2 Dimensional Motion

• Trig Review

• Vectors

• Projectiles

Trigonometry ReviewTrigonometry Review

• Pythagorean TheoremPythagorean Theorem: method : method used to solve for all sides of a used to solve for all sides of a RIGHT triangle. – ONLY works for a RIGHT triangle. – ONLY works for a triangle with one 90triangle with one 90°° angle! angle!

• AA22 + B + B22 = C = C22 C is the C is the hypotenuse hypotenuse

- Side opposite - Side opposite 9090°angle°angle

**Visual Truth****Visual Truth**(next slide)(next slide)

famous 3,4,5 trianglefamous 3,4,5 triangle

3322 + 4 + 422 = 5 = 522

9 + 16 = 259 + 16 = 25

3

4

5

• If you know any 2 sides of If you know any 2 sides of a RIGHT triangle you can a RIGHT triangle you can solve for the thirdsolve for the third

2222 + 6 + 622 = C = C22

4 + 36 = C4 + 36 = C22

40 = C40 = C22

40 = C40 = C6.32 = C6.32 = C

22

66

Solve these triangles Solve these triangles for practicefor practice(Not Drawn To Scale)(Not Drawn To Scale)

77

1010

22

55 44

77

12.12.22 5.4

5.4 8.0

8.0

66

• If you are given the Hypotenuse… the triangle solves this way

a2 + b2 = c2

a2 = c2 – b2

a2= 152 – 32

a2 = 225 – 9a2 = 216a = 216a = 14.7

1515

33

aa

Solve For PracticeSolve For Practice

66 1010

88

1919 88

17.17.22

1515

30302266

Sine, Cosine, TangentSine, Cosine, TangentThese are used to solve for These are used to solve for

angles and sides of triangles.angles and sides of triangles.used for right trianglesused for right triangles

Opposite OOpposite O

Adjacent AAdjacent A

Hypotenuse H Hypotenuse H –– opposite 90°opposite 90°

Theta(symbol for angle) 0Theta(symbol for angle) 0

Depend on angle you are using

OOHH

AA

AA

OO

HH

Relationship of Sine, Relationship of Sine, Cosine & Tangent to a Cosine & Tangent to a

Right TriangleRight Triangle•Sin 0 = oppositeSin 0 = opposite

HypotenuseHypotenuse•Cos 0 = AdjacentCos 0 = Adjacent

HypotenuseHypotenuse

•Tan 0 = OppositeTan 0 = Opposite

AdjacentAdjacent

•Sin 0 = OSin 0 = O

HH

•Cos 0 = ACos 0 = A

HH

•Tan 0 = OTan 0 = O

AA

S O H C A H T O AS O H C A H T O AS O H C A H T O AS O H C A H T O A

We use this to find a We use this to find a SIDESIDE……

Sin 30°= opp/ hyp Sin 30°= opp/ hyp 20/H20/H

Sin 30° = 20/HSin 30° = 20/H

H = 20/ Sin 30°H = 20/ Sin 30°

H = 20/ .5H = 20/ .5

H = 40

2200

HH

30°30°

AA 1111

40°40°

Cos 40°= adj / Cos 40°= adj / hyphyp

Cos 40°= A / 11Cos 40°= A / 11

Cos 40° (11) = Cos 40° (11) = AA

0.766 (11) = A0.766 (11) = A

8.4 = A8.4 = A8.4 = A8.4 = A

Example 2Example 2

We use this to find anWe use this to find an ANGLEANGLE

551010

Sin = opp/ hyp Sin = opp/ hyp 5/105/10Sin = .5Sin = .5

= 30°= 30° = 30°= 30°

Example 2Example 2

44

5533

Cos = adj/ hyp Cos = adj/ hyp 4/54/5

Cos = .8 Cos = .8

= = 36.9°36.9°

= = 36.9°36.9°

Example 3Example 3

33

4455

tan = opp/adj tan = opp/adj

Tan Tan = .75= .75

= 36.9°= 36.9° = 36.9°= 36.9°

Page 118 D

Solve for side or angle in each problem.

•A vector is an A vector is an arrow that arrow that

represents any represents any vector quantity. vector quantity. The arrow shows The arrow shows both both magnitudemagnitude

and and directiondirection

QuantitieQuantitiess

QuantitieQuantitiess

Magnitude Magnitude ONLYONLY

Magnitude Magnitude ONLYONLY

Magnitude Magnitude AND AND

directiondirection

Magnitude Magnitude AND AND

directiondirection

• Vector quantities:Vector quantities:FORCEFORCE

DISPLACEMENTDISPLACEMENTACCELERATIONACCELERATION

VELOCITYVELOCITY

• Magnitude is drawn to scaleMagnitude is drawn to scale• Direction is drawn in the 0Direction is drawn in the 0°°-360-360°°

contextcontext(north,south,east,west)(north,south,east,west) or as or as an interior angle to a major axis.an interior angle to a major axis.

