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UNIT 1 Sources of Errors in Measurement of Physical Quantities
Key unit competence
The learner should be able to identify and explain sources of errors in
measurements and report.
My goals
By the end of this unit, I will be able to:
state and explain types of errors in measurements.
distinguish random and systematic errors.
distinguish between precision and accuracy.
explain the concept of significant figures.
explain the error propagation in derived physical quantities.
explain rounding off numbers.
state the fundamental and the derivate quantities and determine
their dimensions.
choose appropriate measuring instruments.
report measured physical quantities accurately. reduce random and systematic errors while performing experiments.
state correct significant figures of given measurements considering precision required.Physics for Rwanda Secondary Schools Learner’s Book Senior Two
estimate errors on derived physical quantities.
use dimension analysis to verify equations in physics.
suggest ways to reduce random errors and minimise systematic errors.Key concepts
1. How does the precision of measurements affect the precision of
scientific calculations?
2. How can one minimise the errors of measurement?
Vocabulary
Accuracy, uncertainty, precision, random, systematic error, rounding off,
significant figures.
Reading strategy
As you read this section, mark paragraphs that contain definitions of key
terms. Use the information you have learnt to write a definition of each key term in your own words.1.1 Dimensions of physical quantities
1.1.1 Selecting an instrument to use for measuringActivity 1.1: Selecting a measuring instrument
Suppose we have to measure the following quantities:
⚫ The length and width of classroom.
⚫The thickness of paper.
⚫ The diameter of a wire.
⚫ The length of a football pitch.
⚫ Diameter of a small sphere.
⚫ The mass of a stone
⚫The mass of a featherDiscuss these questions:
1. How would you measure each quantity?
2. What would you use to measure each quantity?
3. Where else would measurements be applied in real life?In science, measurement is the process of obtaining the magnitude of a
quantity, such as length or mass, relative to a unit of measurement, such
as a meter or a kilogram. The term can also be used to refer to the result
obtained after performing the process.
The different instruments used differ in sensitivity and therefore, we must
always choose one which is most suitable for measuring the quantity
depending on the sensitivity required for the measurement and on the
order of size of the required measurement. The sensitivity of the measuring
instrument is the smallest reading which one can make with certainty using
the instrument. And the accuracy of the readings made on the instrument depends on its sensitivity.For example, the tape measure is the most suitable instrument for the
measurement of the length of a football field because the order of the
size of the field is within the accuracy which can be obtained from a
tape measure and the tape measure measures up 50m. To measure the
diameter of a wire, you use a micrometer screw gauge because it gives
the accuracy matching the order of the size of the diameter of wire. To
measure the width of a person, a meter rule would be the most suitable judging from the order of the size of a finger.Note that each of the instruments has its own advantages and disadvantages when used. Another important point to note is that we must read the instruments properly in order to get accurate readings.
Inaccurate measurements come about if an inaccurate instrument is used or if the readings are not properly taken from the instrument.1.1.2 Fundamental and derived physical quantities and their dimension
Physical quantities are divided into two categories, those with dimensions
and those that are dimensionless. Physical quantities with dimensions are classified into Fundamental and derived quantities.Each of the seven base quantities used in the SI is regarded as having
its own dimension, which is symbolically represented by a single roman capital letterThe symbols used for the base quantities, and the symbols used to denote
their dimension, are given as follows.
Table 1.1: Base quantities and dimensions used in the SI
The dimensions of the derived quantities are written as products of powers
of the dimensions of the base quantities using the equations that relate the
derived quantities to the base quantities.
In general the dimension of any quantity Q is written in the form of a
dimensional product, dim Q = LaMbTcTdIeNfJg where the exponents a, b,
c, d, e, and g, which are generally small integers that can be positive,
negative or zero, are called the dimensional exponents.
The dimension of a derived quantity provides the same information about
the relation of that quantity to the base quantities as is provided by the SI
unit of the derived quantity as a product of powers of the SI base units.
There are some derived quantities Q for which the defining equation is such
that all of the dimensional exponents in the expression for the dimension
of Q are zero. This is true, in particular, for any quantity that is defined as
the ratio of two quantities of the same kind. Such quantities are described
as being dimensionless, or alternatively as being of dimension one. The
coherent derived unit for such dimensionless quantities is always the
number one, 1, since it is the ratio of two identical units for two quantities of the same kind.
The unit of a physical quantity and its dimension are related, but not
identical concepts. The units of a physical quantity are defined by convention and related to some standard;Example:
Length may have units of meters, centimetres, hectometres, millimetres
or micrometers; but any length always has a dimension of L, independent
of what units are arbitrarily chosen to measure it. The physical quantity,
speed, may be measured in units of metres per second; but regardless of
the units used, speed is always a length divided by a time, so we say that
the dimensions of speed are length divided by time, or simply lt. Similarly,
the dimensions of area are L2 since area can always be calculated as a length times a length.
Two different units of the same physical quantity have conversion factors that relate them.1.1.3 Dimensional analysis
The fact that an equation must be homogenous enables predictions to be made about the way in which physical quantities are related to each other.Examples of the method are given in the table below:
Table 1.2: Dimensional analysis
1.2 Sources of errors in measurement of physical quantities
Activity 1.2: Investigating sources of errors
Take the case of measuring the length and width of an A4 paper.
Material:
⚫A4 paper
⚫ Ruler
Procedure:
Measure the dimensions (width and length) of the given paper using the ruler and record your results.Discovery:
1. Compare your results with those of other learners.
2. Why are your results different?
3. What would you call those differences?
4. Discuss and explain in pairs your results.A measurement is an observation that has a numerical value and unit.
When you measure an object, you compare it with a standard unit. Every
measurement must be expressed by a number and a unit. The Oxford
Dictionary explains the term measure as: “Estimate the size, amount or
degree of (something) by using an instrument or device marked in standard
units or by comparing it with an object of known size”.
In order for a measurement to be useful, a standard measurement must
be used.
Standard measurement is an exact quantity that people agree on to be
used for comparison or as a reference to measure other quantities. We
have three kinds of standards: International standard, Regional standard
and National standard.
The science of measurement is called metrology. It has three branches to
know: Legal metrology, Industrial metrology and Material testing.1.2.1 Types of errors
Activity 1.3: Investigating types of errors
Materials:
⚫Tape measure
⚫Table
Procedure:
⚫Using the tape-measure, measure the length of your table and record the result.
