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Vectors

• 2-D Force &

Motion Problems

• Trig Applications

• Relative Velocities

• Free Body Diagrams

• Vector Operations

• Components

• Inclined Planes

• Equilibrium

Vector Addition

• Tip to tail method

• Parallelogram method

8 N4 N

3 N

Suppose 3 forces act on an object

at the same time. Fnet is not 15 N

because these forces aren’t

working together. But they’re not

completely opposing each either.

So how do find Fnet ? The answer

is to add the vectors ... not their

magnitudes, but the vectors

themselves. There are two basic

ways to add vectors w/ pictures:

Tip to Tail Methodin-line examples

Place the tail of one vector

at the tip of the other. The

vector sum (also called the

resultant) is shown in red. It

starts where the black vector

began and goes to the tip of

the blue one. In these

cases, the vector sum

represents the net force.

You can only add or

subtract magnitudes when

the vectors are in-line!

16 N

20 N

4 N

20 N16 N

12 N

9 N

9 N

12 N

21 N

Tip to Tail – 2 Vectors

5 m

2 m

To add the red and blue displacement vectors first note:

• Vectors can only be added if they are of the

same quantity—in this case, displacement.

• The magnitude of the resultant must be less

than 7 m (5 + 2 = 7) and greater than 3 m

(5 - 2 = 3).

Interpretation: Walking 5 m in

the direction of the blue vector

and then 2 m in the direction

of the red one is equivalent to

walking in the direction of the

black vector. The distance

walked this way is the black

vector’s magnitude.

Place the vectors tip to tail

and draw a vector from the

tail of the first to the tip of

the second.

Commutative Property

As with scalars quantities and ordinary numbers, the

order of addition is irrelevant with vectors. Note that

the resultant (black vector) is the same magnitude

and direction in each case.

(We’ll learn how to find the resultant’s magnitude soon.)

Tip to Tail – 3 Vectors

We can add 3 or more vectors

by placing them tip to tail in

any order, so long as they are

of the same type (force,

velocity, displacement, etc.).

Parallelogram Method

This time we’ll add red & blue by

placing the tails together and

drawing a parallelogram with

dotted lines. The resultant’s tail

is at the same point as the other

tails. It’s tip is at the intersection

of the dotted lines.

Note: Opposite

sides of a

parallelogram are

congruent.

Comparison of Methods

Tip to tail method

Parallelogram method

The resultant has

the same magnitude

and direction

regardless of the

method used.

Opposite of a Vector

v

- v

If v is 17 m/s up and

to the right, then -v

is 17 m/s down and

to the left. The

directions are

opposite; the

magnitudes are the

same.

Scalar Multiplication

x

-2x

3x

Scalar multiplication means

multiplying a vector by a real

number, such as 8.6. The

result is a parallel vector of a

different length. If the scalar

is positive, the direction

doesn’t change. If it’s

negative, the direction is

exactly opposite.

Blue is 3 times longer than red in the

same direction. Black is half as long

as red. Green is twice as long as

red in the opposite direction.½ x

Vector Subtraction

red - blue

blue - red

Put vector tails together and

complete the triangle, pointing to the

vector that “comes first in the

subtraction.”

Why it works: In the first diagram,

blue and black are tip to tail, so

blue + black = red

red – blue = black.

Note that red - blue is the opposite of blue - red.

Other Operations

• Vectors are not multiplied, at least not the

way numbers are, but there are two types of

vector products that will be explained later.

– Cross product

– Dot product

– These products are different than scalar mult.

• There is no such thing as division of vectors

– Vectors can be divided by scalars.

– Dividing by a scalar is the same as multiplying

by its reciprocal.

Comparison of Vectors

15 N

43 m

0.056 km

27 m/s

Which vector is bigger?

The question of size here doesn’t make sense. It’s like

asking, “What’s bigger, an hour or a gallon?” You can

only compare vectors if they are of the same quantity.

Here, red’s magnitude is greater than blue’s, since

0.056 km = 56 m > 43 m, so red must be drawn longer

than blue, but these are the only two we can compare.

