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CHAPTER 1
PHYSICAL QUANTITIES AND VECTORS
1.1 THE NATURE OF PHYSICS
The word ‘physics’ comes from the Greek word which means nature. Physics was conceived as
a study of the natural phenomena around us.
Each theory in physics involves:
a. A few concept or physical quantities
b. Assumptions in order to obtain a mathematical model
c. Procedures to relate mathematical models to actual measurement from experiments
d. Relationships between various physical concepts
e. Experimental proofs to devise explanations to natural phenomena
1.2
BASIC QUANTITIES AND SI UNITS
Physics is based on quantities known as physical quantities. Example: length, mass and time. A
physical quantity is clearly defined with a numerical value and a unit. In this text, we emphasize
the system of units known as SI units, which stands for the French phrase”Le Systeme
International d’Unites”.
1.2.1 Base quantities and SI units
In the International System of Units (SI), six physical quantities are selected as base quantities.
Units for these base quantities are known as base units.
Basic quantity
Base unit
Symbol
Length
Meter
m
Mass
Kilogram
kg
Time
Second
s
Electric current
Ampere
A
Thermodynamic temperature
Kelvin
K
Quantity of matter
Mole
mol
Luminous intensity
Candela
cd
Table 1: Base quantities and their SI base units
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1.2.2 Derived Quantities and Derived Units
Physical quantities other than the base quantities are known as derived quantities.
Derived quantity
Force
Pressure
Energy
Power
Charge
Voltage
Resistance
Capacitance
Inductance
Frequency
Unit and Symbol
Newton, N
Pascal, Pa
Joule, J
Watt, W
Coulomb, C
Volt, V
Ohm, 
Farad, F
Henry, H
Hertz, Hz
In Terms of Base Units
N = kgms-2
Pa = Nm-2 = kgm-1s-2
J = Nm = kgm2s-2
W = Js-1 = kgm2s-2
C = As
V = JC-1 = kgm2s-3A-1
 = VA-1 s = kgm2s-3A-2
F = CV-1 = kg-1m-2s4A2
H = VA-1s = kgm2s-2A-2
Hz = s-1
Table 2: Derived quantities and their units
1.2.3
Other Systems of Units / The Conversion of Units
Sometimes, it is necessary to convert one system of units to another. This is done using the
conversion factors. Some conversion factors between the SI system of units and the C.G.S (cm
gram second) system are shown in table 3. In any conversion, if the units do not combine
algebraically to give the desired result, the conversion has not been carried out properly.
SI to C.G.S
1m = 100 cm
1kg = 1000g
1m2 = 104 cm2
1 m3 = 106 cm3
C.G.S to SI
1 cm = 10-2 m
1 g = 10-3 kg
1 cm2 = 10-4 m2
1cm3 = 10-6 m3
Table 3: Conversion factors
EXAMPLE 1.1
An acre is defined such that 640 acres=1 mi2. How many square meters are in 1 acre?
SOLUTION:
1 mile=1.609 km=1609 m
1
1
1 acre=
mi2=
(1609m) 2=4.05 x 103 m2
640
640
EXAMPLE 1.2:
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A geologist finds that a rock sample has a volume of 2.40 in3. Express this volume in cubic
centimeters and in cubic meters.
Solution:
1 in = 2.54 cm, so V = 2.40 in3 = 2.40(2.54cm) 3
1 cm = 10-2m, so V = 39.3 cm3 = (39.3) (10-2m) 3=3.93 x 10-5 m3
EXAMPLE 1.3
The mass of the parasitic wasp Caraphractus cintus can be as small as 5x10-6 kg. What is this
mass in (a) grams (g), (b) milligrams (mg), and (c) micrograms (μg)
REASONING When converting between units, we write down the units explicitly in the
calculations and treat them like any algebraic quantity. We construct the appropriate conversion
factor (equal to unity) so that the final result has the desired units.
SOLUTION
3
a. Since 1.0 10 grams = 1.0 kilogram, it follows that the appropriate conversion factor is
3
(1.0  10 g)/(1.0 kg)= 1. Therefore,
5x10-6 kg =
1.0 x103 g
x 5x10-6 kg=5x10-3g
1.0kg
3
b.
Since 1.0 10 milligrams = 1.0 gram,
1.0 x103 mg
5x10-3g= 5x10-3g x
=5 mg
1.0 g
c.
Since 1.0 10 micrograms = 1.0 gram,
1.0 x106 g
5x10-3g=5x10-3g x
=5x103g
1.0 g
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EXAMPLE 1.4
An engineering student wants to buy 18 gal of gas, but the gas station has installed new pumps
that are measured in liters. How many liters of gas (rounded off to a whole number) should he
ask for?
SOLUTION 18 gal = (18 gal) 
3.785 L
= 68 L .
1 gal
EXAMPLE 1.5
An automobile speedometer is shown.
(a)
What would be the equivalent scale readings (for each empty box) in kilometers per
hour?
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(b)
What would be the 70-mi/h speed limit in kilometers per hour?
40
50
60
30
70
20
80
10
90
km/h
100
0 mi/h
km/h 0
Speedometer readings
SOLUTION:
(a)
10 mi/h = (10 mi/h) 
1.609 km
= 16 km/h for each 10 mi/h .
1 mi
(b)
70 mi/h = (70 mi/h) 
1.609 km
= 113 km/h .
1 mi
1.3 DIMENSIONS OF PHYSICAL QUANTITIES / DIMENSIONAL ANALYSIS
In Physics, the term dimension is used to refer to the physical nature of a quantity and the type of
unit used to specify it. The dimension of a physical quantity relates the physical quantity to the
base quantities such as:
 Mass (M)
 Length (L)
 Time (T)
 Electric current (I)
 Temperature (θ)
 Quantity of matter (N)
EXAMPLE 1.6:
a)
velocity   displacement   ( L)  LT 1
time
(T )
b) [force] = [mass] x [acceleration]
=M x LT-2 =MLT-2
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USE OF DIMENSIONS
1. To check the homogeneity of physical equations
Homogeneous-the dimension on both sides of an equation must also be the same
EXAMPLE 1.7: Show that this equation s = ut +
Left side:
1 2
at is dimensionally homogeneous.
2
[s] = L
L
.T = L
T
L
[at2]= 2.T2 = L
T
Since all the terms in the equation have the same dimension, the equation is dimensionally
homogeneous.
Right side:
[ut] =
2. To derive a physical equation
Using dimensions, an equation can be derived to relate a physical quantity to the variables that
