Ch. 4 Vectors

Vector & Scalar 2 types of quantities Vector Scalar Describe magnitude & direction Examples of vector quantities… Velocity, force, acceleration, displacement Scalar Describe only magnitude Examples of scalar quantities… mass, time, distance, speed

Distance & Displacement Total length travelled, what your odometer would read for a car trip. Scalar Quantity Displacement Total length from starting point to ending point. Meaning the straight line distance from start to end Vector Quantity B C Distance equals the sum of the magnitudes of A,B,&C Displacement equals magnitude of blue vector A Displacement

How to draw vectors Vector arrows are scaled representations of a vector quantity The direction of the arrow tells us the direction And the length of the arrow shows (combined with the scale) what the magnitude of the vector is Ex. 30 km North (scale 1 cm= 5 km)

Finding the Resultant Vector .. Resultant -- One vector that represents the combination of 2 or more vectors. To find the resultant means to combine (or add) vectors. Any number of vectors can all be combined together to create one resultant vector. Example. Johnny walks 4 km East then 3 km West 4 km East Vector Resultant Vector = 1 km East 3 km West vector

Finding Resultant Procedure 1) Determine a proper scale for drawing 2) Draw first vector to scale 3) Draw 2nd vector to scale, with its tail connected to the head of the first Note, it does not matter which vector is drawn first Draw any other vectors (the third, fourth…etc.) in the same fashion 4) Draw the Resultant from the tail of the first vector to the head of the last to show proper direction 5) Determine the magnitude and direction of the resultant. Determine mathematically OR determine graphically

Resultant Vector = 130 km/hr East Another Example Even though displacement vectors seem to make the most sense, this process works for any type of vectors Ex. A plane is flying with an engine speed of 100 km/hr going East, but it is also encountering a 30 km/hr wind directed towards the East. What is the plane’s resultant velocity?? Scale 1cm=10 km/hr 1 cm Resultant Vector = 130 km/hr East

Determining Magnitude of Resultant mathematically When resultant vector creates a right triangle, you can immediately use pythagorean theorem to determine length of the hypotenuse (resultant) A2 + B2 = C2 with C being the resultant So to solve for C the pythagorean becomes C = A2 + B2

Determining Magnitude of Resultant Graphically (generally don’t need to make perfectly scaled drawings in this class, but do need to set up drawings to better understand problem) Measure length of the resultant on your paper Use scale to bring back into the vector units For example… Scale is 1 cm = 30 m/s You measure resultant to be 5 cm, so this means that the magnitude of your resultant is 5 x 30 or 150 m/s This method involves more measuring and not so much math You should be able to find the magnitude of the Resultant both Mathematically and Graphically!!!!

Vector Combinations in 2-D If vectors are in two different planes (i.e. One going North and another going West) then that would be considered to be in 2 dimensions We can also combine vectors this way. For example. A car drives 30 km East, then 50 km North. What was the cars resultant displacement? (they are basically asking you what was the car’s net movement)…. Drawing an arrow from starting point to ending point Resultant Vector = 58.3 km NorthEast Scale 1 cm= 5 km If found mathematically ..Use pythagorean (A2+B2=C2) theorem to find this R=(302 + 502) R=58.3 If found graphically Resultant should be 11.66 cm long. Then if we use our scale to convert that back into km… 11.66 x 5 = 58.3 km 1 cm

Red arrow should be 10 cm long, purple arrow should be 3 cm long Another 2-D example A plane flying due North with a speed of 200 km/hr encounters a strong eastern wind of 60 km/hr Scale 1 cm= 20 km/hr 1 cm Red arrow should be 10 cm long, purple arrow should be 3 cm long To find GRAPHICALLY.. Resultant will be measured to be 11.44 cm long. Multiply that by 20 to get into km/hr and we get R = 228 km/hr

Components of Vectors R Ry=Rsinθ θ Rx=Rcosθ Now that we know how to find a resultant.. We can also do the reverse process Finding Horizontal and Vertical components of vectors… In other words we are breaking down a vector into its two individual parts Ex. Telling someone how far you went East or how far you went North We can use trig and SOH CAH TOA to determine magnitudes of the 2 components **These two equations are always accurate for x & y components when θ is the angle between the vector and the x-axis R- any vector Ry- the y-component of the vector Rx- the x-component of the vector R Ry=Rsinθ θ Rx=Rcosθ