Vector Components

150 N

Horizontal

component

Vert

ical

com

ponent

A 150 N force is exerted up and to

the right. This force can be

thought of as two separate forces

working together, one to the right,

and the other up. These

components are perpendicular to

each other. Note that the vector

sum of the components is the

original vector (green + red =

black). The components can also

be drawn like this:

Finding Components with Trig

v

v cos

vsin

Multiply the magnitude of the original vector by

the sine & cosine of the angle made with the

given. The units of the components are the

same as the units for the original vector.

Here’s the

correspondence:

cosine adjacent side

sine opposite side

Note that 30.814 + 14.369 > 34. Adding up vector components

gives the original vector (green + red = black), but adding up

the magnitudes of the components is meaningless.

Component Example

34 m/s

30.814 m/s

2514.369 m/s

A helicopter is flying at 34 m/s at 25 S of W (south of west).

The magnitude of the horizontal component is 34 cos 25

30.814 m/s. This is how fast the copter is traveling to the

west. The magnitude of the vertical component is 34 sin 25

14.369 m/s. This is how fast it’s moving to the south.

Pythagorean Theorem

34 m/s

30.814 m/s

2514.369 m/s

Since components always form a right triangle, the

Pythagorean theorem holds: (14.369)2 + (30.814)2 = (34)2.

Note that a component can be as long, but no longer than,

the vector itself. This is because the sides of a right triangle

can’t be longer than the hypotenuse.

Other component pairs

There are an infinite number of component pairs into which a

vector can be split. Note that green + red = black in all 3

diagrams, and that green and red are always perpendicular.

The angle is different in each diagram, as well as the lengths

of the components, but the Pythagorean theorem holds for

each. The pair of components used depends on the geometry

of the problem.

v

v cos

vsin

vv

Component FormInstead of a magnitude and an angle, vectors are often

specified by listing their horizontal and vertical components.

For example, consider this acceleration vector:

53.13

3 m/s2

4m

/s2

a = 10 m/s2 at 53.13 N of W

In component form:

a = -3, 4 m/s2

Some books use parentheses rather

than angle brackets. The vector F =

2, -1, 3 N indicates a force that is a

combination of 2 N to the east, 1 N

south, and 3 N up. Its magnitude is

found w/ the Pythag. theorem:

F = [22 + (-1)2 + 32]1/2 = 3.742 N

Finding the direction of a vector

x = 5, -2 meters is clearly a position to the southeast of a

given reference point. If the reference pt. is the origin, then x

is in the 4th quadrant. The tangent of the angle relative to the

east is given by:

5 m

2 m

tan = 2 m /5 m = tan -1(0.4) = 21.801

The magnitude of x is (25 + 4)1/2 = 5.385 m.

Thus, 5, -2 meters is equivalent to

5.385 m at 21.801 S of E.

Adding vectors in component form

If F1 = 3, 7 N and F2 = 2, -4 N, then the

F1

F2

net force is simply given by:

Fnet = 5, 3 N. Just add the

horizontal and vertical components

separately.

Fnet

F1F2

3 N

7 N

2 N

4 N

Inclined Plane

mg

perpendicular

component

parallel component

A crate of chop suey of mass m is setting on a ramp with angle

of inclination . The weight vector is straight down. The

parallel component (blue) acts parallel to the ramp and is the

component of the weight pulling the crate down the ramp. The

perpendicular component (red) acts perpendicular to the ramp

and is the component of the weight that tries to crush the ramp.

Note: red + blue = black

continued on next slide

Inclined Plane (continued)

mgmg cos

mg sin

The diagram contains two right triangles. is the angle

between black and blue. + = 90 since they are both

angles of the right triangle on the right. Since blue and red are

perpendicular, the angle between red and black must also be

. Imagine the parallel component sliding down (dotted blue) to

form a right triangle. Being opposite , we use sine. Red is

adjacent to , so we use cosine.


mg sin

Inclined Plane (continued)

mg

mg cos

mg sin

The diagram does not represent 3 different forces are acting on

the chop suey at the same time. All 3 acting together at one

time would double the weight, since the components add up to

another weight vector. Either work with mg alone or work with

both components together.

How the incline affects the components

mg

mg cos

mg sin

The steeper the incline, the greater is, and the greater sin

is. Thus, a steep incline means a large parallel component and

a small horizontal one. Conversely, a gradual incline means a

large horizontal component and a small vertical one.

mg

mg sin

Extreme cases: When = 0, the ramp is flat; red = mg; blue = 0.