the quantity is dependent on.
EXAMPLE 1.8:
The period T of a simple pendulum depends on its length l and the acceleration due to gravity g.
T  lxgy 
T=k lxgy
Since the dimensions on both sides of the equation must be the same;
[T] = [k lxgy]  T = Lx(LT-2)y
Equating the indices of T;
-2y = 1 
Equating the indices of L;
x+y=0 
1/2 -1/2
y=-1
2
x- 1 =0 
2
x= 1
2
Hence; T=kl g
So, the value of the constant k can be determined experimentally
EXAMPLE 1.9
The following are dimensions of various physical parameters. Here [L], [T] and [M] denote,
respectively, dimensions of length, time and mass.
Parameter
Distance (x)
Time (t)
Mass (m)
Speed (v)
Acceleration (a)
Dimension
[L]
[T]
[M]
[L]/[T]
[L]/[T]2
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Force (F)
Energy (E)
M[L]/[T]2
M[L]2/[T]2
Which of the following equations are dimensionally correct?
a. F = ma
b. x = (1/2)at3
c. E = (1/2)mv
d. E = max
e. v = (Fx/m)1/2
REASONING AND SOLUTION
a. F = [M][L]/[T]2;
ma = [M][L]/[T]2 = [M][L]/[T]2
so F = ma is dimensionally correct .
b. x = [L];
at3 = ([L]/[T]2)[T]3 = [L][T]
so x = (1/2)at3 is not dimensionally correct .
c. E = [M][L]2/[T]2;
mv = [M][L]/[T]
so E = (1/2)mv is not dimensionally correct .
d. E = [M][L]2/[T]2;
max = [M]([L]/[T]2)[L] = [M][L]2/[T]2
so E = max
is dimensionally correct .
e. v = [L]/[T];
(Fx/m)1/2 = {([M][L]/[T]2)([L]/[M])}1/2 = {[L]2/[T]2}1/2 = [L]/[T]
so v = (Fx/m)1/2 is dimensionally correct .
EXAMPLE 1.10 n
The speed, v of an object is given by the equation v = Ar3  Bt, where t refers to time. What are
the dimensions of A and B?
Solutions:
For the equation v  At 3  Bt , the units of At 3 must be the same as the units of v . So the units of A
must be the same as the units of v t 3 , which would be distance time4 . Also, the units of Bt must
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be the same as the units of v . So the units of B must be the same as the units of v t , which would
be distance time2 .
1.4 MEASUREMENT AND ERRORS
1.4.1 Determination of Uncertainty
In Physics, when we take measurements, there are always uncertainties in the values.
Example: determine the diameter of a pencil:
By using a ruler (mm scale) : 7.5mm
By using a micrometer screw gauge, the reading is 7.55mm.
The difference between the two values is in their uncertainties. The measurement using the
micrometer screw gauge is better because it has a smaller uncertainty. The less uncertainty, the
reading will be more accurate.
How to write the measurement???  (7.55  0.01) mm.
The accuracy of the measurement depends on:
1) The types and quality of the instruments
2) The skill of the person taking the reading
3) The number of trials made in the measurement
Instrument
Meter rule
micrometer screw gauge
Stopwatch
Thermometer
Protractor
Smallest scale division
1 mm or 0.1 cm
0.01mm
0.1s
10C
10
Example
(10.00  0.05)cm
(6.28  0.01)mm
(10.5  0.1)s
(39.0  0.5)0C
(30.0  0.5)0
1.4.2 Types of Errors
The error or uncertainty in a measurement is due to factors such as:
 the way the measurement is made
 the instrument used
 the physical limitations of the observer
There are two common types of errors: 1.
2.
1.
systematic errors
random errors
Systematic errors
These are errors made during an experiment. Sources of the systematic errors are:
(1)
instruments
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(a) zero error of a micrometer screw gauge (e.g. zero setting)
(b) fault in the instrument (e.g.: wrong calibration)
(2)
personal error of the observer/physical limitations of the observer (e.g. reaction
time)
(3)
errors due to the physical conditions of the surroundings
 Systematic errors cannot be reduced or eliminated by taking several readings using the
same method, same instrument and done by same observer
 However, it can be reduced by taking measurement carefully and using different
instruments.
The experimental result is can then be accepted as correct within the error limits.
2.
Random errors
 These are errors which cannot be determined and therefore cannot be controlled during
an experiment or measurement.
 Mistakes made by the observer when taking measurements. The reading obtained may
be larger or smaller than the actual value.
 Examples:
a. error made when reading the scale of an instrument ( eye position)
b. a wrong count of the number of oscillations in vibrating
Sources of the random errors are:
1) instruments
2) personal error of the observer/physical limitations of the observer
3) errors due to the physical conditions of the surroundings
Random errors can be reduced / minimized by taking several readings (multiple trials) and
calculating the mean.
EXAMPLE 1.10
Table below show the diameter of a long wire using a micrometer screw gauge. Six readings
have been obtained from several locations of the wires. Suppose the readings obtained are as
follows:
Reading for diameter 1.25
of wire, d(mm)
1.24
1.23
1.22
1.24
1.25
The average value of the readings is determined by taking the sum and divided by the number of
trials.
1.25  1.24  1.23  1.22  1.24  1.25
 1.24mm
6
The error is then calculated as the sum of the deviations of each reading from the average value
divided by the number of trials.
{1.25  1.24}  {1.24  1.24}  ........  {1.25  1.24}
Error=
 0.01mm
6
The average value of the readings is
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Therefore, the correct reading for this measurement is 1.24  0.01mm
The error here is also called standard deviation.
Calculation of percentage error
When the length of the rod is written as l = (67.55  0.05) cm,
 l=  0.05 is known as the absolute error.
l
0.05
l
is known as the relative error where

