2 Dimensional Motion Trig Review Vectors Projectiles

Trigonometry Review Pythagorean Theorem: method used to solve for all sides of a RIGHT triangle. – ONLY works for a triangle with one 90° angle!Pythagorean Theorem: method used to solve for all sides of a RIGHT triangle. – ONLY works for a triangle with one 90° angle! A 2 + B 2 = C 2 C is the hypotenuseA 2 + B 2 = C 2 C is the hypotenuse - Side opposite 90°angle - Side opposite 90°angle

**Visual Truth** (next slide) famous 3,4,5 triangle = =

If you know any 2 sides of a RIGHT triangle you can solve for the thirdIf you know any 2 sides of a RIGHT triangle you can solve for the third = C = C 2 40 = C 2 40 = C 40 = C 6.32 = C 2 6

Solve these triangles for practice (Not Drawn To Scale)

If you are given the Hypotenuse… the triangle solves this way a 2 + b 2 = c 2 a 2 = c 2 – b 2 a 2 = 15 2 – 3 2 a 2 = 225 – 9 a 2 = 216 a = 216 a = a

Solve For Practice

Sine, Cosine, Tangent  These are used to solve for angles and sides of triangles.  used for right triangles Opposite O Adjacent A Hypotenuse H – opposite 90° Theta(symbol for angle) 0 Depend on angle you are using

O H A A O H

Relationship of Sine, Cosine & Tangent to a Right Triangle Sin 0 = oppositeSin 0 = opposite Hypotenuse Hypotenuse Cos 0 = AdjacentCos 0 = Adjacent Hypotenuse Hypotenuse Tan 0 = OppositeTan 0 = Opposite Adjacent Adjacent Sin 0 = OSin 0 = O H Cos 0 = ACos 0 = A H Tan 0 = OTan 0 = O A

S O H C A H T O A

We use this to find a SIDE… Sin 30°= opp/ hyp 20/H Sin 30° = 20/H H = 20/ Sin 30° H = 20/.5 H = H 30°

A 11 40° Cos 40°= adj / hyp Cos 40°= A / 11 Cos 40° (11) = A (11) = A 8.4 = A Example 2

We use this to find an ANGLE 5 10 Sin = opp/ hyp 5/10 Sin =.5 = 30° = 30°

Example Cos = adj/ hyp 4/5 Cos =.8 = 36.9° = 36.9°

Example tan = opp/adj Tan =.75 = 36.9° = 36.9°

Page 118 D Solve for side or angle in each problem.

A vector is an arrow that represents any vector quantity. The arrow shows both magnitude and directionA vector is an arrow that represents any vector quantity. The arrow shows both magnitude and direction

QuantitiesQuantities Magnitude ONLY Magnitude AND direction

Vector quantities:Vector quantities:FORCEDISPLACEMENTACCELERATIONVELOCITY Magnitude is drawn to scaleMagnitude is drawn to scale Direction is drawn in the 0°-360° context (north,south,east,west) or as an interior angle to a major axis.Direction is drawn in the 0°-360° context (north,south,east,west) or as 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 Vectors We can use the Pythagorean Theorem to find the resultant vector of two vectors that form a 90°. We can find angles of the resultant by using sin, cos and tangent functions.We can use the Pythagorean Theorem to find the resultant vector of two vectors that form a 90°. We can find angles of the resultant by using sin, cos and tangent functions.

Keep In mind when solving and giving answersKeep In mind when solving and giving answers interior angle with East would be Subtract 60 from Example: o0o0o0o 90 o 30 o The direction of the vector is 60°The direction of the vector is 60°

Problem: 1. An air plane flying toward 0° at 90k/hr is being blown toward 90° at 50km/hr. What is the resultant velocity and direction of the plane? V p = 90km/hr, 0° V w = 50km/hr, 90° a 2 + b 2 = c 2 tan 0 = op/adj = c kph = C tan 0 = 90/50 tan 0 = = 60.9° direction= 29.1° 50 90

Practice Problems p

Additional Practice 40° 15m 20m 5m Find magnitude and direction for object to return to origin

Solution 9.7 m 62 0 north of west =  28 0 west of north m 

Components of Vectors  We have seen 2 or more vectors represented by a single resultant vector. – It is also possible to begin with one vector and think of it as a resultant of 2 vectors.  These 2 vectors are known as components of the vector. – We will often refer to them as the vertical component and horizontal component.

Finding component vectors is called VECTOR RESOLUTIONFinding component vectors is called VECTOR RESOLUTION Consider this sled…Consider this sled… 30 FhFhFhFh FvFvFvFv F 58 N

We find F v and F h by using trig… Sin 0 = F v /F F v = F sin 0 F v = 58N(sin 30°) F v = 58N(0.5) F v = 29N Cos 0 = F h / F F h = F cos 0 F h = 58N (cos 30°) F h = 58N(.8660) F h = 50.2N

Practice p

Resolving vectors 1. How fast must a truck travel to stay beneath an airplane that is moving 105 km/h at an angle of 25  to 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 15  incline. 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) V y = 5.7 m/s V x = 21.3m/s 4) d v = 5m d h = 0m

The sign +/- for F v and F h can be found from the diagram…The sign +/- for F v and F h can be found from the diagram…

As the component becomes more perpendicular to the direction of motion, the decreasing component approaches zeroAs the component becomes more perpendicular to the direction of motion, the decreasing component approaches zero REMEMBER