When = 90, the ramp is vertical; red = 0; blue = mg.

Inclined Plane - Pythagorean

Theorem

mg

mgcos

mgsin

(mg sin)2 + (mg cos )2 = (mg)2 (sin2 + cos2)

= (mg)2 (1) = (mg)2

Let’s show that the

Pythagorean theorem

holds for components on

the inclined plane:

Inclined Plane: Normal Force

mgmgcos

mgsin

N = mg cos

Recall normal force is perpen-

dicular to the contact surface. As

long as the ramp itself isn’t

accelerating and no other forces

are lifting the box off the ramp or

pushing it into the ramp, N

matches the perpendicular

component of the weight. This

must be the case, otherwise the

box would be accelerating in the

direction of red or green.

N > mg cos would mean the box

is jumping off the ramp.

N < mg cos would mean that the

ramp is being crushed.

Net Force on a Frictionless Inclined Plane

mgmg cos

mg sin

N = mg cos

With no friction, Fnet = mg + N

= mg cos + mg sin + N

= mg sin.

(mg cos + N = 0 since their

magnitudes are equal but they’re

in directions opposite. That is, the

perpendicular component of the

weight and the normal cancel out.)

Therefore, the net force is the

parallel force in this case.

Acceleration on a Frictionless Ramp

mg

mg cos

mg sin

Here Fnet = mg sin = ma. So, a = gsin. Since sin has

no units, a has the same units as g, as they should. Both

the net force and the acceleration are down the ramp.

Incline with friction at equilibrium

mg

mg cos

mg sin

fs = mg sin

N = mg cos

At equilibrium Fnet = 0, so all forces

must cancel out. Here, the normal

force cancels the perpendicular

component of the weight, and the

static frictional force cancels the

parallel component of the weight.


Incline with friction at equilibrium (cont.)

mg

mg cos

mg sin

fs = mg sin

N = mg cos

fs s N = s mgcos. Also,

fs = mgsin (only because we

have equilibrium). So,

mgsin s mg cos .

Since the mg’s cancel and

tan = sin / cos, we have

s tan.


Incline with friction at equilibrium (cont.)

mg

mg cos

mg sin

fs = mg sin

N = mg cos

Suppose we slowly crank up the

angle, gradually making the ramp

steeper and steeper, until the box is

just about to budge. At this angle,

fs = fs,max = s N = s mgcos .

So now we have

mgsin = s mg cos,

and s = tan.

An adjustable ramp is a convenient

way to find the coefficient of static

friction between two materials.

(Neither of these quantities have units.)

Acceleration on a ramp with friction

mg

mg cos

mg sin

fk = kmg cos

N = mg cos

In order for the box to budge,

mg sin must be greater than

fs,max which means tan must be

greater than s. If this is the

case, forget about s and use k.

fk = kN = kmgcos.

Fnet = mg sin - fk = ma.

So, mg sin - kmg cos = ma.

The m’s cancel, which means a

is independent of the size of the

box. Solving for a we get:

a = gsin - kgcos. Once

again, the units work out right.

Parallel applied force on ramp

mg mg cos

mg sin

fk

N

FA

In this case FA and mg sin are

working together against friction.

Assuming FA + mg sin > fs,max

the box budges and the 2nd Law tells

us FA + mgsin - fk = ma.

Mass does not cancel out this time.

If FA were directed up the ramp,

we’d have acceleration up or

down the ramp depending on

the size of FA compared to mg

sin. If FA were bigger, friction

acts down the ramp and a is up

the ramp.

Non-parallel applied force on ramp

mgmg cos

mg sin

fk

N

Suppose the applied force acts on

the box, at an angle above the

horizontal, rather than parallel to

the ramp. We must resolve FA

into parallel and perpendicular

components (orange and gray)

using the angle + .

FA serves to increase acceleration

directly and indirectly: directly by

orange pulling the box down the

ramp, and indirectly by gray

lightening the contact force with

the ramp (thereby reducing

friction).

FA

FA cos( + )

FAsin( + )


Non-parallel applied force on ramp (cont.)

mg

mg sin

fk

N

FA

FA cos( + )

FA sin ( + )

Because of the perp. comp. of

FA, N < mg cos. Assuming

FA sin(+) is not big enough

to lift the box off the ramp,

there is no acceleration in the

perpendicular direction. So,

FA sin( + ) + N = mg cos.

