1. Vectors and Scalars Chapter 8

2. What is a Vector Quantity? A quantity that has both Magnitude and a Direction in space is called a Vector Quantity.

3. Examples of Vector Quantities Displacement Velocity Momentum Acceleration Force Electric Field Strength Magnetic Flux Density These are the ONLY Vector Quantities on your course. All other quantities you meet are Scalar Quantities

4. What is a Scalar Quantity? A quantity that has Magnitude only is called a Scalar Quantity. It has no direction in space.

5. Examples of Scalar Quantities Distance Work Time Power Volume Temperature Pressure Electric current

6. How do I represent a Vector quantity on a diagram? An amount of a vector quantity is represented on a diagram by an Arrow. The Length of the arrow represents its Magnitude. The Direction in which the arrow points shows its Direction.

7. What is the Resultant of two Vectors? The Resultant of two Vectors is that single vector that when acting alone has the same effect as the other two vectors.

8. The combined effect of a displacement of 5 m East and a displacement of 10 m East is a displacement of 15 m East. The combined effect of a force of 2 N and a force of 4 N acting on an object in the same direction is a force of 6 N acting on the object in the same direction. Finding the Resultant: Vectors in same direction

9. If the vectors are in opposite directions: The Magnitude of the Resultant is found by subtracting the magnitude of the smaller from the magnitude of the bigger. The Direction of the Resultant is the direction of the bigger. Finding the Resultant: Vectors in the opposite direction

10. Finding the resultant of two non-collinear vectors If the vectors are not in the same straight line we use the Parallelogram Law to find their resultant.

11. State the Parallelogram Law If two vectors, drawn tail to tail, are the adjacent sides ab and ad of the parallelogram abcd, then the diagonal from a to c of this parallelogram is their Resultant.

12. Magnitude of resultant: = length of arrow ac = = 5 m s -1 (Using Pythagoras' Theorem) Direction of resultant: =  o with direction of motion of ship Where tan  = 3/4   = 36.87 o A ship moves with a constant velocity of 4 m s -1. A man walks across the ship at right angles to the direction of motion of the ship with a velocity of 3 m s -1. Find the resultant velocity of the man in magnitude and direction.

13. Experiment to Find the Resultant of two Forces Use the equipment shown. Adjust the size and direction of the three forces until the knot in the thread remains at rest. Take the reading on each newton balance, ​ F ​ 1 ​, ​ F ​ 2 ​ and ​ F ​ 3 ​. If we want the resultant of the two forces ​ F ​ 1 ​ and ​ F ​ 2 ​ its magnitude is the reading on the third balance ( ​ F ​ 3 ​ ). The direction of the resultant is in the opposite direction to ​ F ​ 3 ​.

14. x is the Resultant of c and d. c and d are two Perpendicular Components of x. Resolving a Vector into Components Resolving a Vector into Components is expressing it in terms of two other vectors so that it is the Resultant of these two. These two are called Components of the given vector. We need only study resolving a vector into components that are at right angles to each other. These are called Perpendicular Components.

15. A 100 N force acts on a cart in the direction shown. This force can be resolved into a horizontal component ( x ) and a vertical component ( y ). Each component represents the complete effect of the100 N in its direction. To prevent the cart from moving horizontally a force of 86.6 N acting to the left is required. If the weight of the cart is less than 50 N it will be lifted off the ground.

16. A vector of magnitude v has perpendicular components x and y. v makes an angle  with the component x The magnitudes of the components are: x = v cos  and y = v sin  Calculating the Perpendicular Components

17. Find the vertical and horizontal components of a vector of magnitude 20 N acting at 60 o to the horizontal. The diagram shows the components. Horizontal component = x = 20 cos 60 o = 10 N Vertical component = y = 20 sin 60 o = 17.32 N

18. A stone of weight 50 N rests on the sloped roof of a house. The roof is inclined at 20 o to the horizontal. Resolve the weight of the stone into components parallel and perpendicular to the roof. Component of weight perpendicular to roof = 50 cos 20 o = 46.98 N Component of weight parallel to roof = 50 sin 20 o = 17.10 N The diagram shows the weight vector and its 2 components. Take care that you draw these in correctly.