Components of Vectors Components can also be modeled graphically as before but should be calculated mathematically If a vector is going directly N, E, S, or W , then it has only one component If N or S then only vertical component If E or W then only horizontal component Horizontal and Vertical components do not affect each other !! POSITIVES AND NEGATIVES STILL INDICATE DIRECTION They act separately For instance …. Thrown or launched objects Gravity ***Boat crossing a river*** Monkey / Gun Example Bullet Example

Which boat reaches the opposite shore first??? 100 m 100 m 10 m/s East 10 m/s East 15 m/s Water current South NO CURRENT

Answer Both boats reach at same time because only the horizontal component affects how long it takes to reach the opposite shore The boat with current will move faster, take a different path, and go a greater distance…. BUT still reaches the opposite shore at the same time B/c Horizontal components ARE NOT affected by Vertical components of Velocity

Finding resultants of 2-D vectors that are not perpendicular (more difficult 2-D Problems) Need to break each vector into components Add components then create new right triangle to find magnitude of Resultant (R) R R R

More difficult 2-d Examples 45 N at 80º NE 20 N at 30º NW Find Resultant of these 2 force vectors Make chart and break down vectors into components Find sum in x-direction, sum in y-direction Use pythagorean to combine these to find ‘R’ Vector x y 1 45cos80= +7.81 N 45sin80= +44.31 N 2 20cos30= -17.32 N 20sin30= +10 N Sum -9.51 54.31 R Now use horizontal and vertical sums to determine resultant magnitude A2 + B2 = C2 (-9.512 + 54.312 =R2) R= 55.13 N

Direction of Resultant To find direction of Resultant Use Tan-1 Tanθ=opp/adj = (y/x) Θ = Tan-1(y/x) This will always work, for the angle between the vector and x-axis. R θ

Direction of Resultant When finding the resultant it is important to include direction Usually resultants will not be directly in one of the four (N,E, S, W) directions Use ‘in between’ directions… NE, NW, SE, or SW But we need to include an angle to give a exactly specific direction So it could read as 30º N of W

Practice Drawing Vectors Find the resultant of these two vectors… 30 km at 55˚ NE and 20 km at 35˚ SE (move protractor on screen to check for correct angle) 35˚ 55˚ 20˚

15 m/s 42⁰ SW 20 m/s and 25⁰ SE (move protractor on screen move arrows to correct angle)

Equilibrium Problems If an object is in equilibrium ΣFx = 0 and ΣFy=0 Referred to as the 1st Condition of Equilibrium In other words, all forces must cancel out… so we can say ΣFx = 0 = F1x + F2x +F3x…. And ΣFy=0 = F1y + F2y + F3y … Example

Force Vectors--- ex. Sign hanging from 2 wires Sign is at rest (in equilibrium) so we can say ΣFx = 0 = T1x + T2x And ΣFy=0 = Fg + T1y + T2y Force of the Wire’s Tension (T1) (T1y) (T2) (T2y) JOE’S DINER (T1x) (T2x) Force of Sign’s Weight (Fg)

Friction / Incline Problems When an object is on an incline, the Normal Force (FN) is no longer equivalent to the objects weight Objects weight (Fg)is still down, but it now has 2 components. FP - Parallel force… force pulling it down the ramp Force pushing object down ramp, can be treated in same way as applied force has been If drawn like below Fp = Fgsinθ (can be seen simply w/ trig identities) FN - Normal Force … perpendicular to the ramp Amount of FN affects frictional force Higher θ of ramp, lower FN is and b/c of this friction will also be lower Friction will be directed opposite the motion/intended motion If drawn like below, FN = Fgcosθ Fk **In this example ΣF = Fp + Fk = ma Since Fp and Fk are only forces in the plane of motion. θ of ramp is equal to θ between FN and Fg Plugging in friction eq. this becomes ΣF = Fp + (-μkFN) = ma Fp Fg FN θ θ

Applied Force Problem A) If there is no friction and the man is pulling sled with a Force of 15 N, what is the acceleration of the 10 kg sled? B) If there exists a coefficient of kinetic friction of .14 between the sled and ground what the acceleration be?

