II. VECRTORS
MODULE 2 SCALARS AND VECTORS
This module differentiates scalar quantities from vector quantities. It also discusses how
vector quantities are represented graphically.
In this module, the students must be able to:
1. Differentiate scalar from vector quantities;
2. Represent vector quantities and define a resultant vector;
3. Define resultant vector and determine the resultant vector using graphical and analytical
method; and
4. Find the components of a resultant vector.
2.1. SCALARS AND VECTORS
Physical quantities are either scalars or vectors.
A scalar quantity is a quantity that is completely described by a magnitude.
Some examples of scalar quantities are the following
o 45 kg, which describes a mass
o 39 min. which tells time
o 44 km, which shows distance
o 37oC, which gives the temperature
One characteristic of scalar quantities is that they add up or subtract like ordinary
numbers. For example the quantity length, l1 = 6 m and another length l2 = 2. The sum of the two
lengths is given as l = l1 + l2= 6 m + 2 m = 8 m. The scalar quantity, length, has a magnitude of 8 m
and has a unit metes represented by m.
A vector quantity is a quantity that is completely described by both magnitude and
direction.
Some examples of vector quantities are the following:
o 20 km/hr E, which tells the velocity of a vehicle(the magnitude is 20 km/hr and the
direction is East)
o 50 N upward, which describes a force of 50 N (magnitude) directed upward
o 1 m / s2 to the right, which expresses the acceleration of a moving object with a
magnitude of 1 m / s2 directed to the right
Vector quantities are important in the study of physics. If scalar quantities follow
ordinary arithmetic rules, vector quantities do not. This is one important characteristic of vectors.
Check- up test
1. Differentiate a scalar quantity from a vector quantity.
2. Which one is a vector quantity? Scalar quantity?
a. 100 m2
f. 120 kph East of Samar
b. 2040 dynes
g. 1000 kg/m3
c. 250 N on the + axis
h. 15 angstroms
d. 209 kgm/s West
i. 55 AMPERES
e. 67 cycle /s
j. 28 mils
f.
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
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2.2. VECTOR REPRESENTATION
Vectors are mathematical representation of physical quantities that involve a magnitude
and a sense of direction. It is represented by a straight line (magnitude) with an arrowhead at one
end to show the direction of the quantity. The length of the line is proportional to the
magnitude of the quantity and the tail represents the origin of the vector. A vector diagram is
a scale drawing of the various forces, velocities, or other vector quantities involved in the motion of
a body.
A displacement, d, of 15 km to the east
Scale: 1 cm : 3 km
15 km = 5 cm
cm
1
2
3
4
5
Directions of vectors may also be specified using the Cartesian coordinate or the direction
guide.
I - quadrant
x +
y x y III - quadrant
II – quadrant
x+
y+
0
N
W
o
x +
y +
IV – quadrant
E
S
Example:
1. Represent the following vector quantities
a. 25 units, along the negative x – axis
c. 18 km/hr East
25 units
18 km/hr
o
o
b. 20 N, 300 with the + x – axis
o
d. 5 m/s, 600 E of S or S 600 E
20 N
300
o
2. Name the following vectors in two ways
A = 458 kph
C = 980 kph
40o
N
43o
W
5 m/s
600
0
65o
70o
B = 492 kph S
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
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E
D = 309 kph
Answer
A = 458 kph 40o W of N = 458 kph 50o N of W
B = 492 kph 65o W of S = 492 kph 25o S of W
C = 980 kph 43o N of E = 980 kph 47o E of N
D = 309 kph 70o S of E = 309 kph 20o E of N
Check- up Test
3. Draw the following vectors using ruler and protractor:
a. 20 m/s2 SE
b. 15 N 250 SW
c. 120 km E 350 S
2.3. RESULTANT VECTOR
A process of combining or adding two or more vectors to give a single vector is called
resultant vector. It is drawn from the origin to the tip of the last vector. This resultant vector may
be determined by using 1) graphical method and 2) analytical method.
2.4. GRAPHICAL METHOD: POLYGON METHOD
This method makes use of a ruler and a protractor and requires scaling. The ruler will
measure the magnitude and the protractor will measure the direction. The scaling process will help
big magnitudes to be reduced to smaller units.
The tail-head method of drawing the vector is a way wherein the second vector is drawn
such that its tail is connected to the arrowhead of the first vector. The resultant is measured from
the tail of the first vector to the head of the last vector drawn.
Example 1
Kikay walks 400 m East, stops to rest and then continues 300 m East. Find the resultant.
Scale: 1 cm : 100 m
S1 = 4 cm = 400 m
o
S2 = 3 cm = 300 m
SR = 7 cm = 700 m
Example 2
Kate runs 600 m East, then turns 400 m North and finally walks 300 m West.
