Transcript Vectors - Urbana School District #116

Vectors
• Vector Operations
• Components
• Inclined Planes
• Equilibrium
• 2-D Force &
Motion Problems
• Trig Applications
• Relative Velocities
• Free Body Diagrams
Vector Addition
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 method
• Parallelogram method
4N
3N
8N
Tip to Tail Method
12 N
in-line examples
9N
21 N
9N
12 N
20 N
Place the tail of one vector
16 N
at the tip of the other. The
vector sum (also called the
resultant) is shown in red. It
20 N
starts where the black vector
16 N
began and goes to the tip of
the blue one. In these
4N
cases, the vector sum
represents the net force.
You can only add or
subtract magnitudes when
the vectors are in-line!
Tip to Tail – 2 Vectors
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).
Place the vectors tip to tail
and draw a vector from the
tail of the first to the tip of
the second.
5m
2m
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.
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
The resultant has
the same magnitude
and direction
regardless of the
method used.
Parallelogram method
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
3x
-2x
½x
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.
Vector Subtraction
red - blue
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
blue - 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
Which vector is bigger?
43 m
15 N
27 m/s
0.056 km
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.
150 N
Horizontal
component
Vertical
component
Vector Components
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 sin 
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.
v

v cos 
Here’s the
correspondence:
cosine  adjacent side
sine  opposite side
Component Example
14.369 m/s
30.814 m/s
25
34 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.
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.
Pythagorean Theorem
14.369 m/s
30.814 m/s
25
34 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
v cos 
v
v sin 


v

v
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.
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:
a = 10 m/s2 at 53.13 N of W
In component form:
4 m/s2
a = -3, 4  m/s2
53.13
3 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:
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.
5m

2m
Adding vectors in component form
If F1 =  3, 7  N and F2 =  2, -4  N, then the
F2
F1
net force is simply given by:
Fnet =  5, 3  N. Just add the
horizontal and vertical components
separately.
2N
Fnet
4N
F1
3N
F2
7N
Inclined Plane
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.
parallel component

Note: red + blue = black
perpendicular
component
mg
continued on next slide
Inclined Plane
(continued)
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 



mg cos 
mg sin 
mg
continued on next slide
Inclined Plane (continued)
mg sin 

mg cos 
mg
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 sin 
mg sin 
mg cos 
mg
mg
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.
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 sin

mg cos
mg
Let’s show that the
Pythagorean theorem
holds for components on
the inclined plane:
(mg sin )2 + (mg cos )2 = (mg)2 (sin2 + cos2 )
= (mg)2 (1) = (mg)2
Inclined Plane: Normal Force
N = mg cos
mg sin
mg cos
mg
Recall normal force is perpendicular 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
 be the case, otherwise the
must
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
With no friction, Fnet = mg + N
= mg cos + mg sin + N
= mg sin.
N = mg cos
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.)

mg cos
mg
Therefore, the net force is the
parallel force in this case.
Acceleration on a Frictionless Ramp
Here Fnet = mg sin = m a. So, a = g sin. 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.
mg sin

mg cos
mg
Incline with friction at equilibrium
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.
mg sin
N = mg cos
fs = mg sin

mg cos
continued on next slide
mg
Incline with friction at equilibrium
fs  s N = s mg cos. Also,
fs = mg sin (only because we
have equilibrium). So,
mg sin  s mg cos .
Since the mg’s cancel and
tan  = sin / cos, we have
mg sin
s  tan.
(cont.)
N = mg cos
fs = mg sin

mg cos
continued on next slide
mg
Incline with friction at equilibrium
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,
N = mg cos
fs = fs, max = s N = s mg cos .
So now we have
mg sin = s mg cos,
and s = tan.
(cont.)
fs = mg sin
mg sin

(Neither of these quantities have units.)
An adjustable ramp is a convenient
way to find the coefficient of static
friction between two materials.
mg cos
mg
Acceleration on a ramp with friction
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.
N = mg cos
fk = k mg cos
fk = kN = k mg cos.
mg sin
Fnet = mg sin - fk = ma.
So, mg sin - k mg cos = ma.

The m’s cancel, which means a
is independent of the size of the
box. Solving for a we get:
a = g sin - k g cos. Once
again, the units work out right.
mg cos
mg
Parallel applied force on ramp
In this case FA and mg sin are
working together against friction.
Assuming FA + mg sin > fs, max
N
the box budges and the 2nd Law tells
us FA + mg sin  - fk = ma.
Mass does not cancel out this time.
mg sin
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.
fk
FA

mg
mg cos
Non-parallel applied force on ramp
Suppose the applied force acts on
the box, at an angle  above the
horizontal, rather than parallel to
FA sin( +  )
the ramp. We must resolve FA
FA
into parallel and perpendicular
N
components (orange and gray)
fk

using the angle  + .

FA cos( +  )
FA serves to increase acceleration mg sin

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
mg
friction).
mg cos
continued on next slide
Non-parallel applied force on ramp
FA sin ( + )
FA
FA cos( +  )
mg sin
N


fk

mg
continued on next slide
(cont.)
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 = k N
= k [mg cos - FA sin( +  )].
Non-parallel applied force on ramp
Assuming the combined force
of orange and blue is enough
to budge the box, we have
FA sin( +  )
FA
Fnet = orange + blue - brown = ma.
FA cos( +  )
mg sin
Substituting, we have
(cont.)
N


fk

FA cos( +  ) + mg sin
- k [mg cos - FA sin( +  )] = ma.
mg
mg cos
Hanging Sign Problem
Support Beam
T2
T1
1
2
mg
continued on next slide
Hanging sign f.b.d.
Free Body Diagram
T2
T1
1
2
mg
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
force
triangle
F = 0 means a closed vector polygon !
net
As long as Fnet = 0, this is
true no matter many forces
are involved.
T2
mg
Vector Equation:
T1 + T2 + mg = 0
T1
continued on next slide
Components
&
Scalar
Equations
Hanging sign equations
T2
T1
T1 sin1
T2 sin2
1
T1 cos1
2
T2 cos2
We use Newton’s 2nd
Law twice, once in
each dimension:
Vertical:
T1 sin1 + T2 sin2 = mg
mg
Horizontal:
T1 cos1 = T2 cos2
Hanging sign sample
Support Beam
T1
T2
35
62
75 kg
Answers:
T1 = 347.65 N
T2 = 606.60 N
Accurately draw all vectors and find T1 & T2.
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.
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:
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.
Tweety:
64 N
38°
43°
Bugs:
95 N
continued on next slide
3 - Way Tug-o-War
Sam:
111 N
68.3384 N
38°
87.4692 N
(continued)
43.6479 N
Tweety: 64 N
43°
46.8066 N
Bugs:
95 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
3 - Way Tug-o-War
(continued)
Next we combine all parallel
43.6479 N vectors by adding or
subtracting:
87.4692 N
46.8066 N 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:
95 N
Sam & Tweety together slightly
overpower Bugs vertically by
16.9863 N
about 17 N. But Sam & Tweety
oppose each other horizontally,
where Sam overpowers Tweety
40.6626 N
by about 41 N.
68.3384 N
continued on next slide
3 - Way Tug-o-War
Fnet = 44.0679 N
(continued)
16.9863 N

40.6626 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.
Now it’s kinematics time:
Using the fact
97.9287 m/s2
22.6689
x = 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.
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?
Answer:
Hint: Find this
angle first.
Curly:
( -203.66 , 2246.22 )
1000 N
in meters
Larry:
150 N
78
93
Moe:
500 N
How to budge 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.
(explanation on next slide)
Big Force
Little Force
How to budge a stubborn mule (cont.)
little force
tree
mule
overhead view
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
tree
T
T
mule
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

tree
T

T
mule
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.
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.)
continued on next slide
Relative Velocities in 1 D
vBG = vBT + vTG
(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).
vBT = 23 m/s
vTG = 15 m/s
vBG = 38 m/s
continued on next slide
Relative Velocities in 1 D
(cont.)
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.
vRS
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.
vSR
Therefore, vRS = - vSR
The moral of the story is that you get the opposite of
a vector if you reverse the subscripts. continued on next slide
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?
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
vBT = 23 m/s vTG = 15 m/s
vGR = - vRG.
vBG = 29 m/s
vRG = 9 m/s
River Crossing
campsite
Current
0.3 m/s
river
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
continued on next slide
there?
River Crossing
(cont.)
campsite
0.3 m/s
0.5 m/s  0.4 m/s
Current 0.3 m/s
river
boat
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
River Crossing: Relative Velocities
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.
continued on next slide
campsite
0.3 m/s
0.5 m/s  0.4 m/s
river
Current 0.3 m/s
River Crossing: Relative Velocities
vWC
vBW
(cont.)
vBW = vel. of boat w/ respect to water

vBC
vWC = vel. of water w/ respect to camp
vBC = vel. of boat w/ respect to camp
Look how they add up:
vBW + vWC = vBC
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 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.
C
b
A
Law of Sines:
a
B
c
sin A
a
=
sin B
=
b
sin C
c
Side a is opposite angle A, b is opposite B, and c is opposite C.
C
Law of Cosines
b
c
A
Law of Cosines:
a
B
a 2 = b 2 + c 2 - 2bc cosA
These two sides
are repeated.
This side is always opposite this angle.
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.
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
Wonder Woman Jet Problem
32
80
(cont.)
32
100
vWG
vWG
10
vWA + vAG = vWG
80
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
Wonder Woman Jet Problem
(cont.)
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
v
So, sin = 195 sin 132 / 458, and   18.45
132
v

80
=
sin
195
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
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
32
165.3694
299.3816
10
52.789
continued on next slide
Wonder Woman: Component Method (cont.)
165.3694
52.789
103.3343
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 .
218.1584 mph
52.789
165.3694
103.3343

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.
Free body diagrams
#1
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.)
F2
v
F1
m
N
floor
F2
F1
fk
Two applied forces; F < mg;
2
coef. of kinetic friction = k
answer:
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.
Free body diagrams #2
Bodies start at rest; m3 > m1 + m2; frictionless
pulley with negligible mass. answer :
Let’s choose clockwise as the + direction.
T1
m1
T1
T1 - m1 g -T2 = m1a
m2:
T2 - m2 g = m2a
m3:
m3 g - T1 = m3a
system: m3 g - m1 g - m2 g = (m1 + m2 + m3)a
(Tensions are internal and cancel out.)
m1 g T2
m3
T2
m2 g
m1:
So, a = (m3 - m1 - m2)g / (m1 + m2 + m3)
m2
m3 g
If masses are given, find a first with last
equation and substitute to find the T ’s.
Free body diagrams #3
answer:
Note: T1 must be > T2 otherwise m2 couldn’t accelerate.
T2 - m3 g = m3 a
T1 - T2 - k m2 g = m2 a
m1 g - T1 = m1 a
system: m1 g - km2 g - m3 g = (m1 + m2 + m3) a
v
m1 > m3
m2
T1
T2
m3
m3 g
k
N
T2
T1
m1
fk
m2 g
m1 g
Free body diagrams #4
answer:
D
v
m
Rock falling down in a pool of water
m
mg
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.
Free body diagrams #5
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
R
cotton
Fe
candy
mg
mg
answer:
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 acceleration 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
The boxes are
not sliding;
coefficients of
static friction are
given.
2
N1
f1
m2 g
m1
1
#6 a
answer: There is no friction acting on m2.
It would not be in equilibrium otherwise.
T = m3 g = f1  1 N1 = 1(m1 + m2) g
f1’s reaction pair acting on table is not shown.
2 is extraneous
m2
info in this
problem, but not
in the next slide.
m1
T
T
N2
m1 g
m3
m3
m2
m2 g
m3 g
Free body diagrams
Boxes accelerating
(clockwise); m1 &
m2 are sliding;
coef’s of kinetic
friction given.
2
N1
f2
m1
1
answer: There is friction acting on m2 now.
It would not be accelerating otherwise.
m3 g - T = m3 a; f2 = m2 a; T - f1 - f2 = m1 a,
where f1 = 1 N1 = 1(m1 + m2) g
and f2 = 2 N2 = 2 m2 g.
m2
Note: f2 appears
twice; they’re
reaction pairs.
v
m1
T
f1
m2 g
#6 b
T
N2
m1 g
m2
m2 g
m3
f2
m3 g
Free body diagrams
Boxes moving clockwise
at constant speed.
#7
answer: Constant velocity is the same as
no velocity when it comes to the 2nd Law.
Since a = 0, m2 g = T = m1 g sin + fk = m1 g sin + k m1 g cos
 m2 = m1 (sin + k cos ). This is the relationship
between the masses that must exist for equilibrium.
N
T
v
fk
k
m1 g

T
m2 g
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
#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 s.
answer:
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.
N
v
fs
mg
Fnet = fs
So, ma = fs  fs, max = s m g
Thus, a  s g.
Free body diagrams
F
N
fk

m
F sin
v
k
ground
#9
answer:
Note:  is measured with
respect to the vertical here.
Box does not get lifted up off
the ground as long as
F cos  mg. If F cos  mg,
then N = 0.
Box budges if F sin > fs, max
= s N = s (mg - F cos ).
While sliding,
F sin - k (mg - F cos ) = ma.
mg
Dot Products
First recall vector addition in component form:
 x 1 , y 1 , z 1  +  x 2 , y 2 , z 2  =  x 1 + x 2 , y 1 + y 2, z 1 + z 2 
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:
 x 1 , y 1, z 1    x 2, y 2, z 2  =
x1 x2 + y1 y2 + z1 z2
Note that the dot product of two vectors is a scalar!
Ex:  -2, 3, 10  N

 1, 6, -5  m = -2 + 18 - 50 = -34 N m
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
angle between
a
a  b = a b cos, where 
is the
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b = ba
• A dot product can be performed on two vectors of the
same dimension, no matter how big the dimension.
Unit Vectors in 2-D
Any vector can be written as the sum of its components.
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 
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
A shorthand for the unit vector
 1, 0  is i.
 0, 1  is j.
Thus, v =  -3, 4  = -3 i + 4 j
Unit Vectors in 3-D
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:
v =  7, -5, 9  =  7, 0, 0  +  0, -5, 0  +  0, 0, 9 
= 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)
Unit Vectors in 3-D
z
P
k
1
j
1
(cont.)
y
1
i
The x-, y-, and z-axes are
x
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
-1.8 i – 1.4 j + 1.2 k will extend from the origin to P.
Determinants
To take a determinant of a 2  2 matrix,
 3  2  multiply diagonals and subtract. The

A = 
determinant of A is written | A | and it
4
11

 equals 3(11) - 4(-2) = 33 + 8 = 41.
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.
a
b
c
d
e
f
g
h
i
Minor of a:
e
h
f
i
Minor of b:
d
g
f
i
Minor of c:
d
e
g h
cont. on next slide
Determinants
(cont.)
By definition,
a
b
c
d
e
f
g
h
i
= a (Minor of a) - b (Minor of b) + c (Minor of c)
e
= a
h
f
i
-
d
b
g
f
i
+ c
d
e
g h
= a (e i - h f ) - b (d i - g f ) + c (d h - g e)
Determinants can be expanded about any row or column.
Besides cross products, determinants have many other
purposes, such as solving systems of linear equations.
Let
Cross Products
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.
i
v1  v2 =
j
k
x1 y1 z1
x2 y2 z2
= (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.
Right hand rule
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 a  b.
a b
It can be proven that the magnitude of
a b is given by:
|a b | = a bsin
where

a and b.
is the angle between

b
a
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.
3. Dot product
definition:
Cross product
definition:
 x1, y1, z1    x2, y2, z2 
i
v1  v2 =
j
= x1 x2 + y1 y2 + z1 z2
k
x1 y1 z1
x2 y2 z2
4. a  b = a b cos, and |a b | = a bsin