VECTORVECTORSS v

Fp

Scalar quantities – quantities that can be completely described by a number with the

appropriate units. (They have magnitude only. )Such as length, mass, speed, energy, temperature…

{notation : Δd, m, v, E, to …}

Vector quantities – quantities that require both a magnitude AND a direction for their

complete description. Such as displacement, velocity, force… {Notation : Δd, v, F… }

In a diagram, vectors are represented by arrowswhich point in the direction of the vector and whose length represents the magnitude.

TRIGONOMETRY REFRESHERTRIGONOMETRY REFRESHERANATOMY OF A RIGHT TRIANGLE

Hypotenuse (HYP)

θReference angle

(θ, THETA)

Right angle Adjacent (ADJ)

Opposite(OPP)

TRIGONOMETRY REFRESHERTRIGONOMETRY REFRESHERTRIGONOMETRIC FUNCTIONS

θ

HYPOPP

ADJ

cos ADJ

HYPcos ADJ

HYP tan OPP

ADJtan OPP

ADJ

SOH CAH TOA

sin OPP

HYP

TRIGONOMETRY REFRESHERTRIGONOMETRY REFRESHERIII

IIIIV

All are +Sin is +

Cos is +Tan is +

0o

90o

180o

270o

360o

For a vector, V, when θ is to the positive x-axis :

θ

V

Vx

Vy

sin yVV

cos xVV

tan y

x

VV

Resolving vectors into their Resolving vectors into their componentscomponents

{finding the x- and y components of a vector}{finding the x- and y components of a vector}

To find the x-component

To find the y-component

cos xVV

sin yVV Vy = V sin θ

Vx = V cos θ

Sample problemsSample problems Resolve this velocity into its components.Resolve this velocity into its components.

V = 15 m/s at 60.0o

V = 15 m/s

θ = 60.0o

GIVEN V

θ

Vy

Vx

Vy = V sin θ

Vy = 15 m/s sin 60.0o

Vx = V cos θ

Vx = 15 m/s cos 60.0o

Vx = + 7.5 m/s Vy = + 13 m/s

Sample problemsSample problems Resolve this velocity into its components.Resolve this velocity into its components.

V = 80.0 m/s at 140.0o

θV

Vy

Vx

V = 80.0 m/s

θ = 140.0o

GIVEN

Vy = V sin θ

Vy = 80.0 m/s sin 140.0o

Vy = + 51.4 m/s

Vx = V cos θ

Vx = 80.0 m/s cos 140.0o

Vx = -61.3 m/s

GRAPHIC RESOLUTIONGRAPHIC RESOLUTIONUse a ruler and protractor to draw the two previous vectors to scale. Graph paper is helpful in establishing axes but it is not needed to complete this assignment.

Draw the perpendicular to the x-axis and label the components.

Measure the components in cm and use your scale to convert back to the appropriate units.

GRAPHIC RESOLUTIONGRAPHIC RESOLUTIONFor V = 15 m/s at 60.0o, if I let 1 cm = 0.5 m/s ……

I draw V 30 cm long at 60.0o Vy

Vx

Draw the perpendicular to the x-axis

Label the components

Measure the components and convert

V

θ

xm scmV cm1550 0 5

1. ( ). / = +7.750 m/s

y

m s

cmV cm26 25 0 5

1. ( ). / = +13.12 m/s

15 301

0 5m s cmcm

m s/ ( )

. /

Your turn

Do both vectors and submit

TRIGONOMETRIC VECTOR TRIGONOMETRIC VECTOR ADDITIONADDITION

(FINDING RESULTANTS)(FINDING RESULTANTS)1. Resolve vectors into x- and y-components.

2. Add all of the x-components to find the x-component of the resultant.

3. Add all of the y-components to find the y-component of the resultant.

4. Use tan-1 (arctan) to find the direction, θR.

The signs of the components determine the quadrant.