Scale: 1 cm : 100 m
Sol’n: S1 = 600 m E = 6 cm E
S2 = 400 m N = 4 cm N
S3 = 300 m W = 3 cm W
N
S3 = 3 cm
SR = 5 cm 540 NE
= 5 cm(100m / 1 cm) 54o NE
= 500 m 54o NE
540
W
o
S
S1 = 6 cm
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
Page 18
S2 = 4 cm
E
The resultant displacement is the vector drawn from the origin to the tip of the last vector S 3.
Notice that a polygon has been formed when the resultant, SR, resultant displacement, has been
drawn where SR = 500 m 540 NE
Check – up Test
Draw vector diagrams to solve each problem.
3. A car moves 20 km West, and turns 30 km North. What is the total displacement of the car?
4. A hiker moves with a velocity of 10 m/s, East, then 15 m/s South and finally 20 m/s West. What
is his resultant velocity?
2.5. VECTOR RESOLUTION
The resultant vector has the same effect as the original vectors. Given a single vector you can
get two new vectors in directions that are perpendicular to each other. These two new vectors
are called components of the single vector. The process of finding the magnitudes of the
components in certain directions is called vector resolution. Trigonometry can be used to find
the magnitudes of these perpendicular vector components.
The following are the basic trigonometric functions
opposite
c = hypotenuse
sin  
hypotenuse
adjacent
b = opposite
cos  

hypotenuse
opposite
a = adjacent
tan  
adjacent
The Pythagorean Theorem
c2 = a2 + b2
c  a 2  b2
Example
A boy pulls a toy car by means of a rope that exerts a force of 60 N at an angle of 30 0
along the level road. What are the horizontal and vertical components of the force exerted
by the boy?
Given: F = 60 N, 300 (resultant force)
Find:
a. Fx (horizontal component of the force)
b. Fy (vertical component of the force)
Solution:
y
F = 60 N
Fy =?
30
x
Fx = ?
a. Cos 300 = Fx / F;
Fx = F cos 300 = 60 N cos 300 = 52.2 N
b. Sin 300 = Fy / F;
Fy = F sin 300 = 60 N sin 300 = 30 N
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
Page 19
Check – up Test
5. A box is pulled across a cemented floor by a rope that makes an angle of 35 0 with the floor.
What is the component of the force parallel to the floor if a 100 n force is exerted on it?
6. The wind blow toward 500 with a velocity of 50 kph. What is the vertical
component of the wind’s velocity?
2.6. ANALYTICAL METHOD OF ADDING VECTORS
2.6.1. Adding Two Vectors in the Same Direction or Opposite Directions
Add the magnitude of two or more vectors acting in the same direction and copy the
direction.
The resultant of 6 m/s South and 60 m/s south is 66 m/s South.
Subtract the magnitude of two or more vectors having opposite directions and copy the
direction of the greater value.
The resultant of 39 N East and 25 N south is 14 N East.
2.6.2. Adding Vectors At Right Angle to Each Other
When two vectors are acting at right angle with each other, Pythagorean Theorem is
applied in determining the magnitude of the resultant and tangent function for its
direction.
Example: A person walks 10. 0 km north and then 4.0 km west, find his displacement.
Given:
S1 = 10.0 km north
S2 = 4.0 km west
Req’d: SR resultant displacement
Solution:
Finding the magnitude
SR2 = S12 + S22
= (10.0 km)2 + (4.0km)2
SR = √ ((10.0 km)2 + (4.0km)2
= 10.8 km
S2
SR
θ
S1
o
Finding the direction
The angle θ between SR and S1 is specified by
Tan θ = SR / S1 = 4.0 km / 10.0 km = 0.40
θ = 220
Answer: SR = 10.8 km at θ = 220 west of north.
Check – up Test.
7. A prodigal ants walks 2 cm East and then 5 cm South. What is the displacement of the ant?
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
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2.6.3. Adding Vectors Other Than a Right Angle
If two vectors are acting other then 900 with each other, use Cosine law in
determining the magnitude of the resultant and sine law for its direction.
Example: A ship, leaving its port, sails due north for 45 km and then 60 km in a direction 600
east of north. Determine the location of the ship relative to its port.
Given:
S1 = 45 km N
S2 = 60 km 600 NE
Req’d: SR resultant displacement
Solution:
Using Cosine Law:
SR2 = S12 + S22 - 2 S1S2cos θ
S1 α
o
S2
1200
SR
SR = √ ((45 km)2 + (60km)2 - 2(45km)(60km)cos1200
= 91.24 km
sin 120 o sin 
Using Sine Law:

SR
S
0
sin α = S2 sin 120 / SR
α = sin -1(60 km sin 1200 / 91.24km)
= 34.720
Answer: SR = 91.24 km at α = 34.720 NE
2.6.4. Adding Two or More Vectors Using the Component Method
Use component method when two or more vectors are acting at different angles.
( X - components and Y- components)
Example An ant crawls on a table top. It moves 20 cm/s east, turns 30 cm/s 400 north of
east and finally moves 25 cm/s north. What is the ant’s velocity? What is its displacement
if it travels in 3 seconds?
