9/24 Vectors!!!!

9/24 Vectors!!!!

Pick up Problem SetsPhysics HW 11 & 12 2D KinematicsAP Vectors More Practice

Tests are not graded. But I did sleep in a Holiday Inn last night.Bonus Opportunity

9/25 Vectors!!!! Pick up a piece of graph paperYesterday you picked upPhysics HW 11 & 12 2D KinematicsAP Vectors More Practice

Tests are graded, but not entered.Bonus Opportunity

9/26 Vectors!!!!

Yesterday we worked thorough ex 3 and 5. Today you will complete the pirate vector labYou will work in pairs. Supplies: Book cover Metric ruler Protractor Colored Pencils or markers Lab directions There are 12 clues Test Corrections if below a 70 AP Bonus opportunities

9/29 Vectors!!!!

Friday we did the pirate lab. See me if you were absent.Today we will look at the Girl Scout problem (this is typically

referred to as the river boat problem) and we will consider Force as a vector (push or pull)

Test Corrections if below a 70 AP Bonus opportunities

9/30 Vectors!!!!

Get the calculator that matches your blue tag on your deskMonday we did the Girl Scout problem (ex 6) and two Forces at right angles (ex 7)Today we will look at components, multiple vectors and learn the component method.If time we will look at problems that require correction of motion (ie 15,16, & 19 on HW

#12)

Test Corrections if below a 70 (This mon-thur) AP Bonus opportunities

Pirate Vector LabSolve graphically (using a metric ruler and a protractor)Read the story and follow the directions exactly.Additional instructions: you mustInclude arrows and units for each vectorDraw your resultant in a different colorDetermine resultant magnitude and direction graphically (using a metric ruler and a protractor)Record answer (with units) and secret rule on blank area of your map. Box your answer.Rubric:Name-Scale 1

Pirate Vector Lab Rubric:

1 Name-Scale

Vectors and Scalars

AP Physics B

Scalar

A SCALAR is ANY quantity in physics that has MAGNITUDE, but NOT a direction associated with it.

Magnitude – A numerical value with units.

Scalar Example

Magnitude

Speed 20 m/s

Distance 10 m

Age 15 years

Heat 1000 calories

Vector

A VECTOR is ANY quantity in physics that has BOTH MAGNITUDE and DIRECTION.

Vector Magnitude & Direction

Velocity 20 m/s, N

Acceleration 10 m/s/s, E

Force 5 N, West

Faxv

,,,Vectors are typically illustrated by drawing an ARROW above the symbol. The arrow is used to convey direction and magnitude.

Applications of Vectors

VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add them.

Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started?

54.5 m, E 30 m, E+

84.5 m, E

Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.

Applications of Vectors

VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT.

Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started?

54.5 m, E

30 m, W-

24.5 m, E

Non-Collinear VectorsWhen 2 vectors are perpendicular, you must use

the Pythagorean theorem.

95 km,E

55 km, N

Start

Finish

A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT.The hypotenuse in Physics

is called the RESULTANT.

The LEGS of the triangle are called the COMPONENTS

Horizontal Component

Vertical Component kmc

c

bacbac

8.10912050

5595Resultant 22

22222

BUT……what about the direction?In the previous example, DISPLACEMENT was asked for

and since it is a VECTOR we should include a DIRECTION on our final answer.

NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE.

N

S

EW

N of E

E of N

S of W

W of S

N of W

W of N

S of E

E of S

N of E

BUT…..what about the VALUE of the angle???Just putting North of East on the answer is NOT specific enough

for the direction. We MUST find the VALUE of the angle.

N of E

55 km, N

95 km,E

To find the value of the angle we use a Trig function called TANGENT.

30)5789.0(

5789.095

55

1

Tan

sideadjacent

sideoppositeTan

109.8 km

So the COMPLETE final answer is : 109.8 km, 30 degrees North of East

What if you are missing a component?Suppose a person walked 65 m, 25 degrees East of North. What

were his horizontal and vertical components?

65 m25

H.C. = ?

V.C = ?

The goal: ALWAYS MAKE A RIGHT TRIANGLE!

To solve for components, we often use the trig functions sine and cosine.

EmCHopp

NmCVadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,47.2725sin65..

,91.5825cos65..

sincos

sinecosine

Example 1A bear, searching for food wanders 35 meters east then 20 meters north.

Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.

35 m, E

20 m, N

12 m, W

6 m, S

- =23 m, E

- =14 m, N

23 m, E

14 m, N

3.31)6087.0(

6087.23

14

93.262314

1

22

Tan

Tan

mR

The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST

R

Example 2A boat moves with a velocity of 15 m/s, N in a river which

flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

15 m/s, N

8.0 m/s, W

Rv

1.28)5333.0(

5333.015

8

/17158

1

22

Tan

Tan

smRv

The Final Answer : 17 m/s, @ 28.1 degrees West of North

Example 3A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate

the plane's horizontal and vertical velocity components.