Remember, N is what a scale

would read if placed under the

box, and a scale reads less if a

force lifts up on the box. So,

N = mg cos - FA sin( + ),

which means fk = kN

= k [mg cos - FA sin( + )].continued on next slide

Non-parallel applied force on ramp (cont.)

mg mg cos

mg sin

fk

N

FA

FA cos( + )

FAsin( + )

Assuming the combined force

of orange and blue is enough

to budge the box, we have

Fnet = orange + blue - brown = ma.

Substituting, we have

FA cos(+ ) + mgsin

- k [mg cos - FA sin(+ )] = ma.

Support Beam

mg

1

T1

2

T2


Hanging Sign Problem

mg

1

T1

2

T2Since the sign is not

accelerating in any

direction, it’s in

equilibrium. Since

it’s not moving

either, we call it

Static Equilibrium.

Thus, red + green + black = 0.


Hanging sign f.b.d.Free Body Diagram

mg

T1

T2

Vector Equation:

T1

+ T2

+ mg = 0


As long as Fnet

= 0, this is

true no matter many forces

are involved.

Hanging sign force triangleFnet

= 0 means a closed vector polygon !

T2

mg

T1

1 2

T1cos1 T2cos2

T1 sin1

T2sin2

T1 cos1 = T2 cos2

Horizontal:

Vertical:T1sin1 + T2 sin2 = mg

We use Newton’s 2nd

Law twice, once in

each dimension:

Hanging sign equationsComponents & Scalar Equations

Support Beam

35

T1

62

T2

75 kg

Answers:

Accurately draw all vectors and find T1 & T2.

T1 = 347.65 N

T2 = 606.60 N

Hanging sign sample

Vector Force Lab SimulationGo to the link below. This is not exactly the same as the

hanging sign problem, but it is static equilibrium with three

forces. Equilibrium link

1. Change the strengths of the three forces (left, right, and

below) to any values you choose. (The program won’t

allow a change that is physically impossible.)

2. Record the angles that are displayed below the forces.

They are measured from the vertical.

3. Using the angles given and the blue and red tensions, do

the math to prove that the computer program really is

displaying a system in equilibrium.

4. Now click on the Parallelogram of Forces box and write a

clear explanation of what is being displayed and why.

http://www.walter-fendt.de/ph11e/equilibrium.htm

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°


3 - Way Tug-o-War (continued)

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



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.


16.9863 N

40.6626 N

Fnet = 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 (final)

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 fact

x = v0 t + 0.5a t2

= 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.

3 - Way Tug-o-War 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 budge a stubborn mule

Big Force

Little Force

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. (explanation on next slide)

How to budge a stubborn mule (cont.)

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.

Relative Velocities in 1 DSchmedrick and his dog, Rover, are goofing around on a train.

Schmed can throw a fast ball at 23 m/s. Rover can run at

9 m/s. The train goes 15 m/s.


Question 1: If Rover is sitting beside the tracks with a radar

gun as the train goes by, and Schmedrick is on the train

throwing a fastball in the direction of the train, how fast does

Rover clock the ball?

vBT = velocity of the ball with respect to the train = 23 m/s

vTG = velocity of the train with respect to the ground = 15 m/s

vBG = velocity of the ball with respect to ground = 38 m/s

This is a simple example, but in general, to get the answer we

add vectors: vBG = vBT + vTG (In this case we can simply

add magnitudes since the vectors are parallel.)

Relative Velocities in 1 D (cont.)

• Velocities are not absolute; they depend on the motion of

the person who is doing the measuring.

• Write a vector sum so that the inner subscripts match.

• The outer subscripts give the subscripts for the resultant.

• This trick works even when vectors don’t line up.

• Vector diagrams help (especially when we move to 2-D).

vBG = vBT + vTG

vBT = 23 m/s vTG = 15 m/s

vBG = 38 m/scontinued on next slide

Question 2: Let’s choose the positive direction to be to the

right. If Schmedrick is standing still on the ground and Rover

is running to the right, then the velocity of Rover with respect

to Schmedrick = vRS = +9 m/s.

