This document discusses vector concepts including: - Two methods for adding vectors graphically: the tip-to-tail method and parallelogram method. - Decomposing a vector into perpendicular components using trigonometric functions. - Analyzing forces on an inclined plane by decomposing weight into components parallel and perpendicular to the plane. - The normal force on an object on an inclined plane equals the perpendicular component of the object's weight.

1. Vectors
• 2-D Force &
Motion Problems
• Trig Applications
• Relative Velocities
• Free Body Diagrams
• Vector Operations
• Components
• Inclined Planes
• Equilibrium

2. Vector Addition
• Tip to tail method
• Parallelogram method
8 N
4 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:

3. Tip to Tail Method
in-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 N
16 N
12 N
9 N
9 N
12 N
21 N

4. 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).
5 m
2 m
blue + red
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.

5. Commutative Property
blue + red
red + blue
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.)

6. 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.).
blue + green + red

7. 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.

8. Comparison of Methods
red + blue
Tip to tail method
Parallelogram method
The resultant has
the same magnitude
and direction
regardless of the
method used.

9. 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.

10. 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

11. 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.

12. 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.

13. 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.

14. Vector Components
150 N
Horizontal
component
Vertical
component
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:

15. Finding Components with Trig
θ
v
v cosθ
vsinθ
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

16. 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
25°14.369 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.

17. Pythagorean Theorem
34 m/s
30.814 m/s
25°14.369 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.

18. 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.
θ
v
v cosθ
vsinθ v
v
α
β
v cosβ
vsinβ
vcosα
vsin
α

19. Component Form
Instead 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°
a
=
10
m
/s
2
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

20. 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 mθ
tanθ = 2 m /5 m ⇒ θ = 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.

21. 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
F1
F2
3 N
7 N
2 N
4 N

22. Inclined Plane
m
θ
mg
perpendicular
component
parallel component
A crate of chop suey of mass m is setting on a ramp with angle
of inclination θ. 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

23. Inclined Plane (continued)
m
θ
mgmg cosθ
mg sinθ
θ
The diagram contains two right triangles. α is the angle
between black and blue. α + θ = 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
θ. Imagine the parallel component sliding down (dotted blue) to
form a right triangle. Being opposite θ, we use sine. Red is
adjacent to θ, so we use cosine.
α
continued on next slide
mg sinθ

24. Inclined Plane (continued)
m
θ
mg
mg cosθ
mg sinθ
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.

25. How the incline affects the components
m
mg
mg cosθ
mg sinθ
The steeper the incline, the greater θ is, and the greater sinθ
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.
m
mg
mgcosθ
mg sinθ
Extreme cases: When θ = 0, the ramp is flat; red = mg; blue = 0.
When θ = 90°, the ramp is vertical; red = 0; blue = mg.

26. Inclined Plane - Pythagorean
Theorem
m
θ
mg
mgcosθ
mgsinθ
(mg sinθ)2
+ (mg cosθ)2
= (mg)2
(sin2
θ + cos2
θ)
= (mg)2
(1) = (mg)2
Let’s show that the
Pythagorean theorem
holds for components on
the inclined plane:

27. Inclined Plane: Normal Force
m
θ
mgmgcosθ
mgsinθ
N = mgcosθ
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 > mgcosθ would mean the box
is jumping off the ramp.
N < mgcosθ would mean that the
ramp is being crushed.

28. Net Force on a Frictionless Inclined Plane
m
θ
mg
mgcosθ
mgsinθ
N = mgcosθ
With no friction, Fnet = mg + N
= mgcosθ + mgsinθ + N
= mgsinθ.
(mgcosθ + 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.

29. Acceleration on a Frictionless Ramp
m
θ
mg
mgcosθ
mgsinθ
Here Fnet = mgsinθ = ma. So, a = gsinθ. Since sinθ has
no units, a has the same units as g, as they should. Both
the net force and the acceleration are down the ramp.

30. Incline with friction at equilibrium
m
θ
mg
mgcosθ
mgsinθ
fs = mgsinθ
N = mgcosθ
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.
continued on next slide

31. Incline with friction at equilibrium (cont.)
m
θ
mg
mgcosθ
mgsinθ
fs = mgsinθ
N = mgcosθ
fs ≤ µs N = µs mgcosθ. Also,
fs = mgsinθ (only because we have
equilibrium). So,
mgsinθ ≤ µs mgcos θ.
Since the mg’s cancel and
tan θ = sinθ /cosθ, we have
µs ≤ tanθ.
continued on next slide

32. Incline with friction at equilibrium (cont.)
mg
mgcosθ
mgsinθ
fs = mgsinθ
N = mgcosθ
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 = µs N = µs mgcos θ.
So now we have
mgsinθ = µs mg cosθ,
and µs = tanθ.
θ
An adjustable ramp is a convenient
way to find the coefficient of static
friction between two materials.
(Neither of these quantities have units.)