⚫Repeat the same measurement several times and record the results.
⚫Compare your findings.Questions:
1. Are your results the same?
2. (If not) What may have caused the differences?
3. Where do you think errors come from?
Experimental errors are inevitable. In absolutely every scientific
measurement there is a degree of uncertainty we usually cannot eliminate.
Understanding errors and their implications is the only key to correctly
estimating and minimising them.
The experimental error can be defined as: “the difference between the
observed value and the true value” (Merriam-Webster Dictionary).
The uncertainties in the measurement of a physical quantity (errors) in
experimental science can be separated into two categories: random and
systematic.⚫Random errors
Random errors fluctuate from one measurement to another. They may
be due to: poor instrument sensitivity, random noise, random external
disturbances, and statistical fluctuations (due to data sampling or
counting).
A random error arises in any measurement, usually when the observer
has to estimate the last figure possibly with an instrument that lacks
sensitivity. Random errors are small for a good experimenter and taking
the mean of a number of separate measurements reduces them in all
cases.⚫Systematic errors
Systematic errors usually shift measurements in a systematic way.
They are not necessarily built into instruments. Systematic errors can
be at least minimised by instrument calibration and appropriate use of equipment.A systematic error may be due to an incorrectly calibrated instrument,
for example a ruler or an ammeter. Repeating the measurement does not reduce or eliminate the error and the existence of the error may not be detected until the final result is calculated and checked, say
by a different experimental method. If the systematic error is small a measurement is accurate.If you do the same thing wrong each time you make the measurement, your measurement will differ systematically (that is, in the same direction each time) from the correct result.
There are two main causes of error: human and instrument.⚫Human error can be due to mistakes (misreading 22.5cm as 23.0cm) or random differences (the same person getting slightly different readings of the same measurement on different occasions).
For example:⚪ the experimenter might consistently read an instrument incorrectly, or might let knowledge of the expected value of a result influence the measurements (Bias of the experimenter)
⚪ incorrect measuring technique: For example, one might make an incorrect scale reading because of parallax error (reading a scale at an angle)
⚪ failure to interpret the printed scale correctly.⚫ Instrument errors can be systematic and predictable (a clock running fast or a metal ruler getting longer with a rise in temperature). The judgment of uncertainty in a measurement is called the absolute
uncertainty, or sometimes the raw error. For example:
⚪ errors in the calibration of the measuring instruments.
⚪ zero error (the pointer does not read exactly zero when no measurement is being made).
⚪ the instrument is wrongly adjusted. Although random errors can be handled more or less routinely, there is no prescribed way to find systematic errors. One must simply sit down and think about all of the possible sources of error in a given measurement, and then do small experiments to see if these sources
are active. The goal of a good experiment is to reduce the systematic errors to a value smaller than the random errors. For example a meter stick should have been manufactured such that the millimeter markings are located much more accurately than one millimeter.1.2.2 Accuracy and Precision
The terms accuracy and precision are often misused. Experimental precision means the degree of exactness of the experiment or how well the result has been obtained. Precision does not make reference to the true value; it is just a quality attribute to the repeatability or reproducibility of the measurement. Accuracy refers to correctness and means how close the result is to the true value. Accuracy depends on how well the systematic errors are compensated. Precision depends on how well random errors are
reduced.Accuracy is the degree of veracity (“how close to true”) while precision is
the degree of reproducibility (“how close to exact”).
Accuracy and precision must be taken into account simultaneously. All measurements have a degree of uncertainty: no measurement can be perfect!Precision is to 1/2 of the granularity of the instrument’s measurement capability. Precision is limited to the number of significant digits of measuring capability. The precision of a measurement system, also
called reproducibility or repeatability, is the degree to which repeated measurements under unchanged conditions show the same results (degree of exactness).Accuracy is the degree of closeness between a measured value and a true
value. Accuracy might be determined by making multiple measurements
of the same thing with the same instrument, and then calculating the
average for example, a five kilogramme weight could be measured on a
scale and then the difference between five kilogrammes and the measured
weight could be the accuracy. An accuracy of 100% means that the
measured values are exactly the same as the given values.A measurement system can be accurate but not precise, precise but
not accurate, neither, or both. For example, if an experiment contains
a systematic error, then increasing the sample size generally increases
precision but does not improve accuracy. Eliminating the systematic error
improves accuracy but does not change precision.
Fig. 1.1: Accuracy and precision
A measurement system is called valid if it is both accurate and precise.
Uncertainty depends on both the accuracy and precision of the
measurement instrument. The lower the accuracy and precision of
an instrument, the larger the measurement uncertainty is. Often, the
uncertainty of a measurement is found by repeating the measurement
enough times to get a good estimate of the standard deviation of the values.
Physical measurements are never exact but approximate because of
error associated with the instruments (limitations of the measuring
instrument) or which arise when using them (the conditions under which
the measurement is made, and the different ways the operator uses the
instrument). For example, it is possible to have readings taken with great
precision which are not accurate i.e. using an inaccurate instrument of
high sensitivity (precision). For example, if the instrument being used has
a zero error which has not been taken care of, the measurements read
from it are consistently affected by the zero error. Similarly, it is possible to
have readings which are accurate but not very precise. This occurs if one
uses an accurate instrument of low sensitivity (precision).Example of measurement using a ruler
⚫ Using a ruler with 0.5cm marking, we might measure this eraser to be 5.5cm long.
⚫ I f we used a more precise ruler, with 0.1cm markings, then we find the length to be 5.4cm.
⚫ I f we used a micrometer that measured to the nearest 0.01cm,
we may find that the measured length is 5.41cm.The Limit of Accuracy of a Measuring Instrument are ± 0.5 of
the unit shown on the instrument’s scale. It is important to assess
the uncertainty in measurements. One way to do this is to repeat
measurements and average the results. The maximum deviation
from the average is one way to assess uncertainty (although not
the best way). In the following measurements, measure each case
at least three times and take an average. Then record the number
like the following example: Measured times: 5.6s, 6.0s, 6.2s; Average: 5.9s
Experimental time:
1.2.3 Calculations of errors
When combining measurements in a calculation, the uncertainty in the
final result is larger than the uncertainty in the individual measurements.
This is called propagation of uncertainty and is one of the challenges of
experimental physics. As a calculation becomes more complicated, there
is increased propagation of uncertainty and the uncertainty in the value of
the final result can grow to be quite large.
