1. Chapter 3–2: Vector Operations Physics Coach Kelsoe Pages 86–94

2. Objectives Identify appropriate coordinate systems for solving problems with vectors. Apply the Pythagorean theorem and tangent function to calculate the magnitude and direction of a resultant vector. Resolve vectors into components using the sine and cosine functions. Add vectors that are not perpendicular.

3. Coordinate Systems in 2-D In Chapter 2 we discussed the motion of the gecko up the tree by a positive or negative sign. In Chapter 3 we’re going to describe this motion using an arrow pointing in a given direction. The most versatile system for diagramming the motion of an object is using the x- and y-axes at the same time. The addition of another axis not only helps describe motion in two dimensions but also simplifies analysis of motion in one dimension.

4. Coordinate Systems in 2-D Two methods can be used to describe the motion of an object. –In the first approach, the coordinate system can be turned so that the object moves along the y- axis. The problem with this method is that the system must be turned again if the motion changes. –In the second approach, we keep our axes fixed so that the positive y-axis points north and the positive x-axis points east.

5. Coordinate Systems in Two Dimensions One method for diagramming the motion of an object employs vectors and the use of the x- and y- axes. Axes are often designated using fixed directions. In the figure shown here, the positive y-axis points north and the positive x- axis points east.

6. Determining Resultant Magnitude and Direction In the previous section, we learned how to find resultant vectors graphically. The problem with this method is that it is very time consuming and the accuracy depends on how well the diagram is drawn. A more efficient way to find resultant vectors is to use the Pythagorean Theorem. Imagine climbing a pyramid in Egypt. You know the height and width of the pyramid, but you want to know the distance covered in a climb from the bottom to the top.

7. Pyramid Physics The magnitude of the vertical displacement is the height of the pyramid. The magnitude of the horizontal displacement is the distance from one edge of the pyramid to the middle, or half the width. These two vectors are perpendicular. ΔyΔy ΔxΔx d

8. Pythagorean Theorem The Pythagorean theorem states that for any right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides, or legs. So in this illustration, we see that d 2 = Δx 2 + Δy 2 ΔyΔy ΔxΔx d

9. Don’t Forget Direction!!! Once you find the displacement, it is important to know the direction. We can use the inverse tangent function to find the angle, which denotes the direction. We can use sine or cosine functions, but our tangent values are given. ΔyΔy ΔxΔx d θ

10. Determining Resultant Magnitude and Direction The Tangent Function –Use the tangent function to find the direction of the resultant vector. –For any right triangle, the tangent of an angle is defined as the ratio of the opposite and adjacent legs with respect to a specified acute angle of a right triangle. Tan θ = Opposite leg Adjacent leg

11. Sample Problem Finding Resultant Magnitude and Direction An archaeologist climbs the Great Pyramid in Giza, Egypt. The pyramid’s height is 136 m and its width is 2.30 x 10 2 m. What is the magnitude and the direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?

12. Sample Problem Define –Given: Δy = 136 m Δx = ½ (width) = 115 m –Unknown: d = ? θ = ? –Diagram: Choose the archaeologist’s starting position as the origin of the coordinate system, as shown above.

13. Sample Problem Plan –Choose an equation or situation: The Pythagorean theorem can be used to find the magnitude of the archaeologist’s displacement. The direction of the displacement can be found by using the inverse tangent function. –Rearrange the equations to isolate the unknowns: d 2 = Δx 2 + Δy 2 d = √ Δx 2 + Δy 2 tan θ = ΔyΔy ΔxΔx θ = tan -1 ΔyΔy ΔxΔx

14. Sample Problem Calculate Evaluate Because d is the hypotenuse, the archaeologist’s displacement should be less than the sum of the height and half of the width. The angle is expected to be more than 45° because the height is greater than half width. d = √ 115 2 + 136 2 θ = tan -1 136 115 d = 178 m θ = 49.8° d = √ Δx + Δyθ = tan -1 ΔyΔy ΔxΔx

15. Resolving Vectors In the previous example, the horizontal and vertical parts that add up to give the actual displacement are called components. The x component is parallel to the x-axis and the y component is parallel to the y-axis. You can often describe an object’s motion more conveniently by breaking a single vector into two components, a process we call resolving the vector.

16. Resolving Vectors Into Components Consider an airplane flying at 95 km/hr. –The hypotenuse (v plane ) is the resultant vector that describes the airplane’s total velocity. –The adjacent leg represents the x component (v x ), which describes the airplane’s horizontal speed. –The opposite leg represents the y component (v y ), which describes the airplane’s vertical speed.

17. Resolving Vectors Into Components The sine and cosine functions can be used to find the components of a vector. The sine and cosine functions are defined in terms of the lengths of the sides of right triangles. Sine of angle θ = Cosine of angle θ = opposite hypotenuse adjacent hypotenuse

18. Adding Vectors That Are Not Perpendicular Suppose that a plane travels first 5 km at an angle of 35°, then climbs at 10 for 22 km, as shown below. How can you find the total displacement? Because the original displacement vectors do not form a right triangle, you can not directly apply the tangent function or the Pythagorean Theorem. d1d1 d2d2

19. Adding Vectors That Are Not Perpendicular You can find the magnitude and the direction of the resultant by resolving each of the plane’s displacement vectors into its x and y components. Then the components along each axis can be added together. Δx1Δx1 d2d2 d1d1 Δx2Δx2 Δy1Δy1 Δy2Δy2

20. Adding Non-Perpendicular Vectors The goal of resolving vectors is to allow us to have components that are aligned along the x- and y-axis. Once the components are set up on these axes, our math becomes simple– we add our x components together and y components together to find our new x and y components. With our new-found x and y components, we can find our new resultant vector.

21. Sample Problem Adding vectors algebraically A hiker walks 27.0 km from her base camp at 35° south of east. The next day, she walks 41.0 km in a direction of 65° north of east and discovers a forest ranger’s tower. Find the magnitude and direction of her resultant displacement.

22. Sample Problem Solution Select a coordinate system, then sketch and label each vector. –Given d 1 = 27.0 kmθ 1 = -35° d 2 = 41.0 kmθ 2 = 65° Tip: θ 1 is negative because clockwise movement from the positive x-axis is negative. –Unknown d = ? θ = ?

23. Sample Problem Solution Find the x and y components of both vectors. –Separate your displacements. –For Day 1: Δx 1 = d 1 cos θ 1 = (27.0 km)(cos -35°) = 22 km Δy 1 = d 1 sin θ 1 = (27.0 km)(sin -35°) = -15 km –For Day 2: Δx 2 = d 2 cos θ 2 = (41.0 km)(cos 65°) = 17 km Δy 2 = d 2 sin θ 2 = (41.0 km)(sin 65°) = 37 km

24. Sample Problem Solution Find the x and y components of the total displacement. Δx total = Δx 1 + Δx 2 = 22 km + 17 km = 39 km Δy total = Δy 1 + Δy 2 = -15 km + 37 km = 22 km Use the Pythagorean theorem to find the magnitude of the resultant vector. d 2 = (Δx total ) 2 + (Δy total ) 2 d = √(39 km) 2 + (22 km) 2 d = 45 km

25. Sample Problem Solution Use a suitable trigonometric function to find the angle. Tan θ = opposite/adjacent Tan θ = Δy/Δx = 22 km/39 km Tan θ = 0.564 θ = 29° north of east

26. Vocabulary Components of a vector

27. Important Formulas c 2 = a 2 + b 2 sin θ = opposite/hypotenuse cos θ = adjacent/hypotenuse tan θ = opposite/adjacent