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SCALARS
A SCALAR is ANY quantity
in physics that has
MAGNITUDE, but NOT a
direction associated with
it.
Magnitude – A numerical
value with units.
Scalar
Example
Magnitude
Speed
20 m/s
Distance
10 m
Age
15 years
Heat
1000
calories
VECTOR
Vector
A VECTOR is ANY quantity
in physics that has BOTH
MAGNITUDE and
DIRECTION.
   
v , x, a, F
Vectors are typically
illustrated by drawing an
ARROW above the symbol.
The arrow is used to convey
direction and magnitude.
Velocity
Magnitude
& Direction
20 m/s, N
Acceleration 10 m/s/s, E
Force
5 N, West
POLAR NOTATION
Polar notation defines a vector by designating the vector’s
magnitude |A| and angle θ relative to the +x axis. Using
that notation the vector is written:
In this picture we have a force vector with
magnitude 12 Newtons oriented at 210
degrees with the + x axis. It would be
characterized as F = 12N < 210
SCALAR MULTIPLICATION
Multiplying a vector by a scalar
will ONLY CHANGE its magnitude.
Multiplying a vector by
“-1” does not change the
magnitude, but it does
reverse it's direction or in a
sense, it's angle.
Thus if A = 12, Then 2A = 24
-1/2A
Thus if A = 12 , then
-A = -12
If A = 12, then
(-1/2)A = -6
UNIT VECTOR NOTATION
An effective and popular system used in engineering is
called unit vector notation. It is used to denote vectors
with an x-y Cartesian coordinate system.
UNIT VECTOR NOTATION
=3j
j= vector of magnitude “1” in the “y” direction
i = vector of magnitude “1” in the “x” direction
= 4i
The hypotenuse in Physics is
called the RESULTANT or
VECTOR SUM.
The LEGS of the triangle are
called the COMPONENTS
A  4iˆ  3 ˆj
3j
4i
Horizontal Component
Vertical Component
NOTE: When drawing a right triangle that
conveys some type of motion, you MUST draw
your components HEAD TO TAIL.
UNIT VECTOR NOTATION
iˆ - unit vecto r  1 in the  x direction
ˆj - unit vecto r  1 in the  y direction
kˆ - unit vecto r  1 in the  z direction
The proper terminology is to use
the “hat” instead of the arrow. So
we have i-hat, j-hat, and k-hat
which are used to describe any
type of motion in 3D space.
How would you write vectors J and K in
unit vector notation?
J  2iˆ  4 ˆj
K  2iˆ  5 ˆj
APPLICATIONS OF VECTORS
VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add them.
Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his
displacement relative to where he started?
54.5 m, E
+
84.5 m, E
30 m, E
Notice that the SIZE of
the arrow conveys
MAGNITUDE and the
way it was drawn
conveys DIRECTION.
APPLICATIONS OF VECTORS
VECTOR SUBTRACTION - If 2 vectors are going in opposite directions, you SUBTRACT.
Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement
relative to where he started?
54.5 m, E
30 m, W
24.5 m, E
-
NON-COLLINEAR VECTORS
When 2 vectors are perpendicular, you must use the
Pythagorean theorem.
A man walks 95 km, East then 55
km, north. Calculate his
RESULTANT DISPLACEMENT.
c2  a2  b2  c  a 2  b2
55 km, N
c  Resultant  952  552
c  12050  109.8 km
95 km,E
BUT…..WHAT ABOUT THE VALUE OF THE
ANGLE???
Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the
VALUE of the angle.
109.8 km
55 km, N
 N of E
95 km,E
To find the value of the angle we
use a Trig function called
TANGENT.
Tan 
opposite side 55

 0.5789
adjacent side 95
  Tan 1 (0.5789)  30
109.8km@ 30 NofE
95iˆ m  55 ˆj m
EXAMPLE
A boat moves with a velocity of 15 m/s, N in a river which flows with a
velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with
respect to due north.
Rv  82  152  17 m / s
8.0 m/s, W
15 m/s, N
Rv

The Final Answer :
8
Tan   0.5333
15
  Tan 1 (0.5333)  28.1
17 m / s @ 28.1 WofN
 8iˆ m / s  15 ˆj m / s
Section
VECTORS
5.1
Vectors in Multiple Dimensions
If two vectors to be added are at an angle other than 90°, then you
can use the law of Cosines or the law of Sines.
Law of Cosines:
R2 = A2 +B2 – 2ABcosѲ
Where Ѳ is the cosine of the angle between length A and B.
Section
VECTORS
5.1
Vectors in Multiple Dimensions
Law of Sines: R / sin Ѳ = A / sin a = B / sin b
a = angle opposite length A
b = angle opposite length B
SOLVE FOR THE MAGNITUDE
1) 55 m north + 97 m northeast
2) 39 m/s south + 45 m/s southeast + 62 m/s north
Section
5.1
Finding the Magnitude of the Sum of Two Vectors
The steps covered were:
Step 1: Analyze and sketch the Problem

Sketch the two displacement vectors, A and B, and the angle between them.
Step 2: Solve for the Unknown

When the angle is 90°, use the Pythagorean theorem to find the magnitude of
the resultant vector.

