1. Chapter Three
Vectors

2. Vector Fundamentals
A vector quantity has two or more variables which define it.
A scalar quantity only has size (i.e. temperature, time, energy, etc.)
Vectors in physics have magnitude and direction:
head
tail
length – represents the magnitude
Equivalent vectors are parallel with the same magnitude.
Opposite/Negative vector are parallel vectors with the same magnitude but in the opposite direction.

3. Vector Addition
Vector Addition is the combination of two or more vectors into a single vector which has the same effect/results.
Resultant: Results of vector addition Wo

4. Net Velocity Upstream
Upstream: Place vectors head to tail, net result, 5 km/hr upstream
Note: Subtraction is adding the opposite

5. Net Velocity Downstream
Downstream: Place vectors head to tail
Can you explain why in one case, addition results in the difference between magnitudes and the other it is the sum of the magnitudes?

6. What if nonparallel? Commutative Property: Triangle Method:
Parallelogram Method:
Use Law of Sines and Law of Cosines to solve the triangle.
Commutative Property:

7. Components
Any vector can be considered to be the resultant of an infinite number of vectors.
Star is linked to a GSP file demonstrating the infinite number of components and the utility of orthogonal components. File is on desktop in the Physics Folder.

8. X- and Y-Components x y

9. Unit Vectors
Scalar Multiplication: The magnitude of any vector can be multiplied by a scalar.
Unit Vector: A vector with a magnitude of one:
Unit Vector Form: Scalar times a unit vector eg
Unit vectors are often written without the absolute value marks and vector symbols. Contextual understanding.

10. Using Components to Combine Vectors
Given two vectors with the following values:

11. Comparison with the graphical method!B

12. Finding the angle the vector makes with the y-axis?
opp = 2 
adj = 12

13. 3 - Way Tug-o-War
Bugs Bunny, Yosemite Sam, and the Tweety Bird are fighting over a giant 450 g Acme super ball. If their forces remain constant, how far, and in what direction, will the ball move in 3 s, assuming the super ball is initially at rest ?
Sam: N
Tweety: 64 N
38°
43°
To answer this question, we must find a, so we can do kinematics. But in order to find a, we must first find Fnet.
Bugs: 95 N
continued on next slide

14. 3 - Way Tug-o-War…
Sam: N
Bugs: 95 N
Tweety: 64 N
38°
43° N N N N
First, all vectors are split into horiz. & vert. comps. Sam’s are purple, Tweety’s orange. Bugs is already done since he’s purely vertical. The vector sum of all components is the same as the sum of the original three vectors. Avoid much rounding until the end.
continued on next slide

15. 3 - Way Tug-o-War…
Next we combine all parallel vectors by adding or subtracting: = , and = A new picture shows the net vertical and horizontal forces on the super ball. Interpretation: Sam & Tweety together slightly overpower Bugs vertically by about 17 N. But Sam & Tweety oppose each other horizontally, where Sam overpowers Tweety by about 41 N.
95 N N N N N N N
continued on next slide

16. 3 - Way Tug-o-War… N N
Fnet = N 
Find Fnet using the Pythagorean theorem. Find  using trig: tan = N / N. The newtons cancel out, so  = tan-1( / ) = . (tan-1 is the same as arctan.) Therefore, the superball experiences a net force of about 44 N in the direction of about 23 north of west. This is the combined effect of all three cartoon characters.
continued on next slide

17. 3 - Way Tug-o-War…
a = Fnet / m = N / 0.45 kg = m/s2. Note the conversion from grams to kilograms, which is necessary since 1 m/s2 = 1 N / kg. As always, a is in the same direction as Fnet.. a is constant for the full 3 s, since the forces are constant.
Now it’s kinematics time: Using the fact x = v0 t a t = ( )(3) = m  441 m, rounding at the end. 
m/s2
So the super ball will move about 441 m at about 23 N of W. To find out how far north or west, use trig and find the components of the displacement vector.

18. Practice Problem
The 3 Stooges are fighting over a g (10 thousand gram) Snickers Bar. The fight lasts 9.6 s, and their forces are constant. The floor on which they’re standing has a huge coordinate system painted on it, and the candy bar is at the origin. What are its final coordinates?
Hint: Find this angle first.
Curly: 1000 N
Larry: 150 N
78
93
Moe: 500 N
Answer:
( , ) in meters

19. How to move a stubborn mule
It would be pretty tough to budge this mule by pulling directly on his collar. But it would be relatively easy to budge him using this set-up.
Big Force
Little Force
(explanation on next slide)

20. How to move a stubborn mule…
overhead view
tree
mule
little force
Just before the mule budges, we have static equilibrium. This means the tension forces in the rope segments must cancel out the little applied force. But because of the small angle, the tension is huge, enough to budge the mule!
(more explanation on next slide)
little force T
tree T
mule