 7.4  10 4
l
67.55
l
l
l
0.05
 100% is the percentage error where
 100% 
 100%  0.074%
l
l
67.55
A measurement with a smaller relative error (or percentage error) is more accurate than a
measurement with a higher relative error.
Combination of errors (analysis)
This section will focus on how to combine errors or uncertainties made during measurements.
When an error has been made in a measurement, the consequences of the quantities related to the
measured variable will also be affected. Therefore, we have to follow standard rules for
determining the errors contributed to the final result.
Adding and subtracting errors
When the operations involve adding or subtracting quantities, the errors incurred are obtained by
adding the absolute uncertainties of the individual quantity.
d=x+y–z
 d  x  y  z
EXAMPLE 1.11
x = 2.1  0.1
y = 4.4  0.2
z = 3.1  0.1
d = (2.1 + 4.4 – 3.1) = 3.4
 d =  (0.1 + 0.2 + 0.1) = 0.4
 d = 3.4  0.4
This method, however, overestimates the final error.
Errors in multiplying by a scalar
The error is determined by the error of the measured variable. If a quantity x is multiplied by a
scalar k, then,
g = kx
g  kx
Errors in multiplying and dividing quantities
When the operations involve multiplying or dividing quantities, the errors are obtained by adding
the relative uncertainties of the individual quantity.
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(1)
In multiplication: f = xy
f x y


