
We want to find

13. Vector Addition Diagram
Must draw vectors being added head to tail
Resultant (c) goes from tail of
first vector to head of second vector

14. Check Your Learning
Given the following vectors, draw a vector diagram to represent each of the following vector equations (where is an unknown vector in each case):

15. Check Your Learning

16. Adding With Components

17. Example
A person walks 5.0 km east and then 8.0 km in a direction 75° N of E. What is his displacement?

18. Solution

19. Check Your Learning
A person walks 4.0 km south and then 7.2 km in a direction 21o W of N. What was the person’s displacement?

20. Check Your Learning

21. Check Your Learning

22. Vector Subtraction

23. Vector Subtraction

24. Check Your Learning
Given the following vectors, draw a vector diagram to represent each of the following vector equations (where is an unknown vector in each case):

25. Check Your Learning

26. Velocity Vectors

27. Relative Velocity Same inside subscripts V of boat with
respect to water
V of boat with
respect to shore
V of water with
respect to shore
Same inside subscripts

28. Example
A boat that has a speed of 5.0 m/s in still water heads north directly across a river that is 250 m wide. The velocity of the river is 3.5 m/s east.
What is the velocity of the boat with respect to the shore?
How long does it take the boat to cross the river?
How far downstream does the boat land?
What heading (direction) would the boat need in order to land directly across from its starting point?

29. Solution

30. Solution

31. Solution
where
is north

32. Check Your Learning
In the example just completed, how long will it take the boat to cross the river in part (d)? In other words, how long will it take to cross the river when corrective action is taken so that the boat lands directly across from its starting point? How does this answer compare with the time obtained in part (a) of the example?

33. Check Your Learning
It takes more time to cross the river when correcting for the flow of the river than was calculated in part (a) since some of the boat’s velocity is being used to compensate for the river and stop the boat from moving downstream.

34. Module Summary In this module you have learned:
To represent two dimensional vectors using two methods:
components and
magnitude and direction.
How to add and subtract displacement vectors using vector diagrams and components:
When adding vectors, they must be drawn head to tail. The resultant vector goes from the tail of the first vector to the head of the last vector.
Subtracting vectors is the same operation as adding a negative vector, where a negative vector points in the direction opposite the direction of the original vector.
How to use vector diagrams and components to calculate relative velocities. 

35. Module 1.2 – Force Vectors
The concept of two-dimensional vectors will be applied to free body diagrams and Newton’s Laws of Motion from Unit 3. Two-dimensional situations that will be studied include forces acting at an angle and inclined planes.

36. Pulling at an Angle
Must break all force down into horizontal and vertical directions

37. Example
A 52.0 kg sled is being pulled along a frictionless horizontal ice surface by a person pulling a rope with a force of 235 N. The rope makes an angle of 35.0o with the horizontal. What is the acceleration of the sled?

38. Solution

39. Horizontal Forces

40. Vertical Forces

41. Check Your Learning
A 52.0 kg sled is being pulled along a horizontal surface by a person pulling a rope with a force of 235 N. The rope makes an angle of 35.0o with the horizontal. If the coefficient of friction between the sled and the surface is 0.25, What is the acceleration of the sled?

42. Check Your Learning

43. Inclined Planes No friction Acceleration will be parallel to the plane
Choose new coordinate system

44. Force of Gravity Components

45. Perpendicular Forces

46. Parallel Forces

47. Example
A 1200 kg car is on an icy (frictionless) hill that is inclined at an angle of 12o with the horizontal. What is the acceleration of the car down the hill?

48. Check Your Learning
A 1200 kg car is on an icy hill that is inclined at an angle of 12o with the horizontal. As the car starts sliding, the driver locks the wheels. The coefficient of friction between the locked wheels and the icy surface is What is the acceleration of the car down the hill?