When the angle does not equal 90°, use the law of cosines to find the
magnitude of the resultant vector.
Step 3: Evaluate the answer: check units, does it make sense
MAGNITUDE AND DIRECTION
If the direction is not one of the cardinals (north, south, east, west, up down, left
right)
Then you must describe it using an angle from one of the cardinal directions
Example:
How far east does one travel if they move 1356 m/s 67 0 NE?
Section
VECTORS
5.1
Components of Vectors
When the motion you are describing is confined to the surface of
Earth, it is often convenient to have the x-axis point east and the yaxis point north.
When the motion involves an object moving through the air, the
positive x-axis is often chosen to be horizontal and the positive yaxis vertical (upward).
If the motion is on a hill, it’s convenient to place the positive x-axis
in the direction of the motion and the y-axis perpendicular to the xaxis.
Section
Vectors
5.1
Algebraic Addition of Vectors
Two or more vectors (A, B, C, etc.) may be added by first resolving
each vector into its x- and y-components.
The x-components are added to form the x-component of the
resultant: Rx = Ax + Bx + Cx.
Similarly, the y-components are
added to form the y-component of
the resultant:
Ry = Ay + By + Cy.
Repetitive, Tedious, but easy
Section
5.1
VECTORS
Algebraic Addition of Vectors
Because Rx and Ry are at a right angle (90°), the magnitude of the
resultant vector can be calculated using the Pythagorean theorem,
R2 = Rx2 + Ry2.
Section
5.1
QUESTION 1
Jeff moved 3 m due north, and then 4 m due west to his friends house.
What is the magnitude of Jeff’s displacement?
QUESTION 2
Calculate the resultant of
three vectors A, B, C as
shown in the figure.
(Ax = Bx = Cx = Ay = Cy = 1
units and By = 2 units)
EXAMPLE
Johnny and Susie push a box containing a tied up physics teacher on ice towards the
water’s edge. If the mass of teacher and box is 95 kg, and Susie pushes with 150
N [N] while Johnny pushes with 200 N 59˚ east of south, find:
A) net force
B) acceleration
C) how far the box was pushed in 2 minutes
2ND EXAMPLE
What is the acceleration of an object whose initial velocity is 20 m/s [S] and final
velocity is 30 m/s [E]. It took 2 seconds to change from initial to final velocity.
SCALAR MULTIPLICATION
SCALAR MULTIPLICATION is multiplying a vector by a scalar.
Example:
A=8i+7j+9K
2A=16i +14j + 18k
THE SCALAR DOT PRODUCT
In unit vector notation, it looks a little
different. Consider:
The "Dot" product between these is equal to:
THE SCALAR DOT PRODUCT
What is the
SIGNIFICANCE of the dot product?
DOT PRODUCTS IN PHYSICS
Consider this situation: A force F is
applied to a moving object as it
transverses over a frictionless surface for
a displacement, d.
It is |F|Cos θ ! Because it is
parallel to the displacement d
that causes the object to move
In fact if you apply the dot product, you get (|F|Cos θ)d, which happens to be
defined as "WORK" (check your equation sheet!)
A  B  A B cos 
W  F  x  F x cos 
Work is a type of energy and energy DOES NOT
have a direction, that is why WORK is a scalar or
in this case a SCALAR PRODUCT
(AKA DOT PRODUCT).
THE “CROSS” PRODUCT (VECTOR MULTIPLICATION)
Multiplying 2 vectors sometimes gives you a VECTOR quantity which we call the
VECTOR CROSS PRODUCT.
The cross product between A and B produces a
VECTOR quantity. The magnitude of the vector
product is defined as:
Where  is the NET angle between the two
vectors. As shown in the figure.