There are simple rules that can provide a reasonable estimate of the
uncertainty in a calculated result:
1. Absolute and Relative Errors (Uncertainties)
When reading a scale it is standard practice to allow an error of one
half of a scale division (depending on the scale being used and the
operator’s eyesight). But as well as the reading being judged there
is also the zero setting to be judged and this also has an uncertainty
of half of a scale division. So for most instruments the total error for
a measurement is
1 scale division.The case (i), is referred to as the absolute error, the Case (ii), as the
relative error (which is often expressed as a percentage). In some
other cases it is easier to work with absolute rather than relative errors (and vice-versa), so be familiar with both.
In experimental measurements, the uncertainty in a measurement
value is not specified explicitly. In such cases, the uncertainty is
generally estimated to be half units of the last digit specified. For
example, if a length is given as 5.2cm, the uncertainty is estimated to be 0.5mm.
• Operations with errors
When measurements with uncertainties are added or subtracted, add
the absolute uncertainties of either addition or substraction errors in order to obtain the absolute uncertainty of the measurement.
Activity 1.4: Length measurement of a stick
Measure a zero value (starting point) of a meter stick as xo.
Measure the position of the end of the given stick as being x.
Question:
Find the length of that stick.3. Multiplication and Division
Multiplication and Division by a constant
We round the absolute uncertainty to 1 sig fig and match precisions
in our final answer; the relative uncertainty is rounded to the same
sig figs as the answer in the absolute case.
We round the absolute uncertainty to 1 sig fig and match precisions
in our final answer; the relative uncertainty is rounded to the same
sig figs as the answer in the absolute case.1.3 Estimating the uncertainty range of measurement
Repeated measurements allow you to not only obtain a better idea of
the actual value, but also enable you to characterise the uncertainty of
your measurement. Below are a number of quantities that are very useful
in data analysis. The value obtained from a particular measurement is
repeated N times. Often times in lab N is small, usually no more than 5 to
10. In this case we use the formulae below:Table 1. 3: Uncertainty calculation
The average value becomes more and more precise as the number of
measurements N increases. Although the uncertainty of any single
measurement is always, the uncertainty in the mean, it becomes smaller
(by a factor of ) as more measurements are made.Example:
You measure the length of an object five times.
You perform these measurements twice and obtain the two data sets
below.
Table 1. 4: Measurement data
Given the table below, use these measurements recorded in the two data
sets and calculate the mean, the range, the uncertainty measurement, the
uncertainty in the mean and the measured value.
The range, uncertainty and uncertainty in the mean for Data Set 1 are
then:
For Data Set 2 yields the same average but has a much smaller range. Form
groups of 4 to reproduce the average mean, the range, the uncertainty,
uncertainty in measurement and the reported measured length.1.4 Significant figures of measurements
No quantity can be measured exactly. All measurements are approximations.
A digit that was actually measured is called a significant digit. Significant
digits may be shown on measuring devices (rulers, meters, etc.) as tick
marks or displayed digits, although you can’t always be sure. The number of
significant digits is called precision. It tells us how precise a measurement
is— how close to exact. For example if you say that the length of an object
is 0.428 m, you imply an uncertainty of about 0.001m.
If a quantity is written properly, all the digits are significant except place
holding zeroes.
The significant figures (also called significant digits and abbreviated sig
figs, sign. figs or sig digs) of a number are those digits that carry meaning
contributing to its precision. Significant figures in a measurement are the
digits in the measurement which are obtained from the instrument with
certainty together with the first digit which is uncertain (estimate).1.4.1 The rules for identifying significant digits
The rules for identifying significant digits when writing or interpreting
numbers are as follows:
⚫ All non-zero digits are considered significant. For example, 91 has
two significant figures (9 and 1), while 123.45 has five significant
figures (1, 2, 3, 4 and 5).
⚫ Zeros appearing anywhere between two non-zero digits (trapped
zeroes) are significant. Example: 101.12 has five significant
figures: 1, 0, 1, 1 and 2.⚫ Leading zeros (zeroes that precede all non-zero digits) are not
significant. For example, 0.00052 has two significant figures: 5
and 2. Leading zeroes are always placeholders (never significant).
For example, the three zeroes in the quantity 0.002 m are just
placeholders to show where the decimal point goes. They were
not measured. We could write this length as 2 mm and the zeroes
would disappear.
⚫ Trailing zeros (zeros that are at the right end of a number) in a
number containing a decimal point are significant. For example,
12.2300 has six significant figures: 1, 2, 2, 3, 0 and 0. The number
0.000122300 still has only six significant figures (the zeros before
the 1 are not significant). In addition, 120.00 has five significant
figures. This convention clarifies the precision of such numbers;
for example, if a result accurate to four decimal places is given as
12.23 then it might be understood that only two decimal places
of accuracy are available. Stating the result as 12.2300 makes it
clear that it is accurate to four decimal places.⚫ The significance of trailing zeros in a number not containing a
decimal point can be ambiguous. For example, it may not always
be clear if a number like 1300 is accurate to the nearest unit (and
just happens coincidentally to be an exact multiple of a hundred)
or if it is only shown to the nearest hundred due to rounding or
uncertainty. Various conventions exist to address this issue:
⚪ A bar may be placed over the last significant digit; any trailing zeros following this are insignificant. For example, has three significant figures (and hence indicates that the number is accurate to the nearest ten).
⚪ The last significant figure of a number may be underlined; for example, “20000” has two significant figures.
⚪ A decimal point may be placed after the number; for example
“100.” indicates specifically that three significant figures are meant.Generally, the same rules apply to numbers expressed in scientific
notation. For example, 0.00012 (two significant figures) becomes
1.2×10−4, and 0.000122300 (six significant figures) becomes
1.22300×10−4.
In particular, the potential ambiguity about the significance of trailing
zeros is eliminated. For example, 1300 to four significant figures is
written as 1.300×103, while 1300 to two significant figures is written
as 1.3×10 3. Numbers are often rounded off to make them easier to read. It’s easier for someone to compare (say) 18% to 36% than to compare 18.148% to 35.922%.Note:
Zeros at the end of a number but to the left of a decimal, in
this handbook will be treated as not significant for example
1 000 m may contain from one to four significant figures,
depending on precision of the measurement, but in this hand
book it will be assumed that measurements like this have one
significant figure.