Resultant Vector

This is the sum of all given vectors

It is one vector that represents all given vectors.

•Resultant vector in one dimension.

•Resultant Vector in 2 Dimensions.

Adding Perpendicular Adding Perpendicular VectorsVectors

• We can use the We can use the Pythagorean Theorem to Pythagorean Theorem to find the resultant vector find the resultant vector

of two vectors that form a of two vectors that form a 9090°. We can find angles of °. We can find angles of the resultant by using sin, the resultant by using sin,

cos and tangent cos and tangent functions.functions.

• Keep In mind when Keep In mind when solving and solving and giving answersgiving answers

interior angle with East would be 60interior angle with East would be 6000. . Subtract 60 from 90Subtract 60 from 9000. .

Example:Example:

00

9090

180180

270270

00oo

9090oo

3030oo

•The direction of The direction of the vector is 60the vector is 60°°

Problem:Problem:

1. An air plane flying toward 01. An air plane flying toward 0°° at at 90k/hr is being blown toward 9090k/hr is being blown toward 90°° at at 50km/hr. What is the resultant 50km/hr. What is the resultant velocity and direction of the plane?velocity and direction of the plane?

VVpp = 90km/hr, = 90km/hr, 00°°

VVww= 50km/hr, = 50km/hr,

90°90°

VVpp = 90km/hr, = 90km/hr, 00°°

VVww= 50km/hr, = 50km/hr,

90°90°

aa22 + b + b22 = c = c22

tan 0 = tan 0 = op/adjop/adj

aa22 + b + b22 = c = c22

tan 0 = tan 0 = op/adjop/adj

909022 + 50 + 5022 = = cc22

909022 + 50 + 5022 = = cc22

103 kph = C103 kph = C103 kph = C103 kph = C

tan 0 = 90/50tan 0 = 90/50

tan 0 = 1.8tan 0 = 1.8

0 = 60.90 = 60.9°°

tan 0 = 90/50tan 0 = 90/50

tan 0 = 1.8tan 0 = 1.8

0 = 60.90 = 60.9°°

direction= direction=

29.129.1°°

5050

9900

Practice Problems p.91• 1

• 2

• 3

Additional PracticeAdditional Practice

40°

1515mm

20m20m

5m5m

Find Find magnitudemagnitude and and directiondirection for object to for object to

return to originreturn to origin

Solution

• 9.7 m

• 620 north of west = • 280 west of north

• 3320

8.5

4.6

9.7m

Components of VectorsComponents of Vectors

We have seen We have seen 2 or more vectors2 or more vectors represented by a represented by a single resultantsingle resultant vector. – It is also possible to vector. – It is also possible to beginbegin with one vectorwith one vector and think of it as a and think of it as a resultantresultant of 2 vectors of 2 vectors..

These 2 vectors are known as These 2 vectors are known as components of the vector. – We will components of the vector. – We will often refer to them as the vertical often refer to them as the vertical component and horizontal component and horizontal component.component.

• Finding component vectors is Finding component vectors is called VECTOR RESOLUTIONcalled VECTOR RESOLUTION

• Consider this sled…Consider this sled…

30

FFhh

FFvv

FF

58 N

We find FWe find Fvv and F and Fhh by by using trig…using trig…

Sin 0 = FSin 0 = Fv v /F/F

FFvv = F sin 0 = F sin 0

FFvv = 58N(sin = 58N(sin

30°)30°)

FFvv = 58N(0.5) = 58N(0.5)

FFvv = 29N = 29NFFvv = 29N = 29N

Cos 0 = FCos 0 = Fhh/ /

FFFFhh = F cos 0 = F cos 0

FFhh = 58N (cos = 58N (cos

30°)30°)FFhh = 58N(.8660)

= 58N(.8660)

FFhh = 50.2N = 50.2NFFhh = 50.2N = 50.2N

Practice p.94

• 1

• 2

• 3

• 4

Resolving vectors1. How fast must a truck travel to stay

beneath an airplane that is moving 105 km/h at an angle of 25to the ground?

2. What is the magnitude of the vertical component of the velocity of the plane in item 1?

3. A truck drives up a hill with a 15incline. If the truck has a constant speed of 22 m/s, what are the horizontal and vertical components of the truck’s velocity?

