VECTORS

1.Scalars Just a Value This Value is called a Magnitude 2.Vectors

Quantities that have  MAGNITUDE (size or value) AND DIRECTION VECTORS

REPRESENTATION OF VECTOR QUANTITIES VECTORS ARE REPRESENTED BY AN ARROW tip tail

THE ARROW: LENGTH = THE MAGNITUDE OR SIZE OF THE VECTOR THE ARROW’S DIRECTION = IS THE DIRECTION OF THE VECTOR

EXAMPLES OF VECTORS FORCE (a push or a pull) ELECTRIC/MAGNETIC FIELD STRENGTH ACCELERATION TORQUE – twist causing rotation DISPLACEMENT – not distance MOMENTUM – possessed by moving mass VELOCITY – not speed

Mass Time Distance Energy + Everything else that’s not a vector….. These quantities have “NO DIRECTION” EXAMPLES OF SCALARS

Take care here You Can NOT Add them like regular numbers (called Scalars)

VECTOR ADDITION (THE TIP-TO-TAIL METHOD) FINDING THE RESULTANT –The SUM or RESULT of Adding 2 Vectors is called A B

VECTOR ADDITION (THE TIP-TO-TAIL METHOD) FINDING THE RESULTANT A A A B B B

REVIEWING VECTOR ADDITION ADD VECTORS IN ANY ORDER (A+B = B+A) IF VECTORS ARE POINTING IN THE SAME DIRECTION  THIS IS REGULAR ALGABRAIC ADDITION

REVIEWING VECTOR ADDITION POSITION THE TAIL OF ONE VECTOR TO THE TIP OF THE OTHER CONNECT FROM THE TAIL OF THE 1 ST VECTOR TO THE TIP OF THE LAST –THIS IS THE RESULTANT

ADDITION CONTINUED A B A B

MORE VECTOR ADDITION SUPPOSE THE VECTORS FORM A RIGHT ANGLE GRAPHICAL SOLUTIONS CAN ALWAYS BE USED BUT… –HERE IS A MATHEMATICAL SOLUTION…. –THIS SOLUTION USES THE PYTHAGOREAN THEORUM… C 2 = A 2 + B 2

ADDING VECTORS THAT ARE AT RIGHT ANGLES TO EACH OTHER… A = 4 lbs B = 3 lbs R = ?? lbs R 2 = A 2 + B 2 R 2 = R 2 = = 25 R = 5 lbs BUT R = 5 lbs IS ONLY HALF AN ANSWER!! WHY?????

REMEMBER !!! VECTORS HAVE 2 PARTS MAGNITUDE AND DIRECTION !!! HERE’S HOW TO FIND THE DIRECTION

TRIG FUNCTIONS TO REMEMBER

TRIG CALCULATIONS  A = 4 lbs B = 3 lbs R = 5 lbs COS(  ) = 4/5=0.8 SIN(  ) = 3/5= 0.6 USE YOUR CALCULATOR TO FIND THE ANGLE THAT HAS THESE VALUES OF SIN OR COS. Could also use TAN

AT LAST THE ANGLE (THE VECTOR’S DIRECTION) SIN(X) = 0.6  ANGLE (X) = 37 DEGREES COS(X) = 0.8  ANGLE (X) = 37 DEGREES SO, THE OTHER HALF OF OUR ANSWER IS…..

RESULTANT……. 5 lbs 37 degrees NORTH OF EAST  = 37 deg A = 4 lbs B = 3 lbs R = 5 lbs Not NORTHEAST i.e. NE is 45 deg

SUMMARY A VECTOR IS A DIRECTED QUANTITY THAT HAS BOTH A MAGNITUDE AND DIRECTION IF THE ANGLE BETWEEN THE VECTORS IS 0 deg: algebraic addition (MAXIMUM ANS.) 180 deg: algebraic “subtraction” (MINIMUM ANS.) 90 deg: use Pythagorean Theorem to find magnitude and trig functions to find the angle

COULD YOU PASS A QUIZ ON THIS MATERIAL???? NOW? LATER, WITH STUDY? RETURN TO BEGINNING CONTINUE TO VECTOR MATH

Imagine you were asked to mark your starting place and walk 3 meters North, followed by two meters East. Could you answer the following: –How far did you walk? –Where are you relative to your original spot? Vector Concepts used in Physics – Fancy Foot Work

How far did you walk? –This requires a MAGNITUDE ONLY SCALAR QUANTITY called DISTANCE 3m + 2m = 5m Where are you relative to your original spot? –This requires both a MAGNITUDE & DIRECTION VECTOR QUANTITY called DISPLACEMENT Fancy Foot Work

Where are you relative to your original spot? –This requires both a MAGNITUDE & DIRECTION VECTOR QUANTITY called DISPLACEMENT 3m, N 2m, E Displacement: Needs Magnitude & Direction Start End Fancy Foot Work

Magnitude =? The Length of the Hypotenuse s 2 =(3m) 2 + (2m) 2 s = Direction =  East of North Pick your Trig function  =33.7 o, E of N 3m, N 2m, E Displacement = s  E N S W Fancy Foot Work

NOW, measure the angle from the +X axis…… 3m, N 2m, E Displacement = s =  =33.7  = = 90 – 33.7 = 56.3 o Fancy Foot Work

When you multiply a vector by a scalar, it only affects the MAGNITUDE of the vector ** Not the direction** Example: Multiplying a Vector by a Scalar

Component means “part” A vector can be composed of many parts known as components It’s best to break a vector down into TWO perpendicular components. WHY? –To use Right Triangle Trig Vector Components

Introducing Vector V Vector V’s X-Component is its Projection onto the X-axis Vector V’s Y-Component is its Projection onto the Y-axis VxVx VyVy Now we have a Right Triangle Sub Scripts in Action Vector Components