⚫ Do not confuse significant figures with decimal places. For
example, consider measurements yielding 2.46 s, 24.6 s
and 0.002 46 s. These have two, one, and five decimal
places, but all have three significant figures.
⚫ If a number is written with no decimal point, assume
infinite accuracy; for example, 12 means 12.0000….1.4.2 Special rules of calculation with significant figures
The final answer should not be more precise than the least precise
measurement in your data. For example, though your calculator gives an
answer to nine digits, do not give this number of digits in your final answer.
Example: Perform these calculations, following the rules for significant
figures
1. Addition or subtraction: the final answer should have the same
number of digits to the right of the decimal as the measurement
with the smallest number of digits to the right of the decimal.
( ≅Means, approximately equal to)
2. Multiplication or division: the final answer has the same number
of figures as the measurement having the smallest number of
significant figures.
123 × 5.35 = 658.05 = 658
11.2 × 6.8 = 77 (6.8 has the least number of significant figures,
namely two)
2035cm × 12.5m = 20.35m × 12.5m =254.375m2 = 2.54 ×
104 cm2 (it is better to make the conversion to the same units before
doing any more arithmetic)1.5 Rounding off numbers
Activity 1.5: Rounding off a number
Using your ruler; measure the width (w) of a note book seven times
and perform the average as:
Questions:
• Round off your result to 2 decimal places
The concept of significant figures is often used in connection with rounding.
For example, the population of a city might only be known to the nearest
thousand and be stated as 52,000, while the population of a country
might only be known to the nearest million and be stated as 52,000,000.
When we compute with measured figures, we often round off numbers so
that they will show the precision or accuracy that is appropriate.
In rounding off, we drop digits or replace digits with zeros to make numerals
easier to use and interpret. Instead of saying 45,125 people attended the football match last Sunday; we would probably round the value to 45,000 people. When we replace digits with zeros by rounding off, the
zeros are not significant. In rounding off a number, the digits dropped must be replaced by ‘place holding’ zeros. The following rules will be found useful when rounding off figures:⚫ If the first of the digits to be dropped (reading from left to right) is 1, 2, 3 or 4, simply replace all dropped digits with the appropriate number of zeros. For example, 57,384 rounded off to the nearest
thousands becomes 57,000.
⚫ If the first of the digits to be dropped (reading from left to right) is 6, 7, 8 or 9, increase the preceding digit by 1. For e.g., 5,383 rounded off to the nearest hundred becomes 5,400.
⚫ If only one digit is to be dropped and this digit is 5, increase the preceding digit by 1 if it is odd, and leave it unchanged if it is even. Thus, if 685 is to be rounded off to the nearest tens it becomes 680, while 635 rounded off to the nearest tens becomes 640.
⚫ If a decimal fraction is rounded off, zeros should not replace the digits that are to the right of the decimal, because zeros to the right of a decimal are significant. For example, 73.2 rounded off to one
significant figure becomes 70 and not 70.0 to the nearest tens.1.6 Unit 1 assessment
1. The learners listed below measured the density of a piece of lead
three times. The density of lead is actually 11.34 g/cm3. Below
are their results;
a) Rachel: 11.32 g/cm3, 11.35 g/cm3, 11.33 g/cm3
b) Daniel: 11.43 g/cm3, 11.44 g/cm3, 11.42 g/cm3
c) Leah: 11.55 g/cm3, 11.34 g/cm3, 11.04 g/cm3
(i) Whose results were accurate?
(ii) Whose were precise?
(iii) Whose measurements were both accurate and precise?
2. Arrange the following measurements in order of precision beginning with the most precise: 17.04cm; 843 cm; 0.006cm; 342.0cm.
3. Round off to;
a) the nearest unit: 6.8; 10.5; 801.625,
b) the nearest tenth 5.83; 480.625; 0.234; 0.285; 6.58;
36.092,
c) the nearest hundredth: 3.632; 812.097; 0.71d) the nearest thousandth: 0.2827; 0.0066.
e) the nearest tens: 56; 44; 17; 656,
f) the nearest hundreds: 219; 256; 71,550; 930.7,
g) the nearest thousands: 890; 1600; 10 500; 13 856; 5420.5
4. Round off the following measurement so that all have the same
degree of accuracy: 468.5m; 0.00708m; 3.467m; 56.93m; 3.004m
5. Perform the following operations, rounding off each answer to
the proper degree of accuracy:
a) 6.574 + 34.57 =
b) 23.12 × 34.9 =
c) 5.2 – 5.7 =
d) 625/15 =
e) Sqrt(5625) =
6. Round off the numbers below to the shown number of significant
figures in the brackets:
a) 245 086 (4);
b) 406.50 (3)
c) 8 465 (3);
d) 84.25 (2);
12. If an equation is dimensionally correct, does this mean that
the equation must be true? If an equation is not dimensionally
correct, does this mean that the equation cannot be true?
13. Is it possible to add a vector quantity to a scalar quantity? Explain.
14. If A = B, what can you conclude about the components of A and B?Key unit competence
The learner should be able to inteprete and solve problems related to linear motion.My goals
By the end of this unit, I will be able to:
describe and define linear motion.
list examples of linear motion.
explain the difference between instantaneous and average values of speed, velocity and acceleration.
describe and explain the acceleration of a free falling body near
the earth’s surface. Describe the motion of a free falling body.
explain effects of air resistance on moving. Derive equations of linear motion.
describe the conditions applicable to equations of uniformly accelerated motion.
distinguish between linear motion from other motions.
solve problems related to linear motion.
apply the scientific techniques in solving problems related to the motion of bodies moving against gravitational acceleration (in sports, airplanes).
develop and improve on the skills in sketching of graphs for bodies in motion.Key concepts
1. What is needed to describe motion completely?
2. When is an object in motion?
3. Why is distance and displacement different?
4. How do you add and subtract displacements?
5. How are instantaneous speed and average speed different?
Vocabulary
Motion, kinematics, trajectory, position, displacement, speed, average
speed, acceleration, translational motion, average acceleration, accelerated motion.
Reading strategy
Study the Newton's laws of motion and relate them to the situations and
common problems in motion of bodies.2.1 Definition and types of linear motion
Activity 2.1: Linear motion
Take the case of a football player in Amavubi National team, kicking
off a ball in a match as shown in (Fig.2.1).