f
x
y
 x y 

f  f 

x
y


Example: let x = 2.5  0.1
y = 5.0  0.1
f = (2.5 x 5.0) = 12.5
 x y 
   f=  12.5
f  f 

y 
 x
 f = 12.5  0.8
(2)
 0.1 0.1 


 =  0.75
 2.5 5.0 
In division: f= x/y
Example: let x = 25  1.0
y = 5.0  0.1
f = (25  5.0) = 5.0
 1 0.1 
 f=  5  
 =  0.3
 25 5.0 
 f = 5.0  0.3
Errors for other operations
If the quantity to be determined involves a single variable, then the erro is determined by
differentiation method as follows;
If,
y=f(x)
dy
= f’(x)
dx
dy  f ' ( x)dx
y  f ' ( x)x
For example if a function y=f(x) is given;
x
y= 2
x 1
dy x 2  1  x( 2 x)

dx
( x 2  1) 2
1 x2
1 x2
dy=


Δy=
 2 dx
1 x2
2Δ x
1 x2


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Let’s look at some special cases;
(i)Given a variable, x raised to power of n
y  xn
dy
 nx n 1
dx
dy  nx n1dx
 y  nx n1x
This error can be written as fractional error,
y
x
n
y
x
(ii) Trigonometric function;
y  sin x
dy
 cos x
dx
 y  cos xx
(iii)Logarithmic functions;
y  log x
dy 1