Problem… Solve for F v and F h for the red wagon…Solve for F v and F h for the red wagon… 100N 45° Radio Flyer

Special Triangle Cos 45 0 (100) = Sine 45 0 (100) = 70.7N

A wind with a velocity of 40km/hr blows toward 30°. Find…A wind with a velocity of 40km/hr blows toward 30°. Find… A) component of wind towards 90°A) component of wind towards 90° B) Component of wind towards 0°B) Component of wind towards 0°

Solution Given: 40km/hr, 30 0 V 90 =20kph V 0 0 =34.6kph

Independence of Vector Quantities Perpendicular vector quantities are independent of one anotherPerpendicular vector quantities are independent of one another Boat Crossing a River V B = 8m/s V R = 3m/s Resultant Vector= 8.5

The velocity of the boat does NOT affect the velocity of the river!The velocity of the boat does NOT affect the velocity of the river! The velocity of the RIVER does NOT affect the velocity of the boat!The velocity of the RIVER does NOT affect the velocity of the boat! If the river is 80m wide it will take the boat 10 seconds to cross it V= d/t t= d/v 80m/ 8 m / s = 10 sec During those 10 seconds the boat will move 30m downstream V= d/t d= Vt 3m/s (10s) = 30m

River Crossing Questions a)Find the resultant velocity for the boat b)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 width = 100m 6 m/s 2 m/s

ANSWERS! a)A 2 + B 2 = R = R 2 = 6.32m/s b)V=d/t t= d/v t=100/6 = 16.67sec c)V= d/t d= Vt d=2(16.67) = 33.34m d)tan 0 = opp/adj tan 0 = 2/6 tan 0 = = angle measured from bank 90 – = e)Aim upstream or 33.34m

Adding Vectors at Any Angle If we have 2 or more vectors that are NOT at 90° to each other, we can still find the Resultant of them.If we have 2 or more vectors that are NOT at 90° to each other, we can still find the Resultant of them. THREE STEPSTHREE STEPS 1.Each given Vector is resolved into its perpendicular components 2.Add the vertical components together, add the horizontal components together Keep in mind + 3.The sum of these components will give the perpendicular components of the resultant. We solve for the resultant w/ A 2 + B 2 = C 2 and Sin, Cos, Tan.

60°10° 2 1 Identify components F 1V F 1h F 2V F 2h 12N8N

F 1h Cos 10°= F 1h /12N F 1h = (12N)Cos10° =11.8N F 1V Sin10°= F 1V / 12N F 1V =(12N)sin10° = 2.08 F 2h Cos60° = F 2h /8 F 2h = (8N)Cos60° = -4N F 2V Sin60° = F 2V /8 F 2V = (8N) Sin60° = 6.9N

#2 F 1h + F 2h = (-4) = 7.8N F 1V + F 2V = = 8.98N R 8.98N 7.8N 11.9N 49 o

Practice Problem 3C Page

Solution 1) 49m; from the line of scrimmage (7.4 0 from downfield) 2) 7.6km;

Equilibrium A state when the net force acting on an object is zero (all forces cancel out)A state when the net force acting on an object is zero (all forces cancel out) 10N10N we say this is in equilibrium Tug of War… F net = 0…. No acceleration

We say this object is in equilibriumWe say this object is in equilibrium there is NO NET FORCE presentthere is NO NET FORCE present solving this graphically

Equilibrium occurs when an object is NOT moving.Equilibrium occurs when an object is NOT moving. It can also occur when something IS movingIt can also occur when something IS moving Thrust Drag Lift Weight

Equilibrant Force a force that, when applied, produces equilibriuma force that, when applied, produces equilibrium The equilibrant force is equal in size to the resultant force, but is in the opposite directionThe equilibrant force is equal in size to the resultant force, but is in the opposite direction R= 10N 300° 8N A 6N B The resultant for A and B is 10N at 300°

The equilibrant force is 10N at 120° E = 10 N at 120° 6N B 8N A

When Finding the Equilibrant Force… 1.Find the resultant force for all given Vectors 2.Equilibrant force is equal in size but opposite in direction to the resultant

Example 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. Betty100N Bob200N A 2 + B 2 = C = R 2 R= Tan 0 = O/A = 200 /100 0 = 63° Equilibrant force = N at 243°

Continued… Betty 100 N Bob 200 N E = N at 243°

Problem #2 This sign has a weight of 168N and is in equilibrium. Two ropes support this sign and are fixed to a point. Each rope makes an angle of 22° with the top of the sign. Find the forces that exist in the ropes.This sign has a weight of 168N and is in equilibrium. Two ropes support this sign and are fixed to a point. Each rope makes an angle of 22° with the top of the sign. Find the forces that exist in the ropes. SINE 22°22°

Given A B Formula Sin 0 = O/ H Solution Sin 22° = F AV / A Sin 22° = F BV / B = 224N

84N B A F AH = - F BH = + They will cancel F AV + F BV = 168N F AV = F BV (similar triangles) B = F BV / Sin22° = 84 /Sin22° = 224 N

Practice… 1.A 300N sign is hung with two ropes. Both ropes are attached to the same point and make an angle with the horizontal of 40°. The sign is in equilibrium. Find the force that each rope exerts on the sign. 2.Two forces are applied to the same object. Force 1 acts at 0° with 100N, Force 2 acts at 35° with 120N. Find the equilibrant force.