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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 Unit and Symbol In Terms of Base Units

Force Newton, N N = kgms-2

Pressure Pascal, Pa Pa = Nm-2

= kgm-1

s-2

Energy Joule, J J = Nm = kgm2s

-2

Power Watt, W W = Js-1

= kgm2s

-2

Charge Coulomb, C C = As

Voltage Volt, V V = JC-1

= kgm2s

-3A

-1

Resistance Ohm, = VA-1

s = kgm2s

-3A

-2

Capacitance Farad, F F = CV-1

= kg-1

m-2

s4A

2

Inductance Henry, H H = VA-1

s = kgm2s

-2A

-2

Frequency Hertz, Hz 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 C.G.S to SI

1m = 100 cm 1 cm = 10-2

m

1kg = 1000g 1 g = 10-3

kg

1m2

= 104

cm2 1 cm

2 = 10

-4 m

2

1 m3 = 10

6 cm

3 1cm

3 = 10

-6 m

3

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 acre=640

1mi

2=

640

1(1609m)

2=4.05 x 10

3 m

2

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 = 10-2

m, so V = 39.3 cm3 = (39.3) (10

-2m)

3=3.93 x 10

-5 m

3

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

a. Since 1.0 103 grams = 1.0 kilogram, it follows that the appropriate conversion factor is

(1.0 103

g)/(1.0 kg)= 1. Therefore,

5x10-6

kg = kg

gx

0.1

100.1 3

x 5x10-6

kg=5x10-3

g

b. Since 1.0 103 milligrams = 1.0 gram,

5x10-3

g= 5x10-3

g x g

mgx

0.1

100.1 3

=5 mg

c. Since 1.0 106

micrograms = 1.0 gram,

5x10-3

g=5x10-3

g x g

gx

0.1

100.1 6=5x10

3g

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

1 gal = 68 L .

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?

km/h

5060

70

80

90

1000 mi/h

10

20

30

40

km/h 0

Speedometer readings

SOLUTION:

(a) 10 mi/h = (10 mi/h) 1.609 km

1 mi = 16 km/h for each 10 mi/h .

(b) 70 mi/h = (70 mi/h) 1.609 km

1 mi = 113 km/h .

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)

1

)(

)( LTT

L

time

ntdisplacemevelocity

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 + 2

1at

2 is dimensionally homogeneous.

Left side: [s] = L

Right side: [ut] = T

L.T = L

[at2]=

T

L 2.T

2 = L

Since all the terms in the equation have the same dimension, the equation is dimensionally

homogeneous.

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 lxg

y T=k l

xg

y

Since the dimensions on both sides of the equation must be the same;

[T] = [k lxg

y] T = L

x(LT

-2)y

Equating the indices of T; -2y = 1 y = -2

1

Equating the indices of L; x+y=0 x-2

1 =0 x=2

1

Hence; T=kl1/2

g-1/2

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 Dimension

Distance (x) [L]

Time (t) [T]

Mass (m) [M]

Speed (v) [L]/[T]

Acceleration (a) [L]/[T]2

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Force (F) M[L]/[T]2

Energy (E) 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 3

v At Bt , the units of 3

At must be the same as the units of v . So the units of A

must be the same as the units of 3

v t , which would be 4

distance time . 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 2

distance time .

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 Smallest scale division Example

Meter rule 1 mm or 0.1 cm (10.00 0.05)cm

micrometer screw gauge 0.01mm (6.28 0.01)mm

Stopwatch 0.1s (10.5 0.1)s

Thermometer 10C (39.0 0.5)

0C

Protractor 10 (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. systematic errors

2. random errors

1. 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

of wire, d(mm)

1.25 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.

The average value of the readings is mm24.16

25.124.122.123.124.125.1

The error is then calculated as the sum of the deviations of each reading from the average value

divided by the number of trials.