33. Acceleration on a ramp with friction
mg
mgcosθ
mgsinθ
fk = µkmgcosθ
N = mgcosθ
θ
In order for the box to budge,
mgsinθ must be greater than
fs,max which means tanθ must be
greater than µs. If this is the
case, forget about µs and use µk.
fk = µkN = µkmgcosθ.
Fnet = mgsinθ - fk = ma.
So, mgsinθ - µkmgcosθ = ma.
The m’s cancel, which means a
is independent of the size of the
box. Solving for a we get:
a = gsinθ - µkgcosθ. Once

34. Parallel applied force on ramp
mg mgcosθ
mgsinθ
fk
N
θ
FA
In this case FA and mgsinθ are
working together against friction.
Assuming FA + mgsinθ > fs,max
the box budges and the 2nd
Law tells
us FA + mgsin θ - 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
sinθ. If FA were bigger, friction
acts down the ramp and a is up
the ramp.

35. Non-parallel applied force on ramp
mg
mgcosθ
mgsinθ
fk
N
θ
Suppose the applied force acts on
the box, at an angle α above the
horizontal, rather than parallel to
the ramp. We must resolve FA
into parallel and perpendicular
components (orange and gray)
using the angle α + θ.
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
α
FA cos(α + θ)
FAsin(α + θ)
θ
continued on next slide

36. Non-parallel applied force on ramp (cont.)
mg
mgcosθ
mgsinθ
fk
N
θ
FA
α
FA cos(α + θ)
FA sin (α + θ)
θ
Because of the perp. comp. of
FA, N < mgcosθ. Assuming
FA sin(α+θ) is not big enough
to lift the box off the ramp,
there is no acceleration in the
perpendicular direction. So,
FA sin(α + θ) + N = mgcosθ.
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 = mgcosθ - FA sin(α + θ),
which means fk = µk N
= µk [mgcosθ - FA sin(α + θ)].
continued on next slide

37. Non-parallel applied force on ramp (cont.)
mg mgcosθ
mgsinθ
fk
N
θ
FA
α
FA cos(α + θ)
FAsin(α + θ)
θ
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(α + θ) + mgsinθ
- µk [mgcosθ - FA sin(α + θ)] = ma.

38. Support Beam
mg
θ1
T1
θ2
T2
continued on next slide
Hanging Sign Problem

39. mg
θ1
T1
θ2
T2
Since 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.
continued on next slide
Hanging sign f.b.d.
Free Body Diagram

40. mg
T1
T2
Vector Equation:
T1 + T2 + mg = 0
continued on next slide
As long as Fnet = 0, this is
true no matter many forces
are involved.
Hanging sign force triangleFnet = 0 means a closed vector polygon !

41. T2
mg
T1
θ1 θ2
T1cosθ1 T2 cosθ2
T1 sinθ1
T2 sinθ2
T1 cosθ1 = T2 cosθ2
Horizontal:
Vertical:
T1 sinθ1 + T2 sinθ2 = mg
We use Newton’s 2nd
Law twice, once in
each dimension:
Hanging sign equationsComponents & Scalar Equations

42. 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

43. Vector Force Lab Simulation
Go 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.

44. 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

45. 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

46. 95 N
87.4692 N
68.3384 N
46.8066 N
43.6479 N
continued on next slide
3 - Way Tug-o-War (continued)
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.

47. 3 - Way Tug-o-War (continued)
16.9863 N
40.6626 N
Fnet = 44.0679 N
θ
Find Fnet using the Pythagorean theorem. Find θ
using trig: tanθ = 16.9863N / 40.6626N. The
newtons cancel out, so θ = 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

48. 3 - Way Tug-o-War (final)
a = Fnet /m = 44.0679N / 0.45 kg = 97.9287m/s2
. Note the
conversion from grams to kilograms, which is necessary
since 1 m/s2
= 1N/ 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 + 0.5at2
= 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.

49. 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

50. 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)

51. 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)

52. How to budge a stubborn mule (final)
tree mule
little force
T T
Because θ 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.
θ θ

53. Relative Velocities in 1 D
Schmedrick 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.
continued on next slide
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.)

54. 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/s
continued on next slide

55. 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.
Relative Velocities in 1 D (cont.)
continued on next slide
vSR
vRS

56. Relative Velocities in 1 D (cont.)
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

57. 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
continued on next slide

58. 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. θ = tan-1
(0.3 / 0.4) ≈ 36.9°.
θ 0.4 m/s
continued on next slide

59. River Crossing: Relative Velocities
Current
0.3 m/s
campsite
river
0.3 m/s
0.5 m/s
θ 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 θ. 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.
continued on next slide

60. River Crossing: Relative Velocities (cont.)
vWC
vBW
θ
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.

61. 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 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

62. Law of Cosines
Law of Cosines: a2
= b2
+ c2
- 2bccosA
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

63. 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

64. 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°
vWA
vAG
vWG
vWA + vAG = vWG
80°
195 mph
304mph
vWG
80°
32°
100°

65. Wonder Woman Jet Problem (cont.)
195 mph
304mph
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:
θ
sin132° sinθ
v 195
=
So, sinθ = 195sin132°/458, and θ ≈ 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.

