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10/11 do now •2 nd and 3 rd period: 3-1 diagram skills •9 th period: page 113 #1-13

10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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Page 1: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

10/11 do now

• 2nd and 3rd period: 3-1 diagram skills

• 9th period: page 113 #1-13

Page 2: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th 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

Page 3: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

3.2 vector operations

Objectives1. 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.

Page 4: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

• 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 y-axis is considered ________. Movement to the left along the x-axis is considered _______.(positive; negative; positive; negative)

Page 5: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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.

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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

R = ?θ

Page 7: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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.

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Using tangent function to determine a Vector's Direction

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• 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

Page 10: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

Class work

Page 11: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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

Page 12: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

homework

• 3.2 essay and 3.1 essay correction are due• Chapter 3 Project is due

• Tonight’s homework – castle learning

Page 13: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

objectives

• Class work – determine resultant of vectors using Pythagorean Theorem and trigonometry.

• Resolving vectors into components

• Presentation of project.

Page 14: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

Class work

Page 90, Sample problem 3APage 91, practice 3A #1-4

Practice 3A Answers1. a. 23 km; b. 17 km, E2. 45.6 m at 9.5 degrees E of N3. 16 m at 22 degrees to the side

of down field4. 1.8 m at 49 degrees below the

horizontal

Page 15: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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

Page 16: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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.

Page 17: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

objectives

• RESOLVING VECTORS INTO COMPONENTS• Adding vectors that are not perpendicular

• Homework: castle learning

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• 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.

RESOLVING VECTORS INTO COMPONENTS

x

y

AyA

Ax

Appendix A, pp. 947-948

Page 19: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

• 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

All Vectors can be broken into COMPONENTS

Page 20: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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.

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/svx

25ovy 45 m/s

-25o

45 m/s

vx

vy

vx= vcosθ = (45 m/s) cos (-25o) = 41 m/s

vy= vsinθ = (45 m/s) sin (-25o) = -19 m/s

Page 21: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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

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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 answers1. 95 km/h2. 44 km/h3. 21 m/s, 5.7 m/s4. 0 m, 5 m5. 110 m, -53 m6. 19.8 m, -11.7 m7. 2.4 m/s/s, -0.77 m/s/s

Page 23: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

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.

Page 24: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

Adding vectors algebraically1. 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.

5. Use tan-1 function to find the angle the resultant vector makes with the x-axis.

6. Evaluate your answer.

22yx RRR

)(tan 1

x

yR R

R

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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.

35o

65o

d2

d1

R

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Page 27: 10/11 do now 2 nd and 3 rd period: 3-1 diagram skills 9 th period: page 113 #1-13

Class work

• Page 97 – 3C• Section 3-2 review – vector operations

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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