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10/11 do now • 2nd and 3rd period: 3-1 diagram skills • 9th period: page 113 #1-13 Homework – posted online • 10/11/13 Homework – due Tuesday 10/15 • Read text book page 88-97 and write an essay to answer the following questions: 1. What is the best choice for orienting axes in a problem? 2. How to determine the resultant magnitude and direction? 3. How to resolve vectors into components? 4. How to add vectors that not perpendicular? • Be sure to use examples in your essay to clarify ideas. Go over homework 3.2 vector operations Objectives 1. Identify appropriate coordinate systems for solving problems with vectors. 2. Apply the Pythagorean theorem and tangent function to calculate the magnitude and direction of a resultant vector. 3. Resolve vectors into components using the sine and cosine functions. 4. Add vectors that are not perpendicular. • Use your knowledge of sign conventions from Chapter 2 to fill in the blanks below. • Movement to the right along the x-axis is considered _____. Movement downward along the y-axis is considered ______. Movement upward along the yaxis is considered ________. Movement to the left along the x-axis is considered _______.(positive; negative; positive; negative) Coordinate systems in two dimensions • Page 88 figure 3-6 • Question: which set of axes will give the correct answer? • In order to describe motion of an object in two dimensions, we use two axes to describe its movement. Determine resultant magnitude and direction • Example: Joe leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Joe's resulting displacement. 11 km 11 km θ R=? Using Pythagorean theorem to determine the magnitude of the resultant This procedure is restricted to the addition of two vectors that make right angles to each other. Using tangent function to determine a Vector's Direction • Note: The measure of an angle as determined through use of SOH CAH TOA is not always the direction of the vector. R2 = (5.0km)2 + (10km)2 R = 11 km Or at 26 degrees south of west Class work 10/15 do now • A 5.0-newton force and a 7.0-newton force act concurrently on a point. As the angle between the forces is increased from 0° to 180°, the magnitude of the resultant of the two forces changes from a. 0.0 N to 12.0 N b. 2.0 N to 12.0 N c. 12.0 N to 2.0 N d. 12.0 N to 0.0 N homework • 3.2 essay and 3.1 essay correction are due • Chapter 3 Project is due • Tonight’s homework – castle learning objectives • Class work – determine resultant of vectors using Pythagorean Theorem and trigonometry. • Resolving vectors into components • Presentation of project. Class work Page 90, Sample problem 3A Page 91, practice 3A #1-4 Practice 3A Answers 1. a. 23 km; b. 17 km, E 2. 45.6 m at 9.5 degrees E of N 3. 16 m at 22 degrees to the side of down field 4. 1.8 m at 49 degrees below the horizontal 10/16 do now • Which pair of concurrent forces could produce a resultant force having a magnitude of 10. Newtons? 1. 10. N, 10. N 2. 10. N, 30. N 3. 4.7 N, 4.7 N 4. 4.7 N, 50. N 10/17 Do now • Forces F1 and F2 act concurrently on point P, as shown in the diagram below. What is the equilibrant of F1 and F2? Include magnitude and direction. objectives • RESOLVING VECTORS INTO COMPONENTS • Adding vectors that are not perpendicular • Homework: castle learning RESOLVING VECTORS INTO COMPONENTS • In mathematics, the components of a vector are called projections. You draw a coordinate system by the tail of the vector. The x component is the projection of the vector along the x-axis, and the y component is the projection of the vector along the y-axis. y Ay A x Ax Appendix A, pp. 947-948 All Vectors can be broken into COMPONENTS • X-Y system of components: θ is measured from 0o • AX = A cos θ • AY = A sin θ • Example – vi = 5.0 m/s at 30° – vix = 5.0 m/s (cos 30°) = 4.33 m/s – viy = 5.0 m/s (sin 30°) = 2.5 m/s Any vector can be broken into unlimited sets of components depends on the coordinates Example 1. An arrow is shot from a bow at an angle of 25o above the horizontal with an initial speed of 45 m/s. Find the horizontal and vertical components of the arrow’s initial velocity. vx= vcosθ = (45 m/s) cos 25o vy = 41 m/s 25o vx 2. The arrow strikes the target with the speed of 45 m/s at an angle of -25o with respect to the horizontal. Calculate the horizontal and vertical components of the arrow’s final velocity. vx= vcosθ = (45 m/s) cos (-25o) = 41 m/s vy= vsinθ = (45 m/s) sin (-25o) = -19 m/s vx vy -25o What’s wrong? • A skyrocket travels 113 m at an angle of 82.4 degrees with respect to the ground and toward the south. What is the rocket’s horizontal displacement? • vx = v cosθ • vx = (113 m) cos 82.4o • vx = 14.9 m Pd. 4 - homework • Read Page 93 – sample example 3B • Page 94 - Practice 3B • Problem 3A and 3B – choose 2 questions from each set. • Page 94 – practice 3B answers 1. 95 km/h 2. 44 km/h 3. 21 m/s, 5.7 m/s 4. 0 m, 5 m 5. 110 m, -53 m 6. 19.8 m, -11.7 m 7. 2.4 m/s/s, -0.77 m/s/s Adding vectors that are not perpendicular Add the components of the original displacement vectors to find two components that form a right triangle with the resultant vector. Adding vectors algebraically 1. Select a coordinate system, draw a sketch of the vectors to be added, and label each vector. 2. Find the x and y components of all vectors. Ax=AcosθA; Ay=AsinθA; Bx=BcosθB; By=BsinθB 3. Find the x and y component of the resultant vector: Rx=Ax + Bx; Ry=Ay + By 4. Use the Pythagorean theorem to find the magnitude of the resultant vector. R 2 Rx Ry 2 5. Use tan-1 function to find the angle the resultant vector makes with the x-axis. Ry 1 R tan 6. Evaluate your answer. ( ) Rx Sample Example 3C • A hiker walks 25.5 km from her base camp at 35o south of east. On the second day, she walks 41.0 km in a direction 65o north of east at which point she discovers a forest ranger’s tower. Determine the manignitude and direction of her resultant displacement between the base camp and the ranger’s tower. R d2 35o d1 65o Class work • Page 97 – 3C • Section 3-2 review – vector operations 10/11 do now • A car travels down a road at a certain velocity, vcar. The driver slows down so that the car is traveling only half as fast as before. Which of the following is the correct expression for the resultant velocity? a. 2vcar c. -½vcar b. ½ vcar d. -2vcar