From Rover’s perspective, though, he is the one who is still

and Schmedrick (and the rest of the landscape) is moving to

the left at 9 m/s. This means the velocity of Schmedrick with

respect to Rover = vSR = -9 m/s.

Therefore, vRS = -vSR

The moral of the story is that you get the opposite of

a vector if you reverse the subscripts.



vSR

vRS


Question 3: If Rover is chasing the train as Schmed goes by

throwing a fastball, at what speed does Rover clock the ball now?

vBT = 23 m/s vTG = 15 m/s

vBG = 29 m/s

Note, because Rover is chasing the train, he will measure a

slower speed. (In fact, if Rover could run at 38 m/s he’d say the

fastball is at rest.) This time we need the velocity of the ball with

respect to Rover:

vBR = vBT + vTG + vGR = vBT + vTG - vRG = 23 + 15 - 9

= 29 m/s.

Note how the inner subscripts match up again and the outer

most give the subscripts of the resultant. Also, we make use of

the fact that

vGR = -vRG.

vRG = 9 m/s

River Crossing

Current 0.3 m/s

campsite

boat

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?

river


River Crossing (cont.)

Current 0.3 m/s

campsite

boat

river

0.3 m/s

0.5 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.

0.4 m/s


River Crossing: Relative Velocities

Current 0.3 m/s

campsite

river

0.3 m/s

0.5 m/s 0.4 m/s

The red vector is the velocity of the boat with respect to the

water, vBW, which is what your speedometer would read.

Blue is the velocity of the water w/ resp. to the camp, vWC.

Green is the velocity of the boat with respect to the camp, vBC.

The only thing that could vary in our problem was . It had to

be determined so that red + blue gave a vector pointing directly

across the river, which is the way you wanted to go.


River Crossing: Relative Velocities (cont.)

vWC

vBW

vBC

vBW = vel. of boat w/ respect to water

vWC = vel. of water w/ respect to camp

vBC = vel. of boat w/ respect to camp

vBW + vWC = vBC

Look how they add up:

The inner subscripts match; the out ones give subscripts

of the resultant. This technique works in 1, 2, or 3

dimensions w/ any number or vectors.

Law of SinesThe 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 A sin B sin C

a b c= =

Side a is opposite angle A, b is opposite B, and c is opposite C.

A B

C

c

b a

Law of Cosines

Law of Cosines: a2 = b2 + c2 - 2bc 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 -2bc cos A. This term

allows us to work with non-right triangles. Note if A = 90, this

term drops out (cos90 = 0), and we have the normal

Pythagorean theorem.

A B

C

c

b 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 ProblemSuppose 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.)


10

32

vWG

vWA + vAG = vWG80

vWG

80

32

100

Wonder Woman Jet Problem (cont.)

v

132

By the law of cosines v2 = (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 cos132 < 0. The law of sines says:

sin132 sin

v 195=

So, sin = 195sin132 /458, and 18.45

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.

Wonder Woman Problem: Component Method

32

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

165.3694

52.789

299.3816


Wonder Woman: Component Method (cont.)

103

.33

43

165.3694

52.789

103.3343

52.789 165.3694

218.1584 mph

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 .

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.

floor

Free body diagrams #1

mF1

F2

Two applied forces; F2 < mg;

coef. of kinetic friction = k

For the next several slides, draw a free body diagram for each

mass in the set-up and find a (or write a system of 2nd Law

equations from which you could find a.)

v

F1

F2

fk

mg

ma = F1 - fk = F1 - kN

= F1 - k(mg - F2) (to the right). There is not enough

info to determine whether or not N is bigger than F2.

N

answer:


Bodies start at rest; m3 > m1 + m2; frictionless

pulley with negligible mass. answer :

T1

m3g

T1

m1g T2

T2

m2g

Let’s choose clockwise as the + direction.

m1: T1 - m1g -T2 = m1a

m2: T2 - m2g = m2a

m3: m3g - T1 = m3a

system: m3g - m1g - m2g = (m1 + m2 + m3)a(Tensions are internal and cancel out.)

So, a = (m3 - m1 - m2)g / (m1 + m2 + m3)

If masses are given, find a first with last

equation and substitute to find the T ’s.

m1

m3

m2


m2

m1m3

v

k

m1 > m3

m1g

T1

m3g

T2

Note: T1 must be > T2 otherwise m2 couldn’t accelerate.

