Chapter Three

Vectors

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:

headtail

length – represents the magnitude

Vector Fundamentals

Equivalent vectors are parallel with the same magnitude.

Opposite/Negative vector are parallel vectors with the same magnitude but in the opposite direction.

1

2

.

.

n

v

v

v

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

1 2 3 ... nR F F F F

Net Velocity Upstream

Upstream: Place vectors head to tail,

net result, 5 km/hr upstream

b cR v v

R

bv

cv

Note: Subtraction is adding the opposite

Net Velocity Downstream

Downstream: Place vectors head to tail

b cR v v

R

cv

bv

Can you explain why in one case, addition results in the difference between magnitudes and the other it is the sum of the magnitudes?

What if nonparallel?

Commutative Property:

a b b a

a b

R

Triangle Method:

b a

Parallelogram Method:

Components

Any vector can be considered to be the resultant of an infinite number of vectors.

X- and Y-Components

x yA A A

x

y

A

xA

yA

x

y

A A Cos

A A Sin

2 2

x yA A A

1tany

x

A

A

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

ˆ ˆ ˆ, , ,x y z etc

ˆx xA A x

ˆ ˆx yA A x A y

ˆ ˆx x y yA B A B x A B y

ˆx xA A x

ˆ ˆx yA A x A y

ˆ ˆx x y yA B A B x A B y

Unit vectors are often written without the absolute value marks and vector symbols. Contextual understanding.

Using Components to Combine Vectors

ˆ ˆ ˆ ˆ3 4

ˆ ˆ ˆ ˆ5 8

x y

x y

A A x A y x y

B B x B y x y

Given two vectors with the following values:

yx

yx

yBAxBABA yyxx

122

8453

x

y

A

B

Comparison with the graphical method!

2121481444

122 22

.

BA

Finding the angle the vector makes with the y-axis?

ˆ ˆ2 12A B x y

opp = 2

y

x

adj = 12

01 5961

61

122

.tantan adj

opp

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 ?

Bugs: 95 N

Tweety: 64 N

Sam: 111 N

To answer this question, we must find a, so we

can do kinematics. But in order to find a, we must first find Fnet.

38° 43°

continued on next slide

3 - Way Tug-o-War…

Sam: 111 N

Bugs: 95 N

Tweety: 64 N

38° 43°

87.4692 N

68.3384 N

46.8066 N

43.6479 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

95 N

87.4692 N

68.3384 N

46.8066 N

43.6479 N

continued on next slide

3 - Way Tug-o-War…

16.9863 N

40.6626 N

Next we combine all parallel vectors by adding or subtracting: 68.3384 + 43.6479 - 95 = 16.9863, and 87.4692 - 46.8066 = 40.6626. 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.

3 - Way Tug-o-War…

40.6626 N

16.9863 NFnet

= 44.0679 N

Find Fnet using the Pythagorean theorem. Find using trig: tan = 16.9863 N / 40.6626 N. The newtons cancel out, so = tan-1(16.9863 / 40.6626) = 22.6689. (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

3 - Way Tug-o-War…

a = Fnet / m = 44.0679 N / 0.45 kg = 97.9287 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.

22.6689

97.9287 m/s2

Now it’s kinematics time: Using the factx = v0 t + 0.5 a t 2

= 0 + 0.5 (97.9287)(3)2

= 440.6792 m 441 m,rounding at the end.

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.

Practice Problem

The 3 Stooges are fighting over a 10 000 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?

78

Curly: 1000 N

Moe: 500 N

93

Larry: 150 N

Hint: Find this angle first.

Answer:

( -203.66 , 2246.22 ) in meters

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)

How to move a stubborn mule…

overhead viewtree 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!

tree mule

little force

T T

(more explanation on next slide)

How to budge a stubborn mule (final)

tree mule

little force

T T

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.

River Crossing

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?

campsite

Current 0.3 m/s

boatriver

continued on next slide

River Crossing

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.

Current 0.3 m/s

campsite

boatriver

0.3 m/s

0.5 m/s 0.4 m/s

continued on next slide

Law of Sines

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.

Law of Sines: sin sin B

sin Ca b c

= =

A B

C

c

b a

Law of Cosines

a

2 = b

2 + c

2 - 2 b c

cosA

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.

A B

C

cb a

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)

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?

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

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.

Wonder Woman Jet Problem (cont.)

continued on next slide

10

32

vW

A

vAG

vWG

vWA + vAG = vWG

80

195 mph

304 m

ph vWG

8032

100

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 =

So, sin = 195 sin 132 / 458, and 18.45195 mph

304

mp

h v

132

80

This mean the angle between green and the horizontal is 80 - 18.45 61.6

Therefore, from Aqua Man’s perspective, Wonder Woman is flying at 458 mph at 61.6 N of E.

Using the Component Method

32

vW

A =

304 m

phv AG =

195 mph

10

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.

103.3343

304

195

165.3694

52.789

299.3816

continued on next slide

Component Method…

165.3694

304

195103

.3343

52.789

299

.3816

Combine vertical & horiz. comps. separately and use Pythag. theorem. = tan-1(218.1584 / 402.7159) = 28.4452. is measured from the vertical, which is why it’s 10 more than .

165.3694

103.3343

52.789

299

.3816

402

. 71

59

m

ph

218.1584 mph

458.

0100

mph

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 2 separate triangles.)

Regardless of the method, draw a vector diagram! To determine which two vectors add to the third, use the subscript trick.