Fig 2.1: Amavubi football player
⚫ Carefully study the statute of player no. 12 (in yellow and in
opponents (in white). The kick he is about to deliver to the ball,
is linear, circular, backwards or angular?
⚫ Explain different movements of the yellow dressed player on the playground if he has to score!
When a body moves in a straight line, then we say that it is executing linear
motion. When it moves without rotating, it is said to have translational
motion. A car moving down a highway is an example of translational motion.
When a body moves in a straight line, then the linear motion is called
rectilinear motion.
Fig. 2.2: People running on a straight road in Rwamagana
District
Example: An athlete running along a straight track is said to be in rectilinear motion.
When a body moves along a curved path then the motion is called
curvilinear motion. E.g., a planet revolving around its parent star.
Fig. 2.3: A boat moving in curvilinear motion in a river
Examples of Linear Motion
Fig. 2.4: Train running on straight tracks in India
Other motion problems examine the effects of forces such as gravity on
an object’s rectilinear motion. One common example involves shooting
a projectile up into the air. Whether one shoots the object straight up,
perpendicular to the ground, or at an angle, gravity immediately begins
to take effect, slowing the projectile down, and in the case of the angle of
projection, turning the rectilinear path into a curvilinear one.2.2 Equation for uniform acceleration and in one dimension
2.2.1 Acceleration
Activity 2.2: Comparing the velocity change of a marble
Materials:
A marble, Stop watch; an inclined rail with marked strips 1m each
Procedure:
⚫ Arrange the incline plane as shown in Figure 2.6.
⚫ Allow a marble to roll from rest down the rail.
⚫ Time the marble as it moves the first 1 m.
⚫ Time the marble as it moves through the first 2 m.
⚫ Time the marble as it moves the 3 m.Questions:
1. What is the average velocity as the marble moves the first 1m?
2. What is the average velocity of the marble as it moves the second 1 m?
3. What is the average velocity of the marble as it moves the third 1 m?
4. Where is the marble moving fastest?
5. Is the velocity increasing or decreasing?
Fig. 2.5: Movement of a marble on an inclined plane
When the velocity of a body is changing, the body is said to be accelerating.
Acceleration is defined as the rate of change of velocity with time.
Fig. 2.6: A car, modeled as a particle, moving along the x axis
⚫When acceleration is constant both in direction and in magnitude,
the object is said to be undergoing uniformly accelerated motion.
⚫ A body whose velocity is increasing or decreasing in magnitude
is said to be accelerating. A decrease in velocity or slowing down
indicates a retardation or deceleration or negative acceleration.
⚫ A change in the direction of velocity also shows a body accelerating,
even though there is no change of the magnitude of velocity. Hence
a body moving with constant speed in a circle has acceleration.
The SI unit of acceleration is the meter per square second abbreviated
as ms–2, like, velocity, acceleration is a vector quantity.When the object’s velocity and acceleration are in the same direction,
the object is speeding up. On the other hand, when the object’s velocity
and acceleration are in opposite directions, the object is slowing down.
Activity 2.3: Calculating the maximum speed
A car starts from rest and is accelerated uniformly at the rate of
2m/s2 for 6s. It then maintains a constant speed for half a minute. The
brakes are then applied and the vehicle uniformly retarded to rest in
5s . Find the maximum speed reached in km/h and the total distance
covered in meters.2.2.2 Velocity with constant acceleration
By rearranging the equation for acceleration, we can find a value for the final velocity
Do not confuse acceleration and velocity. Acceleration tells us how fast the
velocity changes, whereas velocity tells us how fast the position changes.2.2.3 Displacement with constant acceleration
For an object moving with constant acceleration, the average velocity is
equal to the average of the initial velocity and final velocity;
Example 1. A race car reaches a speed of 42m/s. It immediately then
begins a uniform negative acceleration, using its braking system, and
comes to rest 5.5s later. Find how far the car moves while stopping.
Example 2. A plane starting at rest at one end of a runway undergoes a
constant acceleration of 4.8m/s2 for 15s before takeoff. What is its speed
at takeoff? How long must the runway be for the plane to be able to take off?
This relation is useful when time is not known explicitly
If we know any three of u, v, a, S and t the others can be found from these equations.
These formulae only apply to cases of particles moving under constant
acceleration. If this condition does not apply to the situation under
consideration, then you cannot use these formulae.
Sign Convention
Before we start applying these formulae, let us introduce a sign convention.
Since we are working in one dimension, there are only two directions we
need to worry about. For instance, if we consider motion in a horizontal
direction, the only two directions are left and right. Likewise, if we consider
motion in a vertical direction, the only two directions are up and down.
Mathematically, we can denote the two directions with a sign. The convention that must be used for quantities associated with the body with respective motion below:
Horizontal Motion Vertical Motion
Right is (+). Up is (+).
Left is (-). Down is (-).
For example: If a rocket is moving up at the speed of 10,000m/s, we
can just write the rocket’s velocity as 10,000m/s. If the rocket had been
moving downward, then the sign infront of the 10,000m/s would have
been negative, (-).
Example 1. What is the velocity of an object, at rest, if it experiences a
constant acceleration of 10m/s2 to the right after a period of 3s?
The object will move at a velocity of 30m/s to the right after undergoing a
constant acceleration of 10m/s2 to the right for 3s.
Example 2. As a bus comes to stop, it slows from 9.00m/s to 0.00m/s in
360s. Find the average acceleration of the bus.
Example 3. Consider a ball thrown upward with an initial velocity of 20
m/s. What will its velocity be after 3s if it undergoes a constant acceleration
of a = 10m/s2 downward?
a = –10m/s
Fig. 2.7: Ball thrown upward
Activity 2.4: Calculating the maximum height
A stone is thrown vertically upwards with an initial velocity of 14m/s.
Neglecting air resistance, find;
a) The maximum height reached;
b) The time taken before it reaches the ground. (acceleration due to
the gravity is 9.8m/s2)2.3 Acceleration due to gravity and free fall
Activity 2.5: Gravity
Materials:
⚫ A stop clock.
⚫ Five stones of different masses between 0.5kg → 5kg.
⚫ A long wooden pole.
Procedure:
⚫ Measure out a distance of 2m from the floor of your laboratory but against a pole.
⚫ From the smallest stone to the biggest stone, drop the stones
one by one. Using the stop clock, find out how long each stone takes to reach the floor.
⚫ Repeat this three times for each stone and find out the average time for each stone.