49. Check Your Learning
Perpendicular Forces
Parallel Forces

50. Module Summary In this module you learned that
Force vectors can be broken into components so that dynamics situations can be analyzed using free body diagrams and Newton’s Second Law.
Situations involving inclined planes (ramps) can be analyzed by rotating the coordinate system so that the x-axis is parallel to the ramp and the y-axis is perpendicular to the ramp. The components for the force of gravity can then be given by

51. Module 1.3 – Equilibrium
This module is a continuation of the previous module, which introduced two dimensional force vectors. In this module, force vectors will be incorporated into studies of systems in equilibrium. We will look at both translational equilibrium (not accelerating linearly) and rotational equilibrium (not rotating). This module will also introduce the concept of a torque. Real world situations such as using support cables, cranes, scaffolding, and many other structures require an understanding of force vectors and torques.

52. Translational Equilibrium

53. Translational Equilibrium

54. Equilibrant Force
If the vector sum of all of the forces acting on an object is not zero, there will be a net force in some direction.
There is a single additional force that can be applied to balance this net force. This additional force is called the equilibrant force.
The equilibrant force is equal in magnitude to the sum of all of the forces acting on the object, but opposite in direction.

55. Example
A 20.0 kg sack of potatoes is suspended by a rope. A man pushes sideways with a force of 50.0 N and maintains this force so that the sack is in equilibrium. What is the tension in the rope and what angle does the rope make with the vertical?

56. Solution

57. Check Your Learning
Joe wishes to hang a sign weighing N so that cable A attached to the store makes a 30.0° angle as shown in the picture below. Cable B is attached to an adjoining building and is horizontal. Calculate the necessary tension in cable B.

58. Check Your Learning

59. Torque Even though forces balance, object will spin
The size of a torque depends on two things:
The size of the force being applied (a larger force will have a greater effect)
The distance away from the pivot point (the further away from this pivot, the greater the effect).

60. Torque
where torque is in Nm if force is in N and r is in m

61. Example 1
Suppose that you are trying to open a door that is 70.0 cm wide. You are applying a force of 68 N 10.0 cm from the outer edge of the door, but you are pushing at an angle of 75o from the surface of the door. What torque are you applying on the door?

62. Solution

63. Rotational Equilibrium
When calculating torques, all distances must be measured from the pivot point.

64. Static Equilibrium There are two conditions for static equilibrium:
The sum of the forces is zero (providing translational equilibrium), and
The sum of the torques is zero (providing rotational equilibrium).

65. Static Equilibrium
Equilibrant force must provide both translational and rotational equilibrium
Center of gravity – the point at which we could apply a single upward force to balance the object. For a mass with a uniform distribution of mass (such as a ruler), the center of gravity would be at the geometric center (the middle of the ruler).

66. Example 2
A 2.0 kg board serves as a see-saw for two children. One child has a mass of 30.0 kg and sits 2.5 m from the pivot point. At what distance from the pivot must a 25.0 kg child sit on the other side to balance the see-saw? Assume that the board is uniform and centred over the pivot.

67. Solution
Using the centre of the board as the pivot point

68. Example 3
A 4.0 m platform with a uniform distribution of mass has a 3.2 kg box 0.80 m from the left end. The mass of the platform is 2.0 kg. Calculate the size and location of the required equilibrant force.

69. Solution
Translational Equilibrium
Rotational Equilibrium

70. Check Your Learning
A uniform 1500 kg bridge, 20.0 m long, supports a 2200 kg truck whose centre of mass is 5.0 m from the right support column as shown in the diagram below. Calculate the force on each of the vertical support columns.

71. Check Your Learning
Using the left end as the pivot,

72. Check Your Learning

73. Module Summary In this module you learned that
An object is said to be in static equilibrium if it is in both translational equilibrium and rotational equilibrium.
Translational Equilibrium is achieved when the net force is zero:

74. Module Summary Torque can be calculated using the equation
Rotational Equilibrium is achieved when the net torque is zero: