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Vectors
• Vector Operations
• Components
• Inclined Planes
• Equilibrium
• 2-D Motion &
Force 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:
• Tip to tail method
• Parallelogram method
8N
4N
3N
Tip to Tail Method
12 N
in-line examples
9N
21 N
9N
12 N
20 N
Place the tail of one vector at 16 N
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
4N
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!
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).
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 (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
3x
-2x
½
x
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 + black = red
blue + (red – blue) = red
blue - red
red = red
Note that red - blue is the opposite of blue - red.
Other Operations
• Vectors cannot be multiplied but there are
two types of vector products.
– Cross product
– Dot product
– We’ll use these products in Advanced Physics
• 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).
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
horizontal. The units of the components are the
same as the units for the original vector.
Here’s the
correspondence:
v

v cos 
cosine  adjacent side
sine  opposite side
Component Example
30.814 m/s
14.369
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
30.814 m/s
14.369
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.
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 2 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
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
(continued)
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. That is,
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 the
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  = ma. 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  (since we have
equilibrium). So,
mg sin   s mg cos .
Since the mg’s cancel out and
tan  = sin  / cos , we have
s  tan .
(cont.)
N = mg cos 
fs = mg sin 
mg sin 

mg cos 
continued on next slide
mg
Incline with friction at equilibrium (cont.)
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
fs = mg sin 
mg sin 
s = tan .

(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
N = mg cos 
fs, max which means tan  must be
greater than s. If this is the
case, forget about s and use k.
fk = kN = kmg cos .
Fnet = mg sin  - fk = ma. So,
mg sin  - kmg cos  = ma.
fk = kmg cos 
mg sin 

The m’s cancel, which means a is
independent of the size of the box.
Solving for a we get:
a = g sin  - kg 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 the box budges and
the 2nd Law tells us
FA + mg sin  - 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.
N
fk
FA
mg sin 

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 the
FA sin ( + )
ramp. This considerably complicates
FA
the problem. The first thing we have
N
to do, after making the diagram, is to
fk

resolve FA into parallel and

FA cos ( + )
perpendicular components
mg sin 
(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).
mg
mg cos 
continued on next slide
Non-parallel applied force on ramp
FA sin ( + )
(cont.)
Because of the perp. comp. of
FA
FA, N < mg cos . Assuming
N
FA sin ( + ) is not big enough
f
k

to lift the box off the ramp, there

is no acceleration in the
FA cos ( + )
perpendicular direction. So,
mg sin 
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 object. So,
N = mg cos  - FA sin ( + ),
mg
mg cos  which means f =  N
k
k
= k [mg cos  - FA sin ( + )].
continued on next slide
Non-parallel applied force on ramp
Assuming the combined force
of orange and blue is enough to
budge the box, we have
Fnet = orange + blue - brown = ma.
FA sin ( + )
FA
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
Tweety:
64 N
38°
To answer this question, we must find a, so
we can do kinematics. But in order to find
a, we must first find Fnet.
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
continued on next slide
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.
3 - Way Tug-o-War
(continued)
Next we combine all parallel
vectors by adding or subtracting:
68.3384 + 43.6479 - 95 = 16.9863,
87.4692 N
46.8066 N
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
95 N
overpower Bugs vertically by
about 17 N. But Sam & Tweety
16.9863 N
oppose each other horizontally,
where Sam overpowers Tweety by
about 41 N.
40.6626 N
68.3384 N
43.6479 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
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.
97.9287 m/s2
22.6689
Now it’s kinematics time:
Using the fact
x = v0 t + 0.5 a 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 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.
Larry:
150 N
Curly:
1000 N
78
93
Moe:
500 N
( -203.66 , 2246.22 )
in meters
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 setup 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
(cont.)
vBG = vBT + vTG
• 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 Schmedrick 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’s say the fastball is still
from his point of view.) 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.
vBT = 23 m/s
vBG = 29 m/s
vTG = 15 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 there?
continued on next slide
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 with respect to the shore or 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
a
b
A
Law of Sines:
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.
Law of Cosines
C
b
A
Law of Cosines:
a
c
B
a2 = b2 + c2 - 2 b c cos A
These two sides are repeated.
This side is always opposite this angle.
It matters not 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 (and 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.)
The law of cosines says: 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 really positive, since cos 132 < 0. The law of sines says:
sin 132
v
=
sin 
195
So, sin  = 195 sin 132 / 458, and   18.45
132
v
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.

80
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 vertical comp. of blue. All units are mph.
103.3343
32
165.3694
299.3816
10
continued on next slide
52.789
Wonder Woman: Component Method (cont.)
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  was.
165.3694
103.3343
218.1584 mph
52.789
103.3343

52.789
165.3694
Comparison of Methods
We ended up with same result for Wonder Woman doing it in two
different ways. Each way requires some work. You can only use
the laws of sines & cosines if:
• you’re dealing with exactly 3 vectors. (If you’re adding three
vectors, the resultant makes 4, and this method won’t work
• the vectors form a triangle.
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
floor
Two applied forces; F2 < mg;
coef. of kinetic friction = k
answer:
F2
N
F1
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.
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
T1
m1
m1g
m3
m2g
m2
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.)
T2
T2
m1:
So, a = (m3 - m1 - m2)g / (m1 + m2 + m3)
m3g
If masses are given, find a first with last
equation and substitute above to find T’s.
Free body diagrams #3
answer:
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
m1 > m3
v
m2
T1
T2
m3
m3g
k
N
T2
T1
m1
fk
m2g
m1g
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
candy
mg
Fe
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’re 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
m2g
m1
1
#6 a
answer: There is no friction acting on m2.
It would not be in equilibrium otherwise.
f1  1N1 = 1(m1 + m2)g = T = m3g.
f1’s reaction pair acting on table is not shown.
2 is extraneous
m2
info in this
problem, but not
in the next one.
m1
T
T
N2
m1g
m3
m2
m2g
m3g
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.
m3g - T = m3a; f2 = m2a; T - f1 - f2 = m1a,
where f1 = 1N1 = 1(m1 + m2)g
and f2 = 2N2 = 2m2g.
m2
Note: f2 appears
twice; they’re
reaction pairs.
v
m1
T
f1
m2g
#6 b
T
N1
m1g
m2
m2g
m3
f2
m3g
Free body diagrams #7
Boxes moving clockwise answer: Constant velocity is the same as no
at a constant speed.
velocity when it comes to the 2nd Law.
Since a = 0, m2g = T = m1g sin  + fk = m1g sin  + km1g cos 
m2 = m1 (sin  + k cos  ). This is the relationship between the
masses that must exist for equilibrium.
N
T
v
fk
k
m1g

T
m2g
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
answer:
his stick sneakers is s.
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 = ma
fs = ma
smg  ma
N
v
fs
mg
a  sg
Since fs  fs,max = sN
Free body diagrams
answer:
Note:  is measured with respect
to the vertical here.
F
N
fk

m
F sin 
v
k
ground
mg
#9
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
= sN = s(mg - F cos ).
While sliding,
F sin  - k(mg - F cos ) = ma.