Solutions a. Acceleration is only in the x direction so we can say that ΣFx=ma and ΣFy=0 ΣFx=ma = sum of all individual horizontal forces  only one  Fx So ΣFx=ma = Fx = Fcosθ Or ma=Fcosθ (10kg)(a) = (15N)cos(30º) a= 1.3 m/s2 Now another vertical force must be considered, Fk So ΣFx=ma = Fx + Fk ma = (Fcosθ) + μkFN but now FN is no longer going to be equal to (mg) b/c another vertical force is involved. So use the ΣFy=0 equation to analyze vertical (y) forces and determine FN ΣFy=0 = sum of all vertical forces = Fg +Fy +FN ΣFy=0 = Fg +Fk +FN = mg + Fsinθ + FN Plugging in values 0 = (10)(-9.8) +15sin30+FN 0 = -98N + 7.5N + FN FN= 90.5 N Now plug this into friction equation above ma = (Fcosθ) + μkFN 10(a) =(15cos30) + (.14)(90.5) and solve for ‘a’ a = .03 m/s2 F Fk Fy Fx

Projectile Motion Click below for Video

Projectile Motion Anytime an object is launched and only under the influence of gravity it is considered to be a projectile. Typically moving both horizontally and vertically Acceleration is occurring in the vertical (y)direction Velocity is constant in the horizontal (x) direction Horizontal is staying constant….. Vertical component is changing due to the acceleration of gravity**

Projectile Launched Horizontally vx vx Vy vx Vx- horizontal component vector of velocity Vy-vertical component vector of velocity Vx  Constant Vy  increasing downward GRAVITY is increasing Vy, but does not affect Vx Vy vx Vy vx Vy

Or drawn a little differently… vx Vy vx Vy vx Green Arrows  resultant velocity vector of the ball at that point in its path Vy vx Vy

Object launched upward at an angle of 45⁰ with a velocity of 42.4 m/s. Vy=0 m/s Vy=10 m/s Vx=30 m/s Vx=30 m/s Vy=20 m/s Vx=30 m/s Vy=-10 m/s Vx=30 m/s Vx=30 m/s Vy=-20 m/s Vy=30 m/s Vx=30 m/s Vx=30 m/s Vy=-30 m/s Object launched upward at an angle of 45⁰ with a velocity of 42.4 m/s. Initial Horizontal Velocity = Vx = +30 m/s Initial Vertical Velocity = Vyi = +30

Projectile motion problems P.M. is in 2-D, and each dimension must be treated separately when solving equations During P.M. acc. Is only occuring in y- direction, and motion is constant in x- direction… so we must adjust our kinematics equations. Vyf2=vyi2 + 2gΔy vyf=vyi+gt Δy=vyit + ½ gt2 Δy= ½(vyf + vyi)t Δx=vxt The only acc. Is that of gravity and that only occurs in the y. So ‘a’ becomes ‘g’ and all velocities in acceleration equations have ‘y’ subscripts Motion in y-plane- accelerated Motion in x-plane - constant

Bullet Fired & Dropped at Same time X- components are independent of Y-Components and the acceleration of gravity.. So both bullets fall at the same rate and reach ground at same time

Range and Height Range - how far an object travels horizontally Height - how far an object travels vertically How to produce max height?? Launch at 90⁰ (straight up) How to produce max range?? Launch at 45⁰ Perfect balance of horizontal speed, and time of flight Objects launched at complementary angles will produce equal ranges Ex. 30 ˚ and 60 ˚ height range

Objects launched at complimentary angles will produce equal ranges 80⁰ and 10⁰ (as seen above) are complimentary angles http://www.ngsir.netfirms.com/englishhtm/ThrowABall. htm

Hypothetical straight line path cannonball would follow if there was no Gravity Distances below hypothetical straight line path object is at 1s, 2s, and 3 sec. ***Same distances as if ball was launched horizontally **

Satellites Satellites--- objects that orbit another object Moon orbits Earth Earth and other planets orbit Sun These are all essentially just projectiles If objects fall because of gravity, how come satellits stay orbiting forever?? Or do they? Are they getting closer?? Are they falling?? How???

Projectile being thrown a different speeds, from the top of a mountain above the atmosphere

Satellites are moving so fast that the rate at which they are falling matches the rate at which the center object curves away from it So yes they are falling, but falling around Earth not into it and no if they are orbiting properly they will not be getting closer