5. Use cos θR or sin θR to find the magnitude.

θrefθref

θref θref 0o

360o

270o

180o

90o

θR = θrefθR = 180o - θref

θR = 180o + θrefθR = 360o - θref

θR

θRθR

θR

III

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ADVANCE TO SAMPLE PROBLEM

SAMPLE PROBLEM (from sheet)SAMPLE PROBLEM (from sheet)

V1 = 320 N at 240.0oV1 = 320 N at 240.0o V2 = 35 N at 30.0oV2 = 35 N at 30.0o

R = V1 + V2Given

V1 = 320 N

V2 = 35 N

θ1 = 240.0o

θ2 = 30.0o

θ2θ1

V1

V2

Rx = V1x + V2x

R = V1 + V2

Rx = V1 cos θ1 + V2 cos θ2

Rx = 320 N cos 240.0o + 35 N cos 30.0o

Rx = (-160 N) + (+30 N)

Rx = -130 N

Find the x-component

From

R = V1 + V2

Ry = V1y + V2y

Ry = V1 sin θ1 + V2 sin θ2

Ry = 320 N sin 240.0o + 35 N sin 30.0o

Ry = (-280 N) + (+18 N)

Ry = -260 N

Find the y-component

From

Rx = -130 N

Ry = -260 N

R = Rx + Ry

Rx

RyRy

R

θRDraw the components and resultant

Find the direction, θR

Tan RRR

y

x

Using

R

y

x

R

R 1

tan ( )

1 260130tan ( )N

N

Rx

RyRy

R

θR

Your calculator only gives youthe reference angle to the x-axis

θref = 63o

θref

Since the x- and y-components are both negative,R is in Quadrant III 63o beyond 180o on the x-axis.

θR = 180o + 63o

θR = 243o

FromR

NN

1 260

130tan ( )

Rx

RyRy

R

θR

θref

Find the magnitude, R

--- using the sin θR

R N

o

260

243sin= 290 N

R

yRR

sin

R Ry

R

sin

Rx

RyRy

R

θR

θref

Find the magnitude, R

--- using the cos θR

R N

o

130

243cos= 290 N

RxRR

cos

R Rx

R

cos

Rx

RyRy

R

θR

θref

Find the magnitude, R

--- using pythagorean theorem

2 2 2

R R Rx y

R R Rx y 2 2

R N N 2 2

130 260( ) ( )

R = 290 N

Rx

RyRy

R

θR

θref

And so, the final answer is

R = 290 N at 243o

Do all of the practice problems on the sheetto prepare yourself for your next quiz.

GRAPHIC VECTOR ADDITIONGRAPHIC VECTOR ADDITIONWhen adding vectors graphically :

1. Draw the vectors to scale.2. Move one of the vectors so that it is “head-to-toe” with the other. (be sure to maintain its magnitude and direction)

3. Draw the resultant from the origin to the head of the last vector you moved.

4. Measure the resultant and use your scale to convert back to the appropriate units.

5. Use a protractor to measure the direction.

GRAPHIC VECTOR ADDITIONGRAPHIC VECTOR ADDITION

Consider a plane flying with a velocity of 220 m/sat 170o in a wind blowing at 45 m/s at 270o.What is the ground velocity of the plane?

The problem is to find the resultant, Vg , of the velocity of the plane, Vp ,in the air and the velocity of the wind, Vw.

Vg =Vp + Vw

Vp = 220 m/s at 170o Vw = 45 m/s at 270o

GRAPHIC VECTOR ADDITIONGRAPHIC VECTOR ADDITION

Vp

VwVwVg

Let 1 cm = 10 m/sVp = 220 m/s (1cm /10 m/s) = 22 cm

Vw = 45 m/s (1cm /10 m/s) = 4.5 cmVg = 21.70 cm (10m/s / 1cm) = 217.0 m/s

The protractor measures 3o beyond 180o so θg =183o

θp

θg

BONUS PROBLEMBONUS PROBLEMA pilot wishes to maintain a ground velocityof 150.0 m/s at 270o and the wind is blowingat 50.0 m/s at 120o. What air speed must hehave and at what bearing (direction)?

Hint : This is actually a vector subtraction.Just as in math, subtraction is the same asadding a negative. A negative vector pointsin the opposite direction. Add or subtract180o to the direction of the vector to be subtracted and then add as usual.

(solve both graphically and trigonometrically to earn maximum credit)

THIS IS THE END OF VECTORS