Given this diagram, find V’s X & Y Components Vx=Vx= Vy=Vy= 5 4 What’s the Magnitude of Vector V?  =38.66 o Vector Components

Now, knowing the magnitude of vector V, verify the V’s X & Y components using Trig o VxVx VyVy V x =? V y =?   Vector Components

Golden Rules of Vector Components –1. If you know the magnitude and direction of vector V to be (V,  ), then you can find V x & V y by V x =Vcos  V y =Vsin  –2. If V x & V y are known, the magnitude of the vector can be found with Pythagorean Theorem: Vector Components

WHAT ARE THE X & Y COMPONENTS OF VECTOR ‘A’? A Ax Ay  Sin  OPP/HYP Sin  Ay/A A y = Asin  =Asin  Cos  ADJ/HYP Cos  Ax/A A x = Acos  =Acos  THESE ARE The VECTOR COMPONENTS OF A Ax Ay

=Asin  AN EXAMPLE Suppose the magnitude of A = 5 and   deg. Find the VALUES of the X & Y components. A Ax Ay  =Acos  =37 = 5 5sin 375(0.6) 3 5cos375(0.8) 4

Example: You’re a pilot & are instructed to go around a massive thunderstorm. The control tower tell you take a detour & follow these 2 paths : 100 km, 45 o & 90 km, 10 o What is the plane’s displacement from where it began it’s detour? Vector Components: Example of the Golden Rule TRY IT ON YOUR OWN

Let V 1 =100 km, 45 o & V 2 = 90 km, 10 o KINEMATICS Vector Components MAGNITUDE DIRECTION

MAKE A ROUGH SKETCH OF THE VECTORS USING THE INFORMATION GIVEN. FIND THE X- AND Y- COMPONENT OF ALL OF THE VECTORS. (sometimes a table of values is helpful) ADD ALL OF THE VECTORS IN THE X-DIRECTION. (check the tips and tails of each vector ----vectors pointing in the same direction are added algebraically; in opposite directions-- this is algebraic subtraction). THIS RESULT WILL GIVE YOU THE X-PART OF THE RESULTANT

ADD ALL THE VECTORS IN THE Y-DIRECTION. AGAIN, FROM YOUR SKETCH, CHECK THE DIRECTION OF EACH Y-VALUE. (ALG.ADD. OR ALG. SUBT.) THIS RESULT WILL GIVE YOU THE Y-PART OF THE RESULTANT …(either pointing up/down or is zero) ROUGHLY SKETCH THE RESULTANT USING THE PYTHAGOREAN THEOREM FIND THE MAGNITUDE OF THE RESULTANT. USING THE ARCTAN FORMULA, FIND THE ANGLE THIS RESULTANT MAKES WITH THE AXES. STATE YOUR ANSWER WITH A MAGNITUDE (including a unit) AND THE DIRECTION.

A couple on vacation are about to go sight-seeing in a city where the city blocks are all squares. They start out at their hotel and tour the city by walking as follows: 1 block East; 2 blocks North; 3 blocks East; 3 blocks South; 2 blocks West;1 block South; 6 blocks East; 8 blocks North; 8 blocks West. WHAT IS THEIR DISPLACEMENT? (i.e., WHERE ARE THEY FROM THEIR HOTEL)? QUESTION #1

1 block East; 2 blocks North; 3 blocks East; 3 blocks South; 2 blocks West;1 block South; 6 blocks East; 8 blocks North; 8 blocks West. USING GRAPH PAPER THEY ARE 6 BLOCKS NORTH OF THE HOTEL ANSWER #1: WHERE ARE THEY FROM THE HOTEL? H

1 block East; 2 blocks North; 3 blocks East; 3 blocks South; 2 blocks West;1 block South; 6 blocks East; 8 blocks North; 8 blocks West. ANOTHER METHOD… SUM THE COMPONENTS IN THE X AND Y DIRECTIONS (THEN USE TRIG AS IF IT WAS A SINGLE VECTOR) N(+)S(-)E(+) W(-) THEY ARE 6 BLOCKS NORTH OF THE HOTEL N-S  Y-axis; E-W  X-axis + +

QUESTION #2 A river flows in the east-west direction with a current 6 mph eastward. A kayaker (who can paddle in still water at a maximum rate of 8 mph) wishes to cross the river in his boat to the North. If he points the bow of his boat directly across the river and paddles as hard as he can, what will be his resultant velocity ? RIVER BOAT

ANSWER #2 6 8 RESULTANT VELOCITY USE PYTHAGOREAN THEOREM  R 2 = R 2 = R 2 = 100 R = 10  = ARCTAN 6/8 = 37 deg. R = 10 mph, 37 deg East of North or 10 mph, bearing 037 deg.

QUESTION #3 RIVER BOAT The kayaker wants to go directly across the river from the North shore to the South shore, again, paddling as fast as he can. At what angle should the kayaker point the bow of his boat so that he will travel directly across the river? What will be his resultant velocity? Desired path to the South shore

ANSWER #3... RIVER BOAT Desired path to the South shore RESULTANT VELOCITY VEL.RIVER VEL.BOAT 8 mph 6 mph R  8 2 = Vr 2 Vr 2 = = 5.3 mph  = arctan = arctan(opp/adj) =arc tan(6/5.3) = arctan 1.1 = 49 deg upstream

QUESTION #4: A B 80 newtons 60 newtons Two forces A and B of 80 and 60 newtons respectively, act concurrently(at the same point, at the same time) on point P. P Calculate the resultant force.

ANSWER #4: A B 80 newtons 60 newtons P RESULTANT  Pythagorean Theorem for right triangle = arctan (80/60) = 53 deg 