Given: v1 = 20 cm/s east
v2 = 30 cm/s 400 NE
v3 = 25 cm/s N
Req’d: vR
Req’d: vR
S = displacement
S = total distance traveled
v2x = 30cos400
v3
O
v2
v2y = 30sin400
400
v1
Solution:
Vector
v1 = 20 cm/s E
v2 = 30 cm/s 400 NE
v3 = 25 cm/s N
X-component
20 cm/s cos 00 =
30 cm/s cos 400 =
25 cm/s cos 900 =
Σvx =
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
Page 21
20
22.98
0
42.98
Y-component
20 cm/s sin 00 = 0
30cm/s sin 400 = 19.28
25 cm/s sin 900 = 25
Σvy = 44.28
For determining the magnitude use Pythagorean Theorem
vR2 = (Σvx)2 + (Σvy)2
vR
2
2
= (42.98) + (44.28)
Σvy
θ
vR = 61.71 cm/s
0
Σvx
For direction use tangent function
θ = tan -1(Σvy / Σvx)
= tan -1(44.28 / 42.98)
= 45.850
Solution
vR = 61.71 cm/s at θ = 45.850 NE.
after 3 seconds
S = 61.71 cm/s ( 3 s) at θ = 45.850 NE.
S
= 185.13 cm at θ = 45.850 NE.
Check – up Test
8. Jib and Jam pull at two ropes attached to a cabinet with forces 80 N and 100 N . What is the
resultant force if they pull at 600 with each other?
9. The driver starts his route by driving 30 km due North, then turns onto a rough road and
continues in a direction 300 NE for 45 km and finally turns onto the highway due East for 50
km. What is his total displacement from the starting point?
10. Determine the resultant and equilibrant of the following forces acting on an object: 40 N north;
50 N 300 east of north; 20 N 600 south of east; 45 N west and 50 N south.
SUPPLEMENTARY PROBLEMS
1. A driver becomes lost and travels 12.0 km west, 5.0 km south, and 8.0 km east. Find the
magnitude and direction of the car’s displacement from the starting place.
2. A woman jogs 4.2 km east and then 2.9 km south at a speed of 8.0 km/h. How much time
would she have saved if she had jogged directly to her destination? In what direction
should she headed for this?
3. Two cars leave a crossroads at the same time, one headed north at 50 mi/h and the other
headed east at 70 mi/h. How far apart are they after 0.5 h? After 2.0 h?
4. The resultant of two perpendicular forces has a magnitude of 40 N. If the magnitude of one
of the forces is 25 N, what is the magnitude of the other force?
5. The horizontal and vertical forces combine to give a resultant force of 10 lb that acts in a
direction 400 above the horizontal. Find the magnitudes of the horizontal and vertical
forces.
6. A woman pushes a 50-N lawn mower with a force of 25 N. If the handle of the lawn mower
is 450 above the horizontal, how much downward force is the lawn mower exerting on the
ground?
7. A horse is towing a barge along a canal with a rope that makes an angle of 250 with the
canal. The horse exerts a force of 500 lb on the rope. How much force is exerted on the
barge in the direction of its motion?
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
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8. The shadow of an airplane taking off moves along the runway at 170 km/h. The sun is
directly overhead and the airplane’s air speed is 200 km/h. Find the angle of which the
airplane is climbing.
9. An airplane is heading southeast when it takes off at an angle of 25 0 above the horizontal
at 200 km/h. (a) What is the vertical component of its velocity? (b) What is the horizontal
component of its velocity? (c) What is the component of the velocity of the plane toward
the south?
10. An airplane whose speed is 150 km/h climbs from a runway at an angle of 20 0 above the
horizontal. What is the altitude 1.00 min after takeoff? How many kilometers does it travel
in a horizontal direction in this period of time?
11. Two billiard balls are rolling on a flat table. One has the velocity components v x = 1.0 m/s,
vy = 3.0 m/s. If both balls started from the same point, what is the angle between their
paths?
12. In going from one city to another, a car whose driver tends to get loss goes 30 mi. north,
50 mi west, and 20 mi in the direction 300 south of east? How far apart are the cities? In
what direction should the car have headed to travel directly from the first city to the
second?
13. The following forces act on an object resting o a level frictionless surface: 10 N to the
north, 20 N to the east, 10 N at an angle of 400 south of east, and 20 N at an angle of 500
west of south. Find the magnitude and direction of the resultant force acting of the object.
14. The vector A has a magnitude of 10 cm and points 37 0 clockwise from the + y-direction.
The vector B has a magnitude of 10 cm and points 37 0 clockwise from the + x-direction.
Find the magnitude and direction of A + B, A – B, and B – A.
15. George walks 4.0 km in one direction and 2.0 km in another direction to end up 5.0 km
from where he started. What is the angle between the two directions?
College Physics Notes. Prepared by : Engr. Marlyn G. Jover
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