⚫ Determine the average speed of each stone after falling for 2m.Questions:
How fast did different objects fall? Grab a tennis ball and a basketball
and drop them from the same height and the same time. What do you
notice? How about a shoe and a ping pong ball?
It is well known that, in the absence of air resistance, all objects dropped
near the Earth’s surface fall toward the Earth with the same constant
acceleration under the influence of the Earth’s gravity (the pull of gravity
which the earth exerts on the falling bodies). It was not until about 1600
that this conclusion was accepted. Before that time, the teachings of the
great philosopher Aristotle (384 – 322 B.C.) had held that heavier objects
fall faster than lighter ones.
Galileo Galilei a famous scientist, who lived in Italy (1564 - 1642), showed
that all bodies, irrespective of their masses and nature, fall towards the
Earth at the same rate. He further explained that the commonly observed
difference in the rate of fall of heavy and light objects was due to the air
resistance these objects experience due to their shapes and hence the
differences vanished if the experiment was performed in vacuum.
Fig. 2.8: Feather and coin experiment
Galileo’s observations were later verified by Sir Isaac Newton when he performed
his famous feather and coin experiments (Fig. 2.9). He observed that the rate of
fall of a feather and a coin in the long glass tubes was the same if the tubes was
evacuated and different when air were allowed inside the tubes.The falling bodies undergo motion with uniform acceleration i.e. as they
fall, their velocity increases by equal steps in equal time intervals. This
acceleration, which the falling bodies have, is called acceleration due to
gravity and is denoted by the letter, ‘g’. The acceleration due to gravity
is the same for all objects, provided where there is very limited or no air resistance.The kind of fall where the bodies fall under the influence of gravity only
regardless of its initial motion is called free fall motion. In free fall, the only
outside force acting on the falling body is the pull of gravity with different masses.The presence of air affects the motion of falling bodies partly through
buoyancy and partly through air resistance. Thus two different objects
falling in air from the same height will not, in general, reach the ground
at exactly the same time. Because air resistance increases with velocity,
eventually a falling body reaches a terminal velocity that depends on its
mass, size, and shape, and it cannot fall any faster than that.Questions
For each stone, use the velocity formula and calculate the value of the
acceleration.
What result have you determined?
What conclusions can you develop in relation to motion under gravity.
Its direction is always downward toward the centre of the Earth. Its
magnitude varies from one place to another. The g is slightly greater at the
poles (g = 9.832 m/s2) than at the equator (g = 9.780 m/s2), since the
Earth is not a perfectly spherical shape. Its magnitude is approximately g = 9.80 m/s2
When a body is dropped then its subsequent motion is downward.
Dropping means that the motion is under a constant acceleration of "g" downwards.
The falling object starts out at a velocity of zero and, with constant
acceleration the value of velocity increases in a simple straight line.
Example 1. Let’s say you are standing next to a cliff and decide to drop a ball. What is the ball’s velocity after 4s?
The negative sign shows this – the motion is downwards.
Example 2. A stone dropped from the top of a building takes 6s to reach
the ground below.
a) What is the height of the building?
b) How far will the stone fall during the fifth second of its falling?
Activity 2.6: Quick check exercises
1. Jason hits volleyball so that it moves with an initial velocity
of 6.0m/s straight upward. If the volley starts from 2.0m
above the floor, how long will it be in the air before it strikes
the floor? Assume that Jason is the last player to touch the
ball before it hits the floor.
2. A stone is thrown upwards with an initial speed of 5m/s.
a) What will its maximum height be?
b) When will it strike the ground?
c) Where will it be in 2s?
3. Suppose that a ball is dropped from a tower 70m high.
How far will it have fallen after 1s, 2s, and 3s? Assume y is
positive downward. Neglect air resistance.
Note: An object thrown downward or upward at a given
location on the Earth and in the absence of air resistance,
all objects fall with the same place acceleration. (Equator), (pole)Example 1. A man fires a stone out of a slingshot directly upwards. The
stone has an initial velocity of 15m/s. How long will it take for the stone to
return to the level he fired it at?
Example 2. A falling body travels 68m in the last second of its free motion:
Assuming that the body started from rest, determine how long it took to
reach the ground and the altitude from which the body fell.
Activity 2.7: Quick check exercises
1. A person throws a ball upward into the air with an initial
velocity of 5.0 m/s. Calculate:
a) How high it goes.
b) How much time it takes for the ball to reach the maximum height.c) How long the ball is in the air before it comes back to his hand.
d) The velocity of the ball when it returns to the thrower’s hand.
e) At what time, the ball passes a point 8.00m above the thrower’s hand.
We are not concerned here with the throwing action, but only
with the motion of the ball after it leaves the thrower’s hand.
2. A stone is dropped from a balloon that is descending at a uniform rate of 0.2m/s when it is 1000m from the ground.
a) Calculate the velocity and position of the stone after 10s
and the time it takes the stone to hit the ground.
b) Solve the same problem as for the case of a balloon rising at a velocity of 0.1m/s.2.4 Determination of G: Use of a simple pendulum
Many things in nature swing in a periodic motion. That is, they vibrate.
One such example is a simple pendulum. If we suspend a mass at the end
of a piece of string, and we allow the mass to swing up and down along
the vertical axis, we have a simple pendulum. Here, the to and fro motion
represents a periodic motion used in times past to control the motion of clocks.
(a) Simple pendulum (b) Motion of a simple pendulum
Fig. 2.9: A Pendulum bob
Physics for Rwanda Secondary Schools Learner’s Book Senior Two
Such oscillatory motion is called simple harmonic motion. It was Galileo
who first observed that the time a pendulum takes to swing back and forth
through small angles depends only on:
The length of the pendulum, the time of this to and fro motion, called
the period, and does not depend on the mass of the pendulum. Another
factor involved in the period of motion is, the acceleration due to gravity
(g), which on the earth is 9.8m/s2 at the equator.
A long pendulum has a greater period than a shorter pendulum.
Fig. 2.10: A Pendulum bob swinging
With the assumption of small angles of projection, the frequency and period
of the pendulum are independent of the initial angular displacement. All
simple pendulums have the same period regardless of their initial angle
(and regardless of their masses of the bobs).
The period, T for a simple pendulum does not depend on the mass or
the initial angular displacement, but depends only on the length, L of the
string and the value of the gravitational field strength, g;
Activity 2.8: Measuring gravity of any given place
⚫ Strings
Referring to the above concept, use the formula given below and try
to answer the questions.