T2 - m3g = m3a T1 - T2 - km2g = m2a m1g - T1 = m1a

system: m1g - km2g - m3g = (m1 + m2 + m3)a

T1T2

fk

m2g

N

answer:


mv

answer:

Rock falling down in a pool of water

mg - D = ma. So, a = (mg - D) / m. Note: the longer the rock

falls, the faster it goes and the greater D becomes, which is

proportional to v. Eventually, D = mg and a becomes zero,

as our equation shows, and the rock reaches terminal

velocity.

D

mg

m


answer:

cotton

candy Fe

A large crate of cotton candy

and a small iron block of the

same mass are falling in air at

the same speed, accelerating

down.

R

mg

R

mg

Since the masses are the

same, a = (mg - R) / m for

each one, but R is bigger for

the cotton candy since it has

more surface area and they

are moving at the same

speed (just for now). So the

iron has a greater accelera-

tion and will be moving faster

than the candy hereafter.

The cotton candy will reach

terminal vel. sooner and its

terminal vel. will be less than

the iron’s.

Free body diagrams #6 a

The boxes are

not sliding;

coefficients of

static friction are

given.

answer:

m1

m3

m2

1

2

m2

There is no friction acting on m2.

It would not be in equilibrium otherwise.

T = m3g = f1 1N1 = 1(m1 + m2)g

f1’s reaction pair acting on table is not shown.

m3g

m3

N1

m1g

T

m2g

m2g

N2

f1

Tm1

2 is extraneous

info in this

problem, but not

in the next slide.

Free body diagrams #6 bBoxes accelerating

(clockwise); m1 &

m2 are sliding;

coef’s of kinetic

friction given.

answer:

m1

m3

m2

1

2

v

m2

There is friction acting on m2 now.

It would not be accelerating otherwise.

m3g - T = m3a; f2 = m2a; T - f1 - f2 = m1a,

where f1 = 1N1 = 1(m1 + m2)g

and f2 = 2N2 = 2m2g.

m3g

N1

m1g

T

m2g

m2g

N2f1

Tm1

f2

f2

Note: f2 appears

twice; they’re

reaction pairs.


k

v

Boxes moving clockwise

at constant speed.answer:

m2g

T

m1g

T

fk

N

Since a = 0, m2g = T = m1g sin + fk = m1gsin + km1g cos

m2 = m1 (sin + k cos ). This is the relationship

between the masses that must exist for equilibrium.

Constant velocity is the same as

no velocity when it comes to the 2nd Law.

Note: sin, cos,

and k are all

dimensionless

quantities, so we

have kg as units

on both sides of

the last equation.

Free body diagrams #8Mr. Stickman is out for a walk. He’s moseying along but picking up

speed with each step. The coef. of static friction between the grass

and his stick sneakers is s.

v

answer:

mg

N

fs

Here’s a case where friction is a good thing. Without it we couldn’t walk. (It’s difficult to walk on ice since s is so small.)We use fs here since we assume he’s not slipping. Note: friction is in the direction of motion in this case. His pushing force does not appear in the free body diag. since it acts on the ground, not him. The reaction to his push is friction.

Fnet = fsSo, ma = fs fs, max = s mg

Thus, a sg.


ground

m

F

k

vfk

mg

F sinN

Note: is measured with

respect to the vertical here.

Box does not get lifted up off

the ground as long as

Fcos mg. If F cos mg,

then N = 0.

Box budges if Fsin > fs, max

= sN = s (mg - Fcos ).

While sliding,

Fsin - k (mg - F cos ) = ma.

answer:

Dot Products

First recall vector addition in component form:

x1, y1, z1 x2, y2, z2+ = x1+ x2, y1+y2, z1+ z2

It’s just component-wise addition.

Note that the sum of two vectors is a vector.

For a dot product we do component-wise multiplication and

add up the results:

x1, y1, z1 x2, y2, z2 = x1x2 + y1y2 + z1z2

Note that the dot product of two vectors is a scalar!

Ex: -2, 3, 10 N 1, 6, -5m = -2 + 18 - 50 = -34 Nm

Dot products are used to find the work done by a force

applied over a distance, as we’ll see in the future.

Dot Product Properties

• The dot product of two vectors is a scalar.

