1. Lecture Outline Chapter 3 Physics, 4 th Edition James S. Walker Copyright © 2010 Pearson Education, Inc.

2. Units of Chapter 3 Scalars Versus Vectors The Components of a Vector Adding and Subtracting Vectors Unit Vectors Position, Displacement, Velocity, and Acceleration Vectors Relative Motion

3. Copyright © 2010 Pearson Education, Inc. Chapter 3 Vectors in Physics

4. Copyright © 2010 Pearson Education, Inc. 3-1 Scalars Versus Vectors Scalar: number with units or quantity with only magnitude. It can be positive, negative or zero. Example is the speed Vector: quantity with both magnitude and direction. Examples are the velocity, the acceleration, the force. How to get to the library: need to know how far and which way If you know only that the library is 0.5 mi from you, it could lie anywhere on a circle of radius 0.5 mi. If, instead, you are told the library is 0.5 mi northwest, you know its precise location.

5. Copyright © 2010 Pearson Education, Inc. 3-2 The Components of a Vector Even though you know how far and in which direction the library is, you may not be able to walk there in a straight line for 0.5 mi directly to the library, since to do so would take you through buildings where there no doors, through people’s backyards, and through all kinds of other obstructions. What you need to do is to resolve displacement vector r between you and the library into east-west and north-south components. In general, to find the components of a vector we need to set up a coordinate system. In two dimensions we choose an origin, O and a positive direction for both the x and the y axes. If the system were three-dimensional, we would also indicate a z-axis

6. Copyright © 2010 Pearson Education, Inc. 3-2 The Components of a Vector Can resolve vector into perpendicular components using a two-dimensional coordinate system: R x = r cos 25 O = (1.50 m)(0.906) = 1.36 m R y = r sin 25 O = (1.50 m)(0.423) = 0.634 m

7. Copyright © 2010 Pearson Education, Inc. 3-2 The Components of a Vector Length, angle, and components can be calculated from each other using trigonometry: Given the magnitude and direction of a vector, find its components: A x = A cos θ A y = A sin θ Given the components of a vector, find its magnitude and direction: A = (A x 2 + A y 2 ) 1/2 θ = tan -1 A y /A x

8. Copyright © 2010 Pearson Education, Inc. 3-2 The Components of a Vector In the Jules Verne novel Mysterious Island, Captain Cyrus Harding wants to find the height of a cliff. He stands with his back to the base of the cliff, then marches straight away from it for 5.0 x 10 2 ft. At this point he lies on the ground and measures the angle from the horizontal to the top of the cliff. If the angle is 34.0 O, (a) how high is the cliff? (b) What is the straight line distance from Captain Harding to the top of the cliff? = 337 ft = 5.0 x 10 2 ft = 34 0 Height of the cliff h = A x tan θ= (5.0 x 10 2 ft) tan 34 0 = 337 ft A = (A x 2 + A y 2 ) 1/2 A = { (337 ft) 2 + (5.0 x 10 2 ft) 2 } 1/2 = 603 ft What angle would Cyrus Harding have found if he had walked 6.0 x 10 2 ft from the cliff to make his measurement? Answer : θ = 29.3 0

9. Copyright © 2010 Pearson Education, Inc. 3-2 The Components of a Vector Signs of vector components: Examples of vectors with components of different signs To determine the signs of a vector’s components, it is only necessary to observe the direction in which they point. For example, in part (a) the x component points in the positive direction; hence A x > 0. Similarly, the y component in part (a) points in the negative y direction; therefore A y < 0.

10. Copyright © 2010 Pearson Education, Inc. 3-2 The Components of a Vector Signs of vector components: For example, if A x = -0.5 m and A y = 1.0 m, what is the angle? Θ = tan -1 (1.0 m/-0.50 m) = tan -1 ( -2.0) = -63 0 This is similar to Figure c and thus the direction angle of A should be between 90 0 and 180 0. To obtain the correct angle, add 180 0 to the calculator’s result: θ = -63 0 + 180 0 = 117 0 => the direction angle for the vector A #3 The vector B has components B x = -2.10 m and B y = -1.70 m. Find the direction angle, θ, for this vector. Solution: tan -1 [(-1.70 m)/(-2.10 m)] = tan-1 (1.70/2.1) = 39.0 0, θ = 39.0 + 180 o = 219 0

11. Copyright © 2010 Pearson Education, Inc. 3-3 Adding and Subtracting Vectors Adding vectors graphically: Place the tail of the second at the head of the first. The sum points from the tail of the first to the head of the last. Adding several vectors Searching for a treasure that is 5 paces north (A), 3 paces east (B), and 4 paces southeast ( C ) of the sycamore tree. The net displacement from the tree to the treasure is D = A + B + C.