Materials:
⚫ A pendulum bob (mass)
⚫ Masses of 10, 20, 30, 40 and 50 gms
⚫ Stop watch
⚫ A paper and a pencil
⚫ Metre rule⚫ Retort standProcedure:
1. The period, T of a simple pendulum (measured in seconds) is
given by the formula:
Questions:
1. From your data what effect does changing the mass have on the period?
2. Would you conclude that Galileo was correct in his observation
that the period of a simple pendulum depends only on the length of the pendulum?
3. On the moon, the acceleration due to gravity is one-sixth that of earth.
What effect, if any, would this have on the period of a pendulum of length, L?
How would the period of the pendulum on the moon differ from an equivalent one on earth?2.5 Unit 2 assessment
1. a) Is an object accelerating if its speed is constant?
b) Is an object accelerating if its velocity is constant?
2. If you know the position vectors of a particle at two points
along its path and also know the time it took to get from one
point to the other, explain how you will determine the particle’s
instantaneous velocity and its average velocity.
3. The average velocity of a particle moving in one dimension has
a positive value. Is it possible for the instantaneous velocity to
have been negative at any time in the interval? Suppose the particle started at the origin x = 0. If its average velocity is
positive, could the particle ever have been in the - x region of the axis?
4. If the average velocity of an object is zero in some time interval,
what can you say about the displacement of the object for that
interval?
5. Can the magnitude of the instantaneous velocity of an object
ever be greater than the magnitude of its average velocity?
6. If the average velocity is non-zero for some time interval, does
this mean that the instantaneous velocity is never zero during this interval? Explain.
7. If the velocity of a particle is non-zero, can its acceleration be zero? Explain.
8. A stone is thrown vertically upward from the top of a building.
Does the stone’s displacement depend on the location of the
origin of the coordinate system? Does the stone’s velocity depend
on the origin? (Assume that the coordinate system is stationary with respect to the building). Explain.
9. If the velocity of a particle is zero, can its acceleration be nonzero? Explain.
10. Two cars are moving in the same direction in parallel lanes along a highway. At some instant, the velocity of car, A exceeds the velocity of car, B. Does this mean that if the acceleration of the car reduces, A is greater than that of car, B? Explain.11. You are standing on top of a cliff and you decide to throw a stone
upward at a speed of. After 40s, you see the stone hit the base of
the cliff. How far down is the base of the cliff? In addition, what
is the velocity of the stone when it reaches the base of the cliff?
12. A feather and a coin are released from the same height at the
same time. Which one reaches the ground first? Explain why
they do not reach the grand at the same time.
13. How long will it take a car to accelerate from 10m/s to 35m/s at
a constant acceleration of 5m/s2 in a straight line.
14. Imagine a ball that is thrown upward with a velocity of 5m/s.
If the ball experiences a downward constant acceleration of
10m/s2, how long will it take for its velocity to reach 25m/s downward?15. Assume there is a jet plane initially moving to the right at 10
m/s. Furthermore, then it accelerates for 90s and ends up with a
speed of 20m/s but moving to the left. Assuming the acceleration
was constant, what is the constant acceleration the jet plane
undergoes?
16. A car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration?
17. An car with an initial speed of 4.3m/s accelerates at the rate
of 3.0m/s2. Find the final speed and the displacement after 5.0s.
18. A ball is thrown upwards with an initial velocity u. After 3s the
velocity of the ball upwards is determined to be 10m/s. Calculate the value of the initial velocity u. Use g = 10m/s2.
19. A car with an initial speed of 23.7km/h accelerates at a uniform rate of 0.92m/s2 for 3.6s. Find the final speed and the displacement of the car during this time.Key unit competence
The learner should be able to explain the effects of friction and its
importance in life.
My goals
By the end of this unit, I will be able to:
explain the nature of frictional force.
explain effects of frictional force.
discuss advantages and disadvantages of frictional forces.
measure static and dynamic coefficients friction.
describe technological applications of frictional force.
identify factors affecting frictional force.
identify methods of reducing friction.
solve problems on frictional force.Key concepts
1. What is needed to describe frictional force?
2. Which factors cause friction?
3. Discuss different types of friction?
4. How is friction advantageous in real life?
5. What are the disadvantages of friction in real life?
6. How can you overcome friction?
7. How can you increase friction?Vocabulary
Frictional force, resistance force, weight, roughness, smoothness,
coefficient of friction, viscosity, air resistance.
Reading strategy
As you read this section, re-read the paragraphs that contain definitions
of key terms. Use all the information you have learnt to write a definition
of each key term in your own words. Then practice more examples and
activities to help you in performing your assessment.Nature of frictional force
Activity 3.1: Experiencing friction whilepushing a deskCarry out the following activity and discuss the results observed.
Fig. 3.1: Pushing a box
1. Push a heavy object (such as the teacher’s desk) gradually
till it moves at a steady speed.
2. Carefully describe what you felt.
3. Try to define friction force in your own words.
4. Discuss friction and find out if it is good or bad giving typical examples.
5. Plan how you would reduce friction between the desk and the floor.
Fig. 3.2: Frictional force on car tyres
Friction is the force that opposes the motion of an object as it moves
across a surface or as it makes an effort to move across it.
For example, if a book slides across the surface of a desk, then frictional
force will act in the opposite direction to motion. Friction results from
the two surfaces being pressed together closely, causing intermolecular
attractive forces between molecules of different surfaces.3.2 Types of frictional forces
Friction is a surface force that opposes motion. The frictional force is
directly related to the normal force which acts to keep two solid objects
separated at the point of contact. There are two broad classifications of
frictional forces: static friction and kinetic friction.
a) Static friction is a force between two bodies in physical contact that
are NOT moving. The bodies are stationary. Mathematically this
force is given by:
3.3 The laws of solids friction
Experimental results on solid friction are summarised in the laws of friction
which state:
a) The frictional force between two surfaces opposes their relative
motion.
b) The frictional force is independent of the area of contact of given
surfaces when the normal reaction is constant.
c) The limiting frictional force is proportional to the normal reaction
for the case of static friction. The frictional force is proportional to
the normal reaction for the case of kinetic (dynamic) friction, and is
independent of the relative velocity of the surfaces.