4. What are the horizontal and vertical components of a cat’s displacement when it has climbed 5 m directly up a tree?

Solutions

• 1) 95.2 km/hr

• 2) 44.4km/hr

• 3) Vy = 5.7 m/s Vx = 21.3m/s

• 4) dv = 5m dh = 0m

• The sign +/- for FThe sign +/- for Fvv and F and Fhh can be found from the can be found from the

diagram…diagram…

•As the component As the component becomes more becomes more

perpendicular to the perpendicular to the direction of motion, the direction of motion, the decreasing component decreasing component

approaches zeroapproaches zero

REMEMBERREMEMBER

Problem…Problem…

• Solve for FSolve for Fvv and F and Fhh for the red for the red wagon…wagon…

100100NN45°45°

Radio FlyerRadio Flyer

Special Triangle

• Cos 450(100) =

• Sine 450(100) =

• 70.7N

•A wind with a velocity A wind with a velocity of 40km/hr blows of 40km/hr blows toward 30°. Find…toward 30°. Find…

•A) component of A) component of wind towards 90°wind towards 90°

•B) Component of B) Component of wind towards 0°wind towards 0°

Solution

• Given: 40km/hr, 300

•V90=20kph

•V00=34.6kph

Independence of Vector Independence of Vector QuantitiesQuantities

• Perpendicular vector quantities are Perpendicular vector quantities are independent of one anotherindependent of one another

Boat Crossing a RiverBoat Crossing a River

VVBB = 8m/s = 8m/s

VVRR= 3m/s= 3m/sResultant Resultant Vector= 8.5

Vector= 8.5

• The velocity of the boat does The velocity of the boat does NOT affect the velocity of the NOT affect the velocity of the

river!river!• The velocity of the RIVER does The velocity of the RIVER does

NOT affect the velocity of the NOT affect the velocity of the boat!boat!

If the river is 80m wide it will If the river is 80m wide it will take the boat 10 seconds to take the boat 10 seconds to

cross itcross itV= d/tV= d/t t= d/vt= d/v 80m/ 880m/ 8mm//ss = 10 sec= 10 sec

During those 10 seconds the boat During those 10 seconds the boat will move 30m downstreamwill move 30m downstream

V= V= d/td/t

d= Vtd= Vt 3m/s (10s)3m/s (10s) = 30m= 30m

River Crossing River Crossing QuestionsQuestions

a) Find the resultant velocity for the boatb) How long will it take to cross the river?c) How far down stream will the boat go?d) At what angle, measured from the

river bank, will the boats resultant velocity be?

e) If the boat driver wanted to go DIRECTLY across the river- how much should he aim the boat upstream?

River River width width

= 100m= 100m

6 m/s6 m/s

2 m/s2 m/s

ANSWERS!ANSWERS!

a)a) AA22 + B + B22 = R = R22 6 622 + 2 + 222 = R = R22 = 6.32m/s = 6.32m/s

b)b) V=d/t t= d/v t=100/6 = 16.67secV=d/t t= d/v t=100/6 = 16.67sec

c)c) V= d/t d= Vt d=2(16.67) = 33.34mV= d/t d= Vt d=2(16.67) = 33.34m

d)d) tan 0 = opp/adj tan 0 = 2/6 tan 0 = opp/adj tan 0 = 2/6 tan 0 = .333333 0 = 18.43 tan 0 = .333333 0 = 18.43 angle measured from bank angle measured from bank 90 – 18.43 = 90 – 18.43 = 71.5771.5700

e)e) Aim upstream Aim upstream 18.4318.4300 or 33.34m or 33.34m

Adding Vectors at Adding Vectors at Any AngleAny Angle

• If we have 2 or more vectors that are If we have 2 or more vectors that are NOT at 90NOT at 90°° to each other, we can still to each other, we can still find the Resultant of them.find the Resultant of them.

• THREE STEPSTHREE STEPS1.1. Each given Vector is resolved into its Each given Vector is resolved into its

perpendicular componentsperpendicular components2.2. Add the vertical components together, Add the vertical components together,

add the horizontal add the horizontal components together components together Keep in mind Keep in mind ++

3.3. The sum of these components will give the The sum of these components will give the perpendicular components of the resultant. We perpendicular components of the resultant. We solve for the resultant solve for the resultant w/ A w/ A22 + B + B22 = C = C22 and Sin, Cos, Tan. and Sin, Cos, Tan.

60°60° 10°10°

22 11

Identify Identify componentscomponents

FF1V1V FF1h1h FF2V2V FF2h2h

12N8N

FF1h 1h Cos 10°= F Cos 10°= F1h 1h /12N F/12N F1h1h= = (12N)Cos10° =11.8N(12N)Cos10° =11.8N

FF1V1V Sin10°= F Sin10°= F1V 1V / 12N / 12N FF1V1V=(12N)sin10° = 2.08=(12N)sin10° = 2.08

FF2h2h Cos60° = F Cos60° = F2h2h/8 F/8 F2h2h = = (8N)Cos60° = -4N(8N)Cos60° = -4N

FF2V2V Sin60° = F Sin60° = F2V2V/8 F/8 F2V2V = (8N) Sin60° = (8N) Sin60° = 6.9N = 6.9N

#2 #2

FF1h1h + F + F2h2h = 11.8 +(-4) = = 11.8 +(-4) = 7.8N7.8N

FF1V1V + F + F2V2V = 2.08 + 6.9 = = 2.08 + 6.9 = 8.98N8.98N RR

8.98N8.98N

7.8N7.8N 11.9N

49o

Practice Problem 3C Page 97

•1.