12. Copyright © 2010 Pearson Education, Inc. 3-3 Adding and Subtracting Vectors Adding Vectors Using Components: 1. Find the components of each vector to be added. 2. Add the x- and y-components separately. 3. Find the resultant vector.

13. Copyright © 2010 Pearson Education, Inc. 3-3 Adding and Subtracting Vectors C = (C x 2 + C y 2 ) 1/2 θ = tan -1 C y /C x

14. Copyright © 2010 Pearson Education, Inc. 3-3 Adding and Subtracting Vectors Subtracting Vectors: The negative of a vector is a vector of the same magnitude pointing in the opposite direction. Here,.

15. Copyright © 2010 Pearson Education, Inc. 3-4 Unit Vectors Unit vectors are dimensionless vectors of unit length. The x unit vector, x, is a dimensionless vector of unit length pointing in the positive direction The y unit vector, y, is a dimensionless vector of unit length pointing in the positive y direction.

16. Copyright © 2010 Pearson Education, Inc. 3-4 Unit Vectors Multiplying unit vectors by scalars: the multiplier changes the length, and the sign indicates the direction. Multiplying a vector by a positive scalar different from 1 will change the length of the vector but leave its direction the same. If the vector is multiplied by a negative scalar its direction is reversed.

17. Copyright © 2010 Pearson Education, Inc. 3-5 Position, Displacement, Velocity, and Acceleration Vectors Position vector points from the origin to the location in question. The displacement vector points from the original position to the final position.

18. Copyright © 2010 Pearson Education, Inc. 3-5 Position, Displacement, Velocity, and Acceleration Vectors Average velocity vector: (3-3) So is in the same direction as.

19. Copyright © 2010 Pearson Education, Inc. 3-5 Position, Displacement, Velocity, and Acceleration Vectors Instantaneous velocity vector is tangent to the path: Find the speed and direction of motion for a rainbow trout whose velocity is v = (3.7 m/s)x + (-1.3 m/s)y. Solution: Speed = v = ((3.7 m/s) 2 + (-1.3 m/s) 2 ) 1/2 = 3.9 m/s θ = tan -1 C y /C x = tan -1 (-1.3 m/s/3.7 m/s) = -19 0, that is 19 0 below the x axis.

20. Copyright © 2010 Pearson Education, Inc. 3-5 Position, Displacement, Velocity, and Acceleration Vectors Average acceleration vector is in the direction of the change in velocity: SI unit: meter per second per second, m/s 2

21. Copyright © 2010 Pearson Education, Inc. 3-5 Position, Displacement, Velocity, and Acceleration Vectors Velocity vector is always in the direction of motion; acceleration vector can point anywhere: Suppose that the initial velocity of the car is v i = (12 m/s)x and that 10.0 s later its final velocity is v f = (-12 m/s)y. Calculate the average acceleration. = (v f - v i ) / 10.0 s = ((-12 m/s)y – (12m/s)x) / 10.0 s= (-1.2 m/s 2 )x + (-1.2 m/s 2 )y

22. Copyright © 2010 Pearson Education, Inc. 3-6 Relative Motion The speed of the passenger with respect to the ground depends on the relative directions of the passenger’s and train’s speeds: A good example of the use of vectors is in the description of relative motion. Suppose, for example, that you are standing on the ground as a train goes by at 15.0 m/s. Inside the train, a free-riding passenger is walking in the forward direction at 1.2 m/s relative to the train. How fast is the passenger moving relative to you? The answer is 1.2 m/s + 15.0 m/s = 16.2 m/s. 2. What if the passenger had been walking with the same speed, but toward the back of the train? The passenger going by with a speed of -1.2 m/s + 15.0 m/s = 13.8 m/s

23. Copyright © 2010 Pearson Education, Inc. 3-6 Relative Motion This also works in two dimensions: Suppose the passenger is climbing a vertical ladder with a speed of 0.20 m/s, and the train is slowly coasting forward at 0.70 m/s. Find the speed and direction of the passenger relative to the ground. V pg = (0.70 m/s)x + (0.20 m/s)y; thus V pg = (().70 m/s) 2 + (0.20 m/s) 2 ) 1/2 = 0.73 m/s θ = tan -1 (0.20/0.70) = 16 0

24. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 3 Scalar: number, with appropriate units Vector: quantity with magnitude and direction Vector components: A x = A cos θ, B y = B sin θ Magnitude: A = (A x 2 + A y 2 ) 1/2 Direction: θ = tan -1 (A y / A x ) Graphical vector addition: Place tail of second at head of first; sum points from tail of first to head of last

25. Copyright © 2010 Pearson Education, Inc. Component method: add components of individual vectors, then find magnitude and direction Unit vectors are dimensionless and of unit length Position vector points from origin to location Displacement vector points from original position to final position Velocity vector points in direction of motion Acceleration vector points in direction of change of motion Relative motion: Summary of Chapter 3