dx x
 y 
x
x
(iv)Exponential function;
y  ex
dy
 ex
dx
dy  e x dx
y  e x x
EXAMPLE 1.12
The external diameter of a pipe D= (25±2) mm, and its internal diameter d=(15±1)mm. What is
the percentage error of
(a) (D-d)
(b) (D+d)
Solution:
(a) (D-d)=(25-15) ±(2+1)mm
= (10±3) mm
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(b) (D+d)= (25+15) ±(2+1)mm
= (40±3) mm
3
x100%
40
=7.5%
Percentage error of (D+d) =
1.4.3 Significant figures
 The number of significant figures (sf) in a quantity is the number of reliably known digits it
contains.
 E.g.: The quantity
 15.2m  3 sf
 0.052m  2 sf
 3.0m/s  2 sf
Rules of zero
1. Zero at the beginning of numbers are not significant. They merely located the decimal
point.
Eg: 0.0254m (3 sf)
2. Zero within the number are significant
Eg: 104.6 (4 sf)
3. Zero at the end of a number after decimal point are significant
Eg: 2705.6 (5 sf)
4. In the whole number without decimal point at the end in one or more zeros
Eg. 500 kg (zero in this case can be significant or not significant-it depends on estimated
digit in your measurement).
To remove this ambiguity we have to use scientific (power of ten) notation
Eg. 5.0 x102 (2 sf)
5.00 x 102 (3 sf)
In general:
1.
The final result of multiplication and / or division should have the same number of
significant figures as the quantity with the least of significant figures used in the
calculation.
2.
The final result of an addition and / or subtraction should have the same number of
decimal places as the quantity with the least number of decimal places used in the
calculation.
EXAMPLE 1.13:
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1. 0.586 x 3.4 =
2. 13.59 x 4.86  2.1 =
3. 157 – 5.689 + 2 =
Answer:
1. 2.0 (2 sf) 2. 31 ( 2 sf) 3. 153 ( 0 dp)
EXAMPLE 1.14
Which of the following has the greatest number of significant figures?
(a) 103.07 (ans),
(b) 124.5,
(c) 0.099 16, or
(d) 5.408 10 5 .
EXAMPLE 1.15
Determine the number of significant figures in the following measured numbers:
(a) 1.007 m; (b) 8.03 cm; (c) 16.272 kg; (d) 0.015 s (microseconds).
SOLUTION: (a) 4 .
(b) 3 .
(c) 5 .
(d) 2 .
EXAMPLE 1.16
In doing a problem, a student adds 46.9 m and 5.72 m and then subtracts 38 m from the result.
(a)
How many decimal places will the final answer have, (1) zero, (2) one, or (3) two? Why?
(b)
What is the final answer?
SOLUTION: (a) Zero , since 38 m has zero decimal place.
(b) 46.9 m + 5.72 m – 38 m = 15 m .
1.5 SCALARS AND VECTORS
1.5.1 DEFINITION
a. Scalars-physical quantities which have only magnitude
Examples: mass, time, length, temperature, density and energy
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b. Vectors-described fully by stating its magnitude and direction. Symbol for vectors are
printed bold
Examples: displacement, s, velocity, v and acceleration, a.
1.5.2
VECTOR ADDITION AND SUBTRACTION
ADDITION OF VECTORS
A. COLINEAR-the vectors point along the same direction
A. COLINEAR-the vectors point along the same line.
Fig.1: (a) and (b) shows the resultant displacement vector.
B. PERPENDICULAR
Fig. 3: The addition of two perpendicular displacement vectors A and B gives the resultant
vector R
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Let say:
A=275m, due east and B=125m, due north
The resultant displacement vector R=A+B
Since the vectors have different directions, the addition in this equation cannot be carried out by
writing R=275m + 125m.
Instead, we take advantage of the fact that the triangle in fig. 3 is a right triangle and use the
Pythagorean Theorem.
So, the magnitude of R is: R= (275m) 2 + (125m) 2 = 302 m
The angle  in fig. 3 gives the direction of the resultant vector.
 =tan-1 (
125m
B
) = tan-1 (
) =24.4°
275m
A
C. NOT PERPENDICULAR
(a) Analytical method
(b) graphical method
ADDING VECTORS USING GEOMETRICAL METHODS
1. Triangle method
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A+B
B
B
A
A
a.
b.
c.
d.
2.
Draw vector A using scale
Draw vector B with its tail starting at the tip of A
Draw resultant/vector sum from tail of A to the tip of B
Measure resultant vector for the length and angle (using protractor)
Parallelogram method
A+B
A
B
a. Methods are about the same as triangle but now A and B are drawn from tail to tail
b. The resultant is the diagonal of the parallelogram
SUBTRACTION OF VECTORS
When a vector is multiplied by -1, the magnitude of the vector remains the same, but the
direction of the vector is reversed
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Fig. 5 (a)
(b)
Vector addition according to C=A+B
Vector subtraction according to A=C-B=C+ (-B)
EXAMPLE 1.18
At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three
directions. As a result, three forces act on the ball; F1, F2 and F3 (see the drawing). The
magnitudes of F1 and F2 are F1=50N and F2 =90N. Using a scale drawing and the graphical
technique, determine
a. the magnitude of F3
b. the angle θ such that the resultant force acting on the ball is zero
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REASONING AND SOLUTION The following figure is a scale diagram of the forces drawn
tail-to-head. The scale factor is shown in the figure. The head of F3 touches the tail of F1,
because the resultant of the three forces is zero.
a. From the figure, F3 must have a

magnitude of 78 N if the resultant
force acting on the ball is zero.
60.0°
F1 = 50.0 N
F3
60.0°

b. Measurement with a protractor
indicates that the angle   34 .
Scale Factor:
F2 = 90.0 N
20.0 N
1.5.3
THE COMPONENTS OF A VECTOR
VECTOR COMPONENTS
Definition-In two dimensions, the vector components of a vector A are two perpendicular vectors
Ax and Ay that are parallel to the x and y axes, respectively, and add together vectorially so that
A=Ax+Ay
 Any vector can be completely described by its components
Fig.6 An arbitrary vector A and its vector components Ax and Ay
 The components are drawn parallel to convenient x and y axes and are perpendicular.
They add vectorially to equal the original vector A;
A=Ax+Ay
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Fig. 7 This alternative way of drawing the vector A and its vector components is completely
equivalent to that shown in Fig 6
 Component vectors of A : Ax and Ay
 Magnitude of the components : Ax = A cos  , Ay = A sin 
 From Pythagorean Theorem,
The magnitude; A = Ax 2+ Ay 2 and
Ay
The direction;  = tan-1 (
)
Ax
SCALAR COMPONENTS
 Scalar components are positive and negative numbers (with units) that are defined as
follows.
Vector components
Ax=8 meters, directed along the +x axis
Ay=10 meters, directed along the –y axis
Scalar components Unit vectors