63.5 m/s

32

H.C. =?

V.C. = ?

SsmCVopp

EsmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,/64.3332sin5.63..

,/85.5332cos5.63..

sincos

sinecosine

Example 4A storm system moves 5000 km due east, then shifts course at 40

degrees North of East for 1500 km. Calculate the storm's resultant displacement.

NkmCVopp

EkmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,2.96440sin1500..

,1.114940cos1500..

sincos

sinecosine

5000 km, E

40

1500 km

H.C.

V.C.

5000 km + 1149.1 km = 6149.1 km

6149.1 km

964.2 kmR

91.8)364.0(

157.01.6149

2.964

14.62242.9641.6149

1

22

Tan

Tan

kmR

The Final Answer: 6224.14 km @ 8.91 degrees, North of East

Practice adding vectors Ex 5 Draw the following to scale:

Indicate scale. Make sure you draw tail to tip for the two vectors. Indicate the resultant with a dotted line. The tip of

the resultant meets the tip of the second vector. You walk 5 m to the north and then 8 m east Determine your resultant displacement graphically. What did you get? I got 9.7m. Determine your resultant displacement using the

Pythagorean formula. Did you get 9.43 m?

Practice adding vectors Ex 5 Draw the following to scale:

Indicate scale. Make sure you draw tail to tip for the two vectors. Indicate the resultant with a dotted line. The tip of

the resultant meets the tip of the second vector. You walk 5 m to the north and then 8 m east Determine your resultant displacement graphically. What did you get? I got 9.7m. Determine your resultant displacement using the

Pythagorean formula. Did you get 9.43 m?

Ex 5

8 m

5 m

Ex 5

“θ” or Theta, is any unknown angle but in this case it is the angle between the two vectors

Use a protractor to determine the angle “θ” of your resultant.

Place the protractor along the axis of the initial vector. Take the reading in reference to the resultant.

Refer to diagram on next slide

Ex 5

θ

5 m

8 m

Ex 5

What value did you get with your protractor? I got 59° This would be described as 59° east of north

since the resultant is east of the north axis. How could you do this mathematically? Use SOH CAH TOA Label the sides in reference to θ. I would suggest using tan since you calculated the

hypotenuse.

Ex 5

θAdj5m

Opp 8m

Ex 5

Using Tan: Tan θ = opp/adj Tan θ = 8m/5m Tan θ = 1.6 THIS IS NOT THE ANGLE!!! Find inverse tan. Enter 2nd tan (1.6) enter You should get 58.0° The direction is still East of

North This is in the same range as what we got with

the protractor.

Example 6 A girl scout elects to swim across the river. The

river is 37.5 meters wide. A current flows downstream at a rate of 0.66 m/s. If she initially swims towards the boy scout camp (directly cross the river) at a rate of 1.73 m/s, how long will it take her to reach the far shore?

Remember what the question asks. How long does it take her to swim across? To solve for time, what do we need to know? Use velocity and displacement but only in

reference to crossing the river.

] 37.5m

1.73 m/s

v = d/t t = d/v t = 37.5m÷1.73m/s t = 21.7 s

] 37.5m

1.73 m/s

Where exactly does the girl scout end up on the far shore?

What do we need to know? To determine displacement we need

velocity and time but only in reference to downstream.

] 37.5m

1.73 m/s

Example 6

To find where she ends up, what is the downstream velocity?

0.66m/s What is the time? 21.7 sec Time is the same for both cross stream and

downstream. 14.3 m downstream

0.66m/s

When working with multiple vectors remember they are independent of one another although they have a net effect.

In the case of the girl scout, her overall (think resultant) velocity and direction changed.

Do you know how to solve for the apparent resultant velocity and direction?

Resultant velocity? Make sure you only use velocity vectors! c2 = 1.732 + 0.662

c = 1.85 m/s Which angle for direction? tan θ = 1.73/0.66 = 2.62… θ = 69.1° vr = 1.85 m/s at 69.1° downstream in respect to shore

0.66m/s

1.73 m/s c

θ

Fix next slide

What angle should the girl scout enter the water upstream to end up at the boy scout camp??

Expand this

0.66m/s

1.73 m/s c

θ

How are force vectors drawn? Tail to tail

Resultant Force

The resultant force of concurrent forces can be calculated using the parallelogram method or using the component method.

We will use the parallelogram method first.

F1 and F2 are: concurrent forces. FR is the Resultant force. FR’ is the equilibrant force. The mathematical method used with such

diagram is called the parallelogram method.