B
A
THE VECTOR CROSS PRODUCT

A
B
A  B  A B sin   12 5 sin 150
A  B  30kˆ
What about the direction???? Positive k-hat???
We can use what is called the RIGHT HAND
THUMB RULE.
•Fingers are the first vector, A
•Palm is the second vector, B
•Thumb is the direction of the cross product.
•Cross your fingers, A, towards, B so that they
CURL. The direction it moves will be either
clockwise (NEGATIVE) or counter clockwise
(POSITIVE)
In our example, the thumb points OUTWARD which is the Z axis and thus our
answer would be 30 k-hat since the curl moves counter clockwise.
CROSS PRODUCTS AND UNIT VECTORS
The cross product between B and A produces a
VECTOR of which a 3x3 matrix is need to
evaluate the magnitude and direction.
iˆ
B  A  Bx
Ax
ˆj
By
Ay
kˆ
Bz
Az
You start by making a 3x3
matrix with 3 columns, one
for i, j, & k-hat. The
components then go under
each appropriate column.
Since B is the first vector it comes first in the matrix
CROSS PRODUCTS AND UNIT VECTORS
iˆ
B  A  Bx
Ax
ˆj
By
Ay
You then make an X in the columns
OTHER THAN the unit vectors you are
working with.
•For “i” , cross j x k
•For “j” , cross i x k
z •For “k” , cross i x j
kˆ
B
Az
Let’s start with the i-hat vector: We cross j x k
iˆ  ( By Az )  ( Bz Ay )
Now the j-hat vector: We cross i x k
ˆj  ( Bz Ax )  ( Bx Az )
Now the k-hat vector: We cross i x j
kˆ  ( Bx Ay )  ( By Ax )
EXAMPLE
iˆ  (4)(5)  (6)( 4)  44
ˆj  (6)(3)  (2)(5)  8
Let’s start with the i-hat vector: We cross j x k
Now the j-hat vector: We cross i x k
kˆ  (2)( 4)  (4)(3)  20
Now the k-hat vector: We cross i x j
The final answer would be:
B  A  44iˆ  8 ˆj  20kˆ
THE SIGNIFICANCE OF THE CROSS PRODUCT
In this figure, vector A has been split into 2
components, one PARALLEL to vector B and
one PERPENDICULAR to
vector B. Notice that the component
perpendicular to vector B has a magnitude
of |A|sin θ
THEREFORE when you find the CROSS PRODUCT, the result is:
i) The MAGNITUDE of one vector, in this case |B| and,
ii) The MAGNITUDE of the 2nd vector's component that runs
perpendicular to the first vector. ( that is where the sine comes
from)
CROSS PRODUCTS IN PHYSICS
The cross product system will also be used in mechanics (rotation) as well
as understanding the behavior of particles in magnetic fields.
A force F is applied to a wrench a
displacement r from a specific point of
rotation (ie. a bolt).
Common sense will tell us the larger r is the
easier it will be to turn the bolt.
But which part of F actually causes the wrench to turn? |F| Sin θ
A  B  A B sin 
 