21. How to budge a stubborn mule (final)
Because  is so small, the tensions must be large to have vertical components (orange) big enough to team up and cancel the little force. Since the tension is the same throughout the rope, the big tension forces shown acting at the middle are the same as the forces acting on the tree and mule. So the mule is pulled in the direction of the rope with a force equal to the tension. This set-up magnifies your force greatly.
little force   T T
tree
mule

22. River Crossing
campsite
Current
0.3 m/s
boat
river
You’re directly across a 20 m wide river from your buddies’ campsite. Your only means of crossing is your trusty rowboat, which you can row at 0.5 m/s in still water. If you “aim” your boat directly at the camp, you’ll end up to the right of it because of the current. At what angle should you row in order to trying to land right at the campsite, and how long will it take you to get there?
continued on next slide

23. t = d / v = (20 m) / (0.4 m/s) = 50 s.  = tan-1(0.3 / 0.4)  36.9.
River Crossing
Current
0.3 m/s
campsite
boat
river
0.5 m/s 
0.4 m/s
Because of the current, your boat points in the direction of red but moves in the direction of green. The Pythagorean theorem tells us that green’s magnitude is 0.4 m/s. This is the speed you’re moving with respect to the campsite. Thus:
t = d / v = (20 m) / (0.4 m/s) = 50 s.  = tan-1(0.3 / 0.4)  36.9.
continued on next slide

24. Law of Sines a A B C c b a sin sin B sin C b c =
The river problem involved a right triangle. If it hadn’t we would have had to use either component techniques or the two laws you’ll also do in trig class: Law of Sines & Law of Cosines. A B C c b a
Law of Sines:
sin
sin B
sin C a b c =

25. Law of Cosines a 2 = b 2 + c 2 - 2 b c cosA a C b c A B
This side is always opposite this angle.
These two sides are repeated.
It doesn’t matter which side is called a, b, and c, so long as the two rules above are followed. This law is like the Pythagorean theorem with a built in correction term of -2 b c cos A. This term allows us to work with non-right triangles. Note if A = 90, this term drops out (cos 90 = 0), and we have the normal Pythagorean theorem.

26. Wonder Woman Jet Problem
Suppose Wonder Woman is flying her invisible jet. Her onboard controls display a velocity of 304 mph 10 E of N. A wind blows at 195 mph in the direction of 32 N of E. What is her velocity with respect to Aqua Man, who is resting poolside down on the ground?
vWA = vel. of Wonder Woman w/ resp. to the air
vAG = vel. of the air w/ resp. to the ground (and Aqua Man)
vWG = vel. of Wonder Woman w/ resp. to the ground (Aqua Man)
We know the first two vectors; we need to find the third. First we’ll find it using the laws of sines & cosines, then we’ll check the result using components. Either way, we need to make a vector diagram.
continued on next slide

27. Wonder Woman Jet Problem (cont.)
80
195 mph
304 mph
vWG
32
100
10
32
vWA
vAG
vWG
vWA + vAG = vWG
The 80 angle at the lower right is the complement of the 10 angle. The two 80 angles are alternate interior. The 100 angle is the supplement of the 80 angle. Now we know the angle between red and blue is 132.
continued on next slide

28. Wonder Woman Jet Problem
By the law of cosines v 2 = (304)2 + (195)2 - 2 (304) (195) cos 132. So, v = 458 mph. Note that the last term above appears negative, but it’s actually positive, since cos 132 < 0. The law of sines says:
sin 132
sin v
195 =
195 mph
304 mph v
132 
80
So, sin = 195 sin 132 / 458, and   18.45
This mean the angle between green and the horizontal is 80   61.6
Therefore, from Aqua Man’s perspective, Wonder Woman is flying at 458 mph at 61.6 N of E.

29. Using the Component Method
This time we’ll add vectors via components as we’ve done before. Note that because of the angles given here, we use cosine for the vertical comp. of red but sine for the vertical comp. of blue. All units are mph.
32
vWA = 304 mph
vAG = 195 mph
10
304
195
52.789
continued on next slide

30. Component Method… 218.1584 mph 195 402.7159 mph 458.0100 mph 304
Combine vertical & horiz. comps. separately and use Pythag. theorem.  = tan-1( / ) = .  is measured from the vertical, which is why it’s 10 more than .
304
195
52.789
52.789
mph
mph
mph 

31. Comparison of Methods
We ended up with same result for Wonder Woman doing it in two different ways. Each way requires some work. You will only want to use the laws of sines & cosines if:
the vectors form a triangle.
you’re dealing with exactly 3 vectors (If you’re adding 3 vectors, the resultant makes a total of 4, and this method would require using separate triangles.)
Regardless of the method, draw a vector diagram! To determine which two vectors add to the third, use the subscript trick.