EX 7 One force of 7.75 N is exerted on point R in the direction of direct south. A second force of 15.50 N is exerted on R in the direction of 90.0 west of south. What is the magnitude and direction of the resultant force in reference to the larger force?

7.75 N

15.50 N

(15.50 N)2 + (7.75 N)2 = FR2

17.3 N = FR

tan θ = 7.75 N / 15.50 Nθ = 26.6° south of west

θ

EX 8 A sled is being pulled up and to the right at an angle of 27.00 to the horizontal. If the person is pulling with a force of 165.0 N, what are the horizontal and vertical components of the force?

165.0 N

Determine Y using sinO= SH = sin(27)(165 N)A = 74.91 N

27ºDetermine X using cosA= CH = cos(27)(165 N)A = 147.02 N

What happens if the concurrent vectors are not at right angles? We will the component method We will solve for resultant in respect to the X

axis in Quadrant I

What are the values for X & Y in each quadrant ?X is

Y is

X is

Y is

X is

Y is

X is

Y is

What are the values for X & Y in each quadrant ?

X is +

Y is +

X is -

Y is -

X is +

Y is -

X is -

Y is +

Ex 8 Two forces act on point Q. The first, of 435 N, heads 18.0° north of east. The second, of 220 N, is directed due west. Determine magnitude and direction of the resultant force using the component method.


435 N18º

220 NWest


435 N18º

220 NWest

180º

Describe any angles outside of quad 1 in reference to x axis of quad 1

Ex 9 Describe any angles outside of quad 1 in reference to x axis of quad 1

magnitude

direction Direction in relation to

QI435 N 18° N of E 18° N of E

220 N W Use 180°

EX 9 Determine the components and total

Vector X component

Cos H

Y component

Sin H

435 N at 18° N of E

220 N at 180°CCW to E

Total

EX 9 Determine the components

Vector X component

Cos H

Y component

Sin H

435 N at 18° N of E 413.71 N 134.42 N

220 N at 180°CCW to E -220.00N 0 N

Total 193.71N 134.42 N

Draw components on graph, using signs (+ or -) to determine quadrant

Calculate FR with total X and Y Calculate Angle in reference to X using tan. (absolute

values)

134.42 N

193.71 N

FR


435 N18º

220 NWest

Example 9

FR = √(134.42 N)2 + (193.71N)2

FR = 235.78 N Tan = 134.42 N/193.71 = 34.76°N of E FINAL ANSWER The FR is 235.78 N at 34.76°N of E

Ex. 10 What is the net force on the rope?

Socks Blue Patches

Ex 10 A true story

Once upon a time 3 dogs (Socks, Blue, and Patches) found an old smelly rope with a mass of 0.65 kg. They all began to tug on it at the same time. Socks pulled on the rope with a magnificent force of 185N at 15° S of W. Blue, thinking he didn’t have to apply as much force as the other smaller dogs, pulled on the rope with a force of 111N at 35° W of N. Little Patches, eager to make up for his diminutive size, pulled on the rope with 65 N at 60° N of E. In what direction did the rope accelerate across the room? What is the moral of this story?

How do we solve this?

What do we need to know to determine direction of the rope?

Net Force on the rope. There’s too many!!!! Diagram the forces on an X-Y coordinate system Describe Angles over 90 in reference to Quad I List knowns and solve for X and Y components Still use cos and sin (respectively) but use angles in

reference to Quad I

111 N: 35 + 90°= 125°

185N: 15°+ 180° = 195°

magnitude

direction Direction in relation to

QI65 N 60° N of E 60° N of E

111 N 35° W of N Use 125°

185 N 15° S of W Use 195

Determine the components

Force Vector X component

Cos H

Y component

Sin H

65 N at 60° N of E



Total



Cos H

Y component

Sin H

65 N at 60° N of E 32.5 N 56.3 N

111 N at 125°CCW to E -63.7 N 90.9 N

185 N at 195°CCW to E -179 N -47.9 N

Total



Cos H

Y component

Sin H

65 N at 60° N of E 32.5 N 56.3 N

111 N at 125°CCW to E -63.7 N 90.9 N

185 N at 195°CCW to E -179 N -47.9 N

Total -210 N 99.3 N

Draw components on graph, using signs (+ or -) to determine quadrant

Calculate FR with total X and Y Calculate Angle in reference to X using tan. (absolute

values)

99 N

210 N

FR

Example 10

FR = √2102 + 99.32

FR = 232 N Tan = 99.3/210 = 25.3°N of W FINAL ANSWER The rope moves with a force of 232 N at

25.3° N of W

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9/24 Vectors!!!!