PHYSICS
MARJORIE MANABAT
SCALAR
• Is a quantity that has magnitude and
usually a unit of measure
• Examples:
– distance
– Speed
– Time
– Energy
– Mass
SCALAR
• Rules of algebraic addition, subtraction,
multiplication and division are used
mathematically
– Example:
– 2 kg + 5 kg = 7 kg
– 10 m +29 m = 39 m
– 28 cm – 5 cm = 23 cm
MOTION
• SCALAR QUANTITY
– Is completely described by its size or
magnitude.
– Distance, mass, volume
• VECTOR QUANTITIES
– Is completely described by its magnitude and
direction
– Velocity, Force, work
VECTOR
• Is a quantity with magnitude, direction and
usually a unit of measure.
• 3 important parts:
– Arrowhead – indicates the direction of the
vector
– Length of the arrow – represents the
magnitude of the vector
– Tail – represents the origin of the vector
VECTOR
• Tail
arrow head
length of the arrow
ADDITION OF VECTORS
•
RESULTANT (R)
– The sum of two or more vectors
1. Vectors acting in the same direction
•
The resultant of 2 vectors acting in the same
direction is a vector whose magnitude is equal to
the sum of their magnitudes and acts in the same
direction as they do
– Example
– 25 km N + 30 km N = 55 km N
ADDITION OF VECTORS
• RESULTANT (R)
– The sum of two or more vectors
– Vectors acting in the opposite direction
• The resultant of two vectors acting in opposite
direction is a vector whose magnitude is the
difference of their magnitudes and which acts in
the direction of the greater vector
• Example
– 3 km/hr E – 6 km/hr W = 3 km/hr W
ADDITION OF VECTORS
•
RESULTANT (R)
– The sum of two or more vectors
3. Vectors acting in any direction
•
GENERAL METHODS OF FINDING
RESULTANT
a. POLYGON METHOD (GRAPHICAL METHOD)
b. PARALLELOGRAM METHOD
c. ANALYTICAL METHOD
ADDITION OF VECTORS
a. POLYGON METHOD (GRAPHICAL)
POLYGON METHOD
(GRAPHICAL)
POLYGON METHOD
(GRAPHICAL)
1. Choose an appropriate scale and frame
of reference for the given vectors
2. Draw the first vector starting from the
origin of the reference frame
3. Draw the second vector from the head of
the first vector and draw the third vector
starting from the head of the second
vector
POLYGON METHOD
(GRAPHICAL)
4. All vectors must be connected in series,
head-to-tail fashion
5. To determine the resultant vector,
connect the tail of the first vector to the
head of the last vector drawn. Then, use
a protractor in measuring its direction.
POLYGON METHOD
(GRAPHICAL)
• SAMPLE PROBLEM:
– F1 = 20 N, E
– F2 = 30 N, 45°North of East
– F3 = 10 N, North
– Find FR
• SCALE = 10 N = 1 cm
PARALLELOGRAM METHOD
• This method is used in determining the
resultant of two vectors. Just draw a
parallelogram with the two given vectors
as the sides. The diagonal of the
parallelogram from the same point of
origin represents the resultant.
ADDITION OF VECTORS
•
ANALYTICAL METHOD
– Vector addition by components
1. Resolve the vectors into their components in
the x and y direction.