• It can be proven that a b = abcos, where is the

angle between a and b.

• The dot product of perpendicular vectors is zero.

• The dot product of parallel vectors is simply the product of

their magnitudes.

• A dot product is commutative:

• A dot product can be performed on two vectors of the

same dimension, no matter how big the dimension.

a b = b a

Unit Vectors in 2-D

The vector v = -3, 4 indicates 3 units left and 4 units up,

which is the sum of its components:

v = -3, 4 = -3, 0 + 0, 4

Any vector can be written as the sum of its components.

Let’s factor out what we can from each vector in the sum:

v = -3, 4 = -3 1, 0 + 4 0, 1 The vectors on the right side are each of magnitude one. For

this reason they are called unit vectors.

A shorthand for the unit vector 1, 0 is i.

A shorthand for the unit vector 0, 1 is j.

Thus, v = -3, 4 = -3 i + 4 j

Unit Vectors in 3-D

v = 7, -5, 9 = 7, 0, 0 + 0, -5, 0 + 0, 0, 9

One way to interpret the vector v = 7, -5, 9 is that it

indicates 7 units east, 5 units south, and 9 units up. v can

be written as the sum components as follows:

= 7 1, 0, 0 - 5 0, 1, 0 + 9 0, 0, 1

= 7 i - 5 j + 9k

In 3-D we define these unit vectors:

i = 1, 0, 0 , j = 0, 1, 0 , and k = 0, 0, 1

(continued on next slide)

Unit Vectors in 3-D (cont.)

x

y

z

1i

j

1

k1

The x-, y-, and z-axes are

mutually perpendicular,

as are i, j, and k. The

yellow plane is the x-y

plane. i and j are in this

plane. Any point in space

can be reached from the origin using a linear combination

of these 3 unit vectors. Ex: P = (-1.8, -1.4, 1.2) so the vector

P

-1.8 i – 1.4 j + 1.2k will extend from the origin to P.

Determinants

To take a determinant of a 22 matrix,

multiply diagonals and subtract. The

determinant of A is written |A | and it

equals 3(11) - 4(-2) = 33 + 8 = 41.

114

23A =

In order to do cross products we will need to find determinants

of 33 matrices. One way to do this is to expand about the 1st

row using minors, which are smaller determinants within a

determinant. To find the minor of an element, cross out its row

and column and keep what remains.

ihg

fed

cba Minor of a:

ih

fe

ig

fd

Minor of b:

hg

ed

Minor of c:

cont. on next slide

Determinants (cont.)

ihg

fed

cba

(Minor of a) - b

ih

fe

ig

fd

(Minor of b) + c

hg

ed

(Minor of c)

By definition,

= a

= a - b + c

= a (e i - h f ) - b (d i - g f ) + c (dh - ge)

Determinants can be expanded about any row or column.

Besides cross products, determinants have many other

purposes, such as solving systems of linear equations.

Cross Products

Let v1 = x1, y1, z1 and v2 = x2, y2, z2 .

By definition, the cross product of these vectors (pronounced

“v1 cross v2”) is given by the following determinant.

v1 v2 = x1 y1 z1

x2 y2 z2

i j k

= (y1 z2 - y2 z1) i - (x1 z2 - x2 z1)j + (x1 y2 - x2 y1)k

Note that the cross product of two vectors is another vector!

Cross products are used a lot in physics, e.g., torque is a

vector defined as the cross product of a displacement vector

and a force vector. We’ll learn about torque in another unit.

ab

ab.

Right hand rule

b

a

ab

A quick way to determine the direction of a cross product is

to use the right hand rule. To find a b, place the knife

edge of your right hand (pinky side) along a and curl your

hand toward b, making a fist. Your thumb then points in the

direction of

It can be proven that the magnitude of

is given by:

absin|ab| =

where is the angle between

a and b.

Dot Product vs. Cross Product

1. The dot product of two vectors is a scalar; the cross product

is another vector (perpendicular to each of the original).

2. A dot product is commutative; a cross product is not. In fact,

ab = -ba.

x1, y1, z1 x2, y2, z2 = x1x2 + y1y2 + z1z2

3. Dot product

definition:

Cross product

definition: v1 v2 = x1 y1 z1

x2 y2 z2

i j k

4. a b = abcos, and absin|ab| =

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