•2.

Solution• 1) 49m; 82.60 from the line of scrimmage

(7.40 from downfield)

• 2) 7.6km; 25.90

EquilibriumEquilibrium

• A state when the net force acting A state when the net force acting on an object is zero (all forces on an object is zero (all forces

cancel out)cancel out)

10N10N 10N10N

we say this is in equilibriumwe say this is in equilibrium

Tug of War… FTug of War… Fnetnet= 0…. No = 0…. No accelerationacceleration

• We say this object is in We say this object is in equilibriumequilibrium

• there is NO NET FORCE presentthere is NO NET FORCE present

33

44

55

solving this solving this graphicallygraphically

• Equilibrium occurs when Equilibrium occurs when an object is NOT moving. an object is NOT moving. • It can also occur when It can also occur when

something IS movingsomething IS moving

ThrustThrustDragDrag

LiftLift

WeighWeight t

Equilibrant ForceEquilibrant Force• a force that, when applied, a force that, when applied,

produces equilibriumproduces equilibrium• The equilibrant force is equal The equilibrant force is equal

in in sizesize to the to the resultantresultant force, force, but is in the but is in the oppositeopposite directiondirection

R= 10NR= 10N

300°300°

8N8NAA

6N6N

BB The resultant The resultant for for AA and and BB is is 10N at 300°10N at 300°

The The equilibrant equilibrant force is 10N force is 10N

at 120°at 120°

The The equilibrant equilibrant force is 10N force is 10N

at 120°at 120°

E = 10 N at E = 10 N at 120°120°

6N6N

BB

8N8N

AA

When Finding the When Finding the Equilibrant Force…Equilibrant Force…

1.1. Find the resultant force Find the resultant force for all given Vectorsfor all given Vectors

2.2.Equilibrant force is equal Equilibrant force is equal in size but opposite in in size but opposite in

direction to the resultantdirection to the resultant

Example problemExample problem• Two children are pulling on a

Christmas package. Betty pulls North at 100N, Bob pulls 200N east. Find the force a THIRD child applies that

puts the package in equilibrium.

BettBettyy

100100NN

BobBob

200N200N AA22 + B + B22 = C = C22 10010022 + 200 + 20022 = = RR22

R= 223.6R= 223.6

Tan 0 = Tan 0 = O/AO/A

= 200 /100= 200 /100

0 = 0 = 63°63°Equilibrant forceEquilibrant force = 223.6 N at = 223.6 N at

243°243°

Continued…

Betty Betty

100 N100 N

Bob Bob

200 N200 N

E = 223.6 N E = 223.6 N

at 243°at 243°

Problem #2• This sign has a weight of 168N and is in This sign has a weight of 168N and is in

equilibrium. Two ropes support this sign and are equilibrium. Two ropes support this sign and are fixed to a point. Each rope makes an angle of 22° fixed to a point. Each rope makes an angle of 22° with the top of the sign. Find the forces that exist with the top of the sign. Find the forces that exist

in the ropes. in the ropes.

SINE

22°22° 22°22°

GivenGiven

AA BB

168168

8484

22222222

FormulaFormula

Sin 0 = O/ Sin 0 = O/ HH

SolutionSolution

Sin 22° = FSin 22° = FAV AV / / AASin 22° = FSin 22° = FBVBV/ / BB= =

224N224N

84N84NBBAA8484

2222 2222

FFAHAH= - = -

FFBHBH = + = +

They will cancelThey will cancel

FFAVAV + F + FBVBV = 168N = 168N

FFAVAV = F = FBVBV (similar (similar triangles)triangles)

B = FB = FBVBV/ / Sin22° = Sin22° =

84 /Sin22°84 /Sin22°= 224 N= 224 N

Practice…Practice…1.1. A 300N sign is hung with two A 300N sign is hung with two

ropes. Both ropes are attached to ropes. Both ropes are attached to the same point and make an angle the same point and make an angle with the horizontal of 40°. The sign with the horizontal of 40°. The sign is in equilibrium. Find the force is in equilibrium. Find the force that each rope exerts on the sign.that each rope exerts on the sign.

2.2. Two forces are applied to the same Two forces are applied to the same object. object. Force 1Force 1 acts at 0° with acts at 0° with 100N, 100N, Force 2Force 2 acts at 35° with acts at 35° with 120N. Find the equilibrant force.120N. Find the equilibrant force.