Ax = A cos θ, Ay = A sin θ
ADDITION OF VECTORS
2. Add the components in the x direction to give
Rx and add the components in the y direction
to give ry. That is
Rx = Ax + Bx + Cx = sum of x-components
Ry = Ay + By + Cy = sum of y-components
ADDITION OF VECTORS
3. Find the magnitude of the resultant R
from the components Rx and Ry. From the
Pythagorean Theorem
R = √(Rx)² + (Ry)²
ADDITION OF VECTORS
4. Determine the direction of the resultant R
using Tan θ = Ry
Rx
θ = inv tan Ry
Rx
SAMPLE PROBLEM
• GIVEN: A = 2 cm, N; B = 3 cm, 20° N of E
C = 5 cm, 40° S or E; D = 4 cm, 40° N or W
Ax = A cos θ = (2cm) (cos 90°) = 0
Ay = A sin θ = (2cm) (sine 90°) = 2cm
Bx = B cos θ = (3cm) (cos 20°) = 2.82.cm
By = B sin θ = (3cm) (sin 20°) = 1.03 cm
SAMPLE PROBLEM
Cx = C cos θ = (5cm) (cos 40°) = 3.83 cm
Cy = C sin θ = (5cm) (sin 40°) = -3.21 cm
Dx = D cos θ = (4cm) (cos 40°) = -3.06 cm
Dy = D sin θ = (4cm) (sin 40°) = 2.57 cm
Rx = Ax + Bx + Cx + Dx
= 0 + 2.82 +3.83 + (-3.06) = 3.59 cm
Ry = Ay + By + Cy + Dy
= 2 + 1.03 + (-3.21) +2.57 = 2.39 cm
SAMPLE PROBLEM
R = √(Rx)² + (Ry) ²
= √(3.59 cm) ² + (2.39 cm) ²
R = 4.31 cm
Tan θ = Ry / Rx = 2.39 cm / 3.59 cm
= 0.6657
θ = inv tan 0.6657 = 33° 39’ 11”
R = 4.31 cm, 33° 39’ 11”
HOMEWORK
• A car goes 5.0 km south, 2.0 km West and
1.0 km North.
– Find the resultant vector using both:
• POLYGON METHOD
• ANALYTICAL METHOD
MOTION
• Defined as a continuous change of
position with respect to a certain reference
point.
– Occurs all around us
– Motion is relative  everything moves
MOTION
• PHYSICAL QUANTITIES:
1. SPEED
–
is a measure of how fast something is
moving
– Distance covered per unit of time
•
•
INSTANTANEOUS SPEED: speed at any
instant; speedometer of a car
AVERAGE SPEED: whole distance covered
divided by the total time of travel
MOTION
•
AVERAGE SPEED = total distance covered
time interval
Ex. We travelled 240 km in 4 hours
Average speed = 240 km / 4 hours
= 60 km/hr
MOTION
2. VELOCITY
– Is a speed in a given DIRECTION
– V= d, dir
t
v = velocity
d, dir = distance in a given direction
(displacement)
t = time interval
int
int
SAMPLE PROBLEM
• Marina swims north for 15 sec. she
covered a distance of 20 m. Compute her
velocity
• ANSWER: 1.33 m/s, NORTH
MOTION
2. VELOCITY
a. CONSTANT VELOCITY
•
•
Motion at constant velocity is motion in a straight
line at constant speed.
Constant direction ~ direction is a straight line
b. CHANGING VELOCITY
•
•
If either the speed or the direction (or both) is
changing, then the velocity is changing
Constant speed and constant velocity are not the
same
Exercise
• An ocean liner going from Miami to Tokyo
stops for a few days in Honolulu. You are on
board and wish to estimate the time it will
take you to get to Tokyo. It is 3700 km from
Miami to Honolulu and 6200 km from
Honolulu to Tokyo. It took you 5 days to
reach Honolulu from Miami. What is the
average velocity of the ocean liner in km/hr?
How long will it take to reach Tokyo from
Honolulu?
MOTION
• ACCELERATION
– Is a measure of how fast the velocity changes
with respect to time
– The body accelerates whenever there is a
change in speed
– a= Δv
or Vf - Vi
t
t
int
int
SAMPLE PROBLEM
• A car starts from rest, ran with a speed of
7 m/s for 3 sec. what is the acceleration?
a = 7 m/s – o m/s
3 sec
• Answer: 2.3 m/s²
TYPES OF MOTION
•
In 2 forms:
1. LINEAR MOTION
•
Refers to the movement of an object along a
straight path
2. CURVILINEAR MOTION
•
•
Is the motion of bodies along curved paths
Ex. Projectile motion and object moving along
circular paths
TYPE OF MOTION
1. LINEAR MOTION
A. Uniformly accelerated motion – along a
straight line,
•
•
A vehicle maintains a constant velocity
A vehicle maintains a uniform change in its
velocity (constant acceleration)
a= Δv
t
int
or Vf - Vi
t
int
TYPE OF MOTION
1. LINEAR MOTION
A. Uniformly accelerated motion
X = Vi t + at²
2
X~displacement
t ~ time
Vi ~ initial velocity
a ~ acceleration
SAMPLE PROBLEM
•
AN AUTOMOBILE IS TRAVELING 65
km/h. IT BRAKES DECELERATE IT at 6
m/s².