F  r  F r sin 
CROSS PRODUCTS IN PHYSICS
A  B  A B sin 
 
F  r  F r sin 
What about the DIRECTION?
Which way will the wrench turn?
Counter Clockwise
Is the turning direction positive or negative?
Positive
Which way will the BOLT move? IN or OUT of the page?
OUT
You have to remember that cross products give you a direction on the
OTHER axis from the 2 you are crossing. So if “r” is on the x-axis and “F” is
on the y-axis, the cross products direction is on the z-axis. In this case, a
POSITIVE k-hat.
FORCES IN TWO DIMENSIONS
An object is in equilibrium when the net force on it is
zero. When in equilibrium the object is motionless or
move with constant velocity.
Equilibrium also occurs when the resultant of three or
more forces = a net force of zero.
The equilibrant force is equal in magnitude but
opposite in direction to the resultant vector.
B
R
B
A
A
A
B
Equilibrant
EXAMPLE PROBLEM
Creating Equilibrium
A 168-N sign is supported in a motionless position by two ropes that each make 22.5 angles
with the horizontal. What is the tension in the ropes?
FA
22.5 
+y
FB
22.5 
Sign
Fg
+x
MOTION ALONG AN INCLINE PLANE
A tilted surface is called an inclined plane.
Objects accelerate down inclined planes
because of an unbalanced force.
The two forces acting upon a crate which is
positioned on an inclined plane (assumed to
be friction-free) are the force of gravity and
the normal force.
- The force of gravity (weight) acts in a
downward direction; yet the normal force acts
in a direction perpendicular to the surface.
Fgx = F‫ =װ‬Fgsin 
Fgy = F= Fgcos
Fg = Weight= mg
EXAMPLE PROBLEM
Components of Weight for an Object on an incline.
A trunk weighing 562 N is resting on a plane inclined
30.0 above the horizontal. Find the components of the
weight force parallel and perpendicular to the plane.
Skiing Downhill
A 62-kg person on skis is going down a hill sloped at 37 .
The coefficient of kinetic friction between the skis and
the snow is 0.15. How fast is the skier going 5.0 s after
starting from rest?
• Force is a vector
• Vectors are made of components
• The components are independent
of each other
Section 1 :Simple Breakdown of Forces
You can break down forces into several components easily. For example, the force F 1can
be broken into two forces: Fx and Fy.
Section 2 .Two Dimensional Forces into One
Jack pushed a box with a force of 30 N at 0 degree and Michael pushed it with a
force of 40 N at 45 degrees. How can we find the net force acting on the box?
The first thing you have to do is to find all forces on
x direction (x axis) only.
30N + (cos 45 * 40) N = 58.3 N. East
Then, you will have to analyze all forces on y direction (y axis).
0N + (sin 45 * 40) N = 28.3 N. North
Combined forces: use Pythagorean Theorem, we can calculate that
N NorthEast
is the magnitude (size) of the combined forces.
Section 3 :One Dimensional Forces into Two
You can also break down forces. For example, Fred pushed a box to the east and Jack pushed it
to the north. If the net force is 100 N to north east by 45 degrees,
the force applied by Fred would be
FFred = cos 45 * 100 = 70.7 N
and the force by Jack is
FJack = sin 45 * 100 = 70.7 N
Section 4 .Forces involving Gravity
This force, called the force of parallel( F//,)causes the box to move forward. F //
can be calculated by Fg * sin x.
To conclude, the mixture of the force of parallel and the force of friction determines
how the box moves.
F// > fr box will slide.
F// = fr box will not slide.
F perpendicular is equal in magnitude to the normal force
F perpendicular = Fg cos (x)
QUESTION
Draw a free body diagram of a skier moving
down an inclined slope
FORCE OF GRAVITY, NORMAL FORCE
What force is making the skier move down
the slope?
What force is pushing the skier into the
ground?
What direction is the normal force?
On an inclined plane, the magnitude of the normal force
between the object and the plane will usually not be
equal to the object’s weight.
PROBLEM 1
If the skier has a mass of 50 kg and the angle of the slope is 30 degrees from
horizontal, find
1)
What is the angle of inclination?
2)
What is the amount of force pushing object into ground? (force perp?)
3)
What is the amount of force pulling the skier down the hill? (force parallel)
PROBLEM #2
Skier from problem moves down 45 ˚hill.
Part 1) What is their acceleration, if hill was considered smooth?
Part 2) What is their acceleration if hill has a coefficient of friction of 0.21?
Part 3) How far down the hill will the skier travel in 33 seconds in part 2
PROBLEM 3
What minimum coefficient of friction value would be needed on the slope for the
skier from problem #2 to be stuck (not accelerate down the slope?)
Section
5.2
FRICTION
Values that determine the amount of friction
Frictional force depends on the materials that the surfaces are
made of.
For example, there is more friction between skis and concrete than
there is between skis and snow.
The normal force between the two objects also matters. The harder
one object is pushed against the other, the greater the force of
friction that results.
Section
5.2
Friction
Static and Kinetic Friction
The coefficients of friction are nearly independent of the area of contact
The forces themselves, Ff and FN, are at right angles to each other. The table here
shows coefficients of friction between various surfaces.
Although all the listed coefficients are less than 1.0, this does not mean that
they must always be less than 1.0.
KINETIC FRICTION
The force of kinetic
friction acts when the
object is in motion
Frk = µk Fn
STATIC FRICTION, ƑS
Static friction acts to keep the
object from moving
If F increases, so does ƒs
If F decreases, so does ƒs
Frs  µs Fn
CALCULATION OF FRICTION
Ff = µ F n
Ff or f = Force of friction
µ = Coefficient of friction
Fn = Normal force
FORCE OF FRICTION PROBLEMS
 Only the normal force determines the amount of
frictional force applied to a moving object
 Generally, the normal force equals the weight of
sliding object except at an angle then it is the
perpendicular component of the weight
 Other forces applied to an object may also effect
friction if they effect the amount of force applied to
ground by object
DETERMINING NORMAL FORCE
What force is dragging suitcase across floor?
What forces effect normal force on suitcase?
The applied force of woman must be divided up into 2 components
One horizontal, which tries to pull suitcase along ground, T //
One vertical, which tries to lift suitcase off ground, Tperp
FN
Force applied by
woman
T//
Tperp
W
DETERMINING NORMAL FORCE
 X components
 Fnet x = Tx – Fr = 0
 Tx = Fr
 Y components
 Fnet y = Ty + FN – W = 0
 FN = W - Ty