66. Wonder Woman Problem: Component Method
32°
vWA=304mph
vAG
= 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.
304
195
103.3343
165.3694
52.789
299.3816
continued on next slide

67. Wonder Woman: Component Method (cont.)
304
195
103.3343
165.3694
52.789
299.3816
103.3343
52.789 165.3694
299.3816
402.7159mph
218.1584 mph
458.0100mph
Combine vertical & horiz. comps. separately and use Pythag.
theorem. α = tan-1
(218.1584/402.7159) = 28.4452°. α is
measured from the vertical, which is why it’s 10° more than θ.
α

68. 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.

69. floor
Free body diagrams #1
m
F1
F2
Two applied forces; F2 < mg;
coef. of kinetic friction = µk
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 - µkN
= F1 - µk(mg - F2) (to the right). There is not enough
info to determine whether or not N is bigger than F2.
N
answer:

70. Free body diagrams #2
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
m1
m3
m2

71. Free body diagrams #3
m2
m1m3
v
µk
m1 > m3
m1g
T1
m3g
T2
Note: T1 must be > T2 otherwise m2 couldn’t accelerate.
T2 - m3g = m3a T1 - T2 - µkm2g = m2a m1g - T1 = m1a
system: m1g - µkm2g - m3g = (m1 + m2 + m3)a
T1
T2
fk
m2g
N
answer:

72. Free body diagrams #4
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

73. Free body diagrams #5
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.

74. Free body diagrams #6a
The boxes are
not sliding;
coefficients of
static friction are
given.
answer:
m1
m3
m2
µ1
µ2
m2
There is no friction acting on m2.
It would not be in equilibrium otherwise.
T = m3g = f1 ≤ µ1N1 = µ1(m1 + m2)g
f1’s reaction pair acting on table is not shown.
m3g
m3
N1
m1g
T
m2g
m2g
N2
f1
T
m1
µ2 is extraneous
info in this
problem, but not
in the next slide.

75. Free body diagrams #6 b
Boxes accelerating
(clockwise); m1 &
m2 are sliding;
coef’s of kinetic
friction given.
answer:
m1
m3
m2
µ1
µ2
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 = µ1N1 = µ1(m1 + m2)g
and f2 = µ2N2 = µ2m2g.
m3g
N1
m1g
T
m2g
m2g
N2
f1
Tm1
f2
f2
Note: f2 appears
twice; they’re
reaction pairs.

76. Free body diagrams #7
m
1
θ
m2
µk
v
Boxes moving clockwise
at constant speed.
answer:
m2g
T
m1g
T
m
1gsin
θ
fk
m
1 gcos
θ
N
Since a = 0, m2g = T = m1g sinθ + fk = m1gsinθ + µkm1gcosθ
⇒ m2 = m1 (sinθ + µk cosθ ). 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: sinθ, cosθ,
and µk are all
dimensionless
quantities, so we
have kg as units
on both sides of
the last equation.

77. Free body diagrams #8
Mr. 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 µs.
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 µs 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 = fs
So, ma = fs ≤ fs, max = µs mg
Thus, a ≤ µs g.

78. Free body diagrams #9
ground
m
θ
F
µk
vfk
mg
F sinθ
Fcosθ
N
Note: θ is measured with
respect to the vertical here.
Box does not get lifted up off
the ground as long as
Fcosθ ≤ mg. If Fcosθ ≥ mg,
then N = 0.
Box budges if Fsinθ > fs, max
= µsN = µs(mg - Fcosθ).
While sliding,
Fsinθ - µk(mg - Fcosθ) = ma.
answer:

79. 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 + y1 y2 + z1z2
Note that the dot product of two vectors is a scalar!
Ex: 〈 -2, 3, 10〉
N
〈 1, 6, -5〉 m = -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.

80. Dot Product Properties
• The dot product of two vectors is a scalar.
• It can be proven that a • b = abcosθ, where θ 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

81. 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

82. 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〉
= 7i - 5j + 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)

83. Unit Vectors in 3-D (cont.)
x
y
z
1
i
j
1
k
1
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
-1.8i – 1.4j + 1.2k will extend from the origin to P.

84. Determinants
To take a determinant of a 2×2 matrix,
multiply diagonals and subtract. The
determinant of A is written |A| and it
equals 3(11) - 4(-2) = 33 + 8 = 41.





 −
114
23
A =
In order to do cross products we will need to find determinants
of 3×3 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

85. 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(ei - hf ) - b(di - gf ) + 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.

86. 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.

87. a× b
a × b.
Right hand rule
φ
b
a
a× b
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:
absinφ|a× b| =
where φ is the angle between
a and b.

88. 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,
a× b = -b× a.
〈 x1, y1, z1〉 〈 x2, y2, z2〉• = x1x2 + y1 y2 + z1z2
3. Dot product
definition:
Cross product
definition: v1 × v2 = x1 y1 z1
x2 y2 z2
i j k
4. a • b = abcosθ, and absinθ|a× b| =