1. How long will it take to stop the car?
2. How far will the car travel after the brakes
are applied?
ANSWER: 3.0 seconds; 27 meters
TYPE OF MOTION
1. LINEAR MOTION
B. FREE FALL
•
•
•
It is the motion of an object under the influence of
gravitational pull only.
Acceleration is constant; also known as
gravitational acceleration or acceleration due to
gravity
On the earth’s surface at sea level, the
magnitude is 9.8 m/s²
TYPE OF MOTION
1. LINEAR MOTION
B. FREE FALL: 3 cases
1) Object dropped – the initial velocity of the object
dropped with respect to the vertical axis is zero
Vi = 0 m/s²
TYPE OF MOTION
1. LINEAR MOTION
B. FREE FALL: 3 cases
2) Object thrown vertically downward
– In this case, the initial velocity is expressed as a
negative value (downward, -Vi)
– Constant acceleration (g = -9.8 m/s²)
TYPE OF MOTION
1. LINEAR MOTION
B. FREE FALL: 3 cases
3) Object thrown vertically upward
– The initial velocity of the object is the result of the initial
force exerted by the launcher.
– As it goes up ~ gravity decreases its speed until speed
becomes zero (0)  have reached the maximum height
and will start to fall.
SAMPLE PROBLEM
• A STONE IS DROPPED FROM THE TOP
OF NEW YORK’S EMPIRE STATE
BUILDING, WHICH IS 450 m HIGH.
NEGLECTING AIR RESISTANCE,
– How long does it take the stone to reach the
ground?
– What is its speed as it strikes the ground?
ANSWER: 9.6 s; 94 m/s
TYPE OF MOTION
2. CURVILINEAR MOTION
A. PROJECTILE MOTION
•
•
Projectile is when an object moves through
space under the influence of earth’s gravitational
force.
Follows a parabolic path ~ “trajectory”
– Horizontal component (X): horizontal velocity remains
constant if friction is negligible
– Vertical component (Y): vertical velocity changes with
time.
TYPE OF MOTION
2. CURVILINEAR MOTION
B. CIRCULAR MOTION
•
•
•
The motion of a body along circular path
All objects that turn around on axis located within
the object is said to be “rotating or spinning”
While an object turns about external axis the
rotational motion is called “revolution”
TYPE OF MOTION
• CURVILINEAR MOTION
– CENTRIPETAL FORCE
• Means “center-seeking” or towards the center
• Depends on the mass (m), the tangential speed (v)
and the radius of curvature ® of the circular
moving object.
NEWTON’S LAW OF MOTION
• FORCE is any push or pull
• FRICTION is a kind of force that acts
between surfaces of materials that are
moving past each other.
– Occurs because of the irregularities in the
surface of sliding objects.
NEWTON’S LAW OF MOTION
1. LAW OF INERTIA
– Every object continues in a state of rest, or
of motion in a straight line at constant
speed, unless it is compelled to change that
state by force exerted upon it.
•
MASS a measure of inertia – the more mass, the
greater its inertia, and the more force it takes to
change its state of motion
NEWTON’S LAW OF MOTION
1. LAW OF INERTIA
– MASS is not volume
•
Volume – is a measure of space
– Cubic centimeter, cubic meter and liters
•
Mass – measure in kilograms
– Ex. Pillow
– Mass is not weight
•
Measure of the amount of material in an object
and depends only on the number of and kind of
atoms that compose it.
– Weight – depends on the object’s location
NEWTON’S LAW OF MOTION
• LAW OF ACCELERATION
– The acceleration produced by a net force on
an object is directly proportional to the
magnitude of the net force, is in the same
direction as the net force, and is inversely
proportional to the mass of the object.
a = F/m
Unit of force is Newton (N), 1N = (kg) (m/s²)
NEWTON’S LAW OF MOTION
3. LAW OF ACTION AND REACTION
– Whenever one object exerts a force on a
second object, the second object exerts an
equal and opposite force on the first object.
•
One force is “action force” while the other force is
called “reaction force”
– Equal in strength and opposite in direction
NEWTON’S LAW OF UNIVERSAL
GRAVITATION
• Things such as leaves, rain and others fall
because of gravity.
• Newton discovered that gravity is universal
– It pulls everything
• “Every object in the universe attracts every other
object with a force directly proportional to each of
their masses and inversely proportional to the
square of the distance separating them”