Error= mm01.06

}24.125.1{........}24.124.1{}24.125.1{

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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

lis known as the relative error where 4104.7

55.67

05.0

l

l

%100

l

l is the percentage error where %074.0%100

55.67

05.0%100

l

l

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

zyxd

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

xkg

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

y

y

x

x

f

f

y

y

x

xff

Example: let x = 2.5 0.1

y = 5.0 0.1

f = (2.5 x 5.0) = 12.5

y

y

x

xff f= 12.5

0.5

1.0

5.2

1.0= 0.75

f = 12.5 0.8

(2) In division: f= x/y

Example: let x = 25 1.0

y = 5.0 0.1

f = (25 5.0) = 5.0

f= 5

0.5

1.0

25

1= 0.3

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)

dx

dy= f’(x)

dxxfdy )('

xxfy )('

For example if a function y=f(x) is given;

y=12 x

x

22

2

)1(

)(12

x

xxx

dx

dy

dy= 2

2

1

1

x

x

2 dx

Δy= 2

2

1

1

x

x

2Δ x

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Let’s look at some special cases;

(i)Given a variable, x raised to power of n

nxy

1 nnxdx

dy

dxnxdy n 1

xnxy n 1

This error can be written as fractional error,

x

xn

y

y

(ii) Trigonometric function;

xdx

dy

xy

cos

sin

xxy cos

(iii)Logarithmic functions;

xdx

dy

xy

1

log

x

xy

(iv)Exponential function;

dxedy

edx

dy

ey

x

x

x

xey 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

Percentage error of (D+d) =40

3x100%

=7.5%

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) .

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) (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

5.408105

0.015s

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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= )275( m2 + )125( m

2 = 302 m

The angle in fig. 3 gives the direction of the resultant vector.

=tan-1

(A

B) = tan-1 (

m

m

275

125) =24.4°

C. NOT PERPENDICULAR

(a) Analytical method (b) graphical method

ADDING VECTORS USING GEOMETRICAL METHODS

1. Triangle method

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a. Draw vector A using scale

b. Draw vector B with its tail starting at the tip of A

c. Draw resultant/vector sum from tail of A to the tip of B

d. Measure resultant vector for the length and angle (using protractor)

2. Parallelogram method

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

A B

A+B

A

A+B

A

B

B

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Fig. 5 (a) Vector addition according to C=A+B

(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 F

1,

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.

b. Measurement with a protractor

indicates that the angle 34 .

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

60.0°

F2 = 90.0 N

F3

20.0 N

Scale Factor:

60.0°

F1 = 50.0 N

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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 = Ax2+ Ay

2 and

The direction; = tan-1

(Ax

Ay)

SCALAR COMPONENTS

Scalar components are positive and negative numbers (with units) that are defined as

follows.

Vector components Scalar components Unit vectors

Ax=8 meters, directed along the +x axis Ax=+8 meters Ax=(+8 meters)

x

Ay=10 meters, directed along the –y axis 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=Axx

+ Ay

y

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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=Axx

+ Ay

y

b. in magnitude-angle form, e.g.; A = Ax2+ Ay

2 and

= tan-1

(Ax

Ay) (relative to x-axis)

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 x component y component

A Ax=(145)sin 20°=49.6 m Ay=(145) cos 20°=136 m

B Bx=(105)cos 35°=86.0 m By=(105m) sin 35.0°=-60.2 m

C Cx=Ax +Bx =135.6 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+Cy

2)1/2

= [(135.6 m) 2

+ (76 m) 2]

1/2=155 m

The angle θ that C makes with the x axis is

=tan-1(Cx

Cy) =tan-1(

m

m

6.135

76) = 29°

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) (130 N) 2.0 N2 2 102

b. The angle is given by

=tan-1 N

N

150

130

=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.

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 = Rx2+ Ry

2

= )564.0(2+ 491.3

2

= 3.536 N

Direction of resultant is given by

Tan θ =Rx

Ry

=-564.0

491.3

= -6.190

θ = 99°10’ to Ox

F2=7N

F3=4N

F1=6N 100°

120°

O