Ax=+8 meters
Ax=(+8 meters)
x

Ay=-10 meters
Ay=(-10 meters)
y
Table 4: An example of vector and scalar components
Fig. 8: The dimensionless unit vectors

x

and
y
have magnitudes equal to 1, and they point in the +x
and +y directions, respectively. Expressed in terms of unit vectors, the vector components of the
vector A are Ax

x

and Ay
y
 Ax and Ay are its scalar components. The vector A is written as


A=Ax + Ay
y
x
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ADDITION OF VECTORS USING COMPONENT METHOD
 Recommended procedure using analytical methods
1. Resolve the vectors to be added into their x and y components. Include directional
signs(plus and minus) in the components
2. Add algebraically, all the x component together and all the y components together
to get the x and y components of the resultant vector.
3. Express the resultant vector using;


a. the component form, e.g.;
A=Ax + Ay
y
x
b. in magnitude-angle form, e.g.;
A=
Ay
) (relative to x-axis)
 = tan-1 (
Ax
Ax 2+
Ay 2 and
DISCUSSION:
THE COMPONENT METHOD OF VECTOR ADDITION
A jogger runs 145 m in a direction 20° east of north (displacement vector A) and then 105 m in a
direction 35° south of east (displacement vector B). Determine the magnitude and direction of
the resultant vector C for these two displacements.
Reasoning
Fig 1.22 (a)The vectors A and B add together to give the resultant vector C. The vector
components of A and B are also shown.
(b)The resultant vector C can be obtained once its components have been found.
Fig. 1.22a shows the vectors A and B, assuming that the y axis corresponds to the direction due
north. Since the vectors are not given in component form, we will begin by using the magnitudes
and directions to find the components. Then, the components of A and B can be used to find the
components of the resultant C. Finally, with the aid of the Pythagorean Theorem and
trigonometry, the components of C can be used to find its magnitude and direction.
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Solution
The first two rows of the following table give the x and y components of the vectors A and B.
Note that the component By is negative, because By points downward, in the negative y direction
in the drawing.
Vector
A
B
C
x component
Ax=(145)sin 20°=49.6 m
Bx=(105)cos 35°=86.0 m
Cx=Ax +Bx =135.6 m
y component
Ay=(145) cos 20°=136 m
By=(105m) sin 35.0°=-60.2 m
Cy=Ay+By =76 m
The third row in the table gives the x and y components of the resultant vector C: Cx=Ax+Bx
and Cy=Ay+By. Part b of the drawing shows C and its vector components. The magnitude of C
is given by the Pythagorean Theorem as
C= (Cx2+Cy2)1/2= [(135.6 m) 2+ (76 m) 2]1/2=155 m
The angle θ that C makes with the x axis is
Cy
76m
) =tan-1(
) = 29°
 =tan-1(
Cx
135.6m
EXAMPLE 1.19
Your friend has slipped and fallen. To help her up, you pull with a force F, as the drawing shows.
The vertical component of this force is 130 N, and the horizontal component is 150 N. Find
(a) the magnitude of F and
(b) the angle 
REASONING AND SOLUTION
a. From the Pythagorean theorem, we have
F  (150 N) 2  (130 N) 2  2.0  10 2 N
b. The angle  is given by
130 N
=tan
-1 150 N
=41o
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EXAMPLE 1.20
Figure below shows three forces F1, F2 and F3 acting on a point O. Calculate the resultant force.
F2=7N
120°
100°
F1=6N
O
F3=4N
Solution:
Sum of components along Ox,
Rx = 6 cos 0° + 7 cos 120° + 4 cos 220°
= (6-3.500-3.064) N
= -0.564 N
Sum of components along Oy
Ry = 6 sin 0° + 7 sin 120° + 4 sin 220°
= (0 + 6.062 – 2.571) N
=3.536 N
Magnitude of resultant force
R = Rx 2+ Ry 2
= (0.564) 2+
= 3.536 N
3.491 2
Direction of resultant is given by
Ry
Tan θ =
Rx
3.491
=0.564
= -6.190
θ = 99°10’ to Ox
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