The symbol μ represents the coefficient of sliding friction between
the two surfaces and depends on the roughness of the surfaces e.g.
for wood on wood μ = 0.4, for steel on steel μ = 0.2
Experimentally for most surface interfaces, the coefficient of kinetic
friction is less than the coefficient of static friction as shown in the following table.Table 3.1: Coefficients of Friction
3.4 Effects of friction
From the definition of friction force, find at least five different places where
friction can be observed. Thus, answer to the following questions:Questions:
Discuss and explain the effects of friction in each of the cases you mention above.
How would you decrease those effects of friction observed?
⚪ Friction tends to generate heat energy.
⚪ Friction reduces the speed of moving objects and therefore causes loss of energy.
⚪ Friction damages sliding surfaces that are in contact.
⚪ Friction causes wear and tear in moving parts.3.5 Advantages and disadvantages of friction
Friction is very important because it enables us to move. If there was no
friction, our feet would slip, just as they do on smooth surface. Friction
also enables us to write, to make fire, the brakes of cars or bicycles use
friction to slow them down. It is also a nuisance because it wears the
soles of our shoes and the car tyres, causes the unnecessary heat and
undesirable noise, lowers efficiency of machines.
In a machine such as a bicycle, friction hinders the wheels from turning
freely. This is also true in other machines like cars, lorries, buses. This
means that they use more fuel in order to move because they have to overcome friction.3.6 Factors affecting friction and how to reduce it
Friction depends upon the nature of the two surfaces in contact and the
degree to which they are pressed together. Experiments show us that the
force of friction between two surfaces depends:
⚫ The nature of the surfaces. Rough surfaces give more friction than smooth ones. So if we want to make a machine in which very little friction acts, we make the surfaces smooth.
⚫ The force pressing the surfaces together. The bigger this force is, the greater the force of friction.
⚫ The type of shape also, where some shapes meet less resistance
than others. The shapes which meet the least resistance are said
to be streamlined.⚫ The size of frictional force also depends the speed of the moving object.
The frictional force or resistance met by an object through air is always
much less than it experiences when moving through a liquid.
The moving machine parts are always oiled or greased in order to reduce
friction. This helps the moving parts to slip more easily over each other.
The liquid is usually oil, which we refer to as a lubricant (reduce friction
by separating two contacting surfaces with an intermediate layer of softer
material). And we call the effect lubrication. The 2nd method involves
reducing the roughness of the surfaces in contact; we can also use a ball or roller bearings.3.7. Other resistance forces
Tensional forces
Activity 3.2: Investigation of tension force
Material:
⚫ Rope
⚫ Six stones of different masses
⚫ Spring balance (Newton meter).
Procedure:
⚫ Measure the mass and the weight of each stone using a spring balance.
⚫ Tie the rope on the stone and hang it on the spring balance fixed on a ceiling and note the reading on the spring balance.
⚫ Change the stone and repeat the observations.
⚫ Discuss in groups about the nature of forces involved in your experiment.Questions:
1. What force did you observe in the experiment?
2. Where may this type of force appear?
Fig. 3.3: Tension force
The tension force T in Fig. 3.3, is the force which is transmitted through a
string, rope, cable or wire which are massless, frictionless, unbreakable,
and unstretchable when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.3.7.1 Normal Forces
Activity 3.3: Investigating the normal force
Materials:
⚫ Five books
⚫ Two wooden blocks
⚫ A meter ruler Procedure:
⚫ Arrange the blocks and the ruler as a bench as follows in Fig 3.4:
Fig. 3.4: Wooden blocks and the ruler laid in the form of a table.
⚫ Lay the five books on the ruler at its middle and note your observations.
⚫ Remove some books and note the change; then discuss the changes.
Questions:
1. What are the forces involved in the experiment?
2. Which force is opposing the books not to break the ruler down?
Fig. 3.5: Normal forces
The normal force is a component of the contact force that is perpendicular
to the surface of contact exerted on an object by the surface (of a table,
wall etc).
Fig. 3.6
The normal force here represents the force applied by the table against the
book that prevents it (the book) from sinking through the table.
The normal force acts as resistive force that balances with the weight
of the body in vertical direction as shown in fig. 3.53.7.2 Air resistance forces
Activity 3.4: Investigating the air resistanceTake the case of the parachute as in fig. 3.7 and answer to the following
questions.
Fig. 3.7: Air resistance in parachutes
1. What is happening to the parachute?
2. Name the forces involved in the downward motion of the parachutist?The air resistance is a special type of frictional force which acts upon
objects as they travel through the air. The force of air resistance is often
observed to oppose the motion of an object. This force will frequently be
neglected due to its negligible magnitude. It is most noticeable for objects
which travel at high speeds (e.g., a skydiver or a downhill skier) or for
objects with large surface areas.3.7.3 Spring Force
Activity 3.5: Investigating the spring force
Materials:
⚫ Spring balance
⚫ Object (Stone (s))
Procedures:
⚫ Hang the spring balance in a fixed position.
⚫ Put the stone of mass, m and notice the change on the spring
balance.
⚫ Change the stone and continue to note your observations.
Questions:
1. What are the forces involved in this system?
2. Which force does the spring exert on the stone?
Fig. 3.8: Spring forces
The spring force (Elastic force) is the force exerted by a compressed or
stretched spring upon any object which restores its original position. An
object which compresses or stretches a spring is always acted upon by a
force the object to its rest or equilibrium position. The spring for Fk acts as a resistive force as shown in the figure 3.8.3.7.4 Applied forces
Activity 3.6 Investigating applied force
Take the case of a man pushing the box as in fig.3.9. and answer the
following questions:
1. What is being done?
2. Which forces are involved in the process?
3. Name the force the man is using to push the box.
Fig. 3.9: Applied forces
An applied force Fp is a force in an action on the body that makes it to
move or tend to move. The fig. 3.9 above shows a person pushing a body
along an inclined plane. The push is the applied force acting upon the
body. The force that resists (opposes) motion in the figure is denoted by f.
Fig. 3.10: Magnetic forces
This type of force acts between poles of magnets (Fig. 3.10). The magnetic force
is either a pull or a push. For example, two magnets can exert a magnetic pull on each other even when separated by a distance of a few centimeters. Like poles repel (push) and unlike poles attract (pull) each other.3.7.6 Gravitational forces
The force of gravity is the force with which the earth, moon, or other
massively large object attracts another object towards itself. All objects
upon earth experience a force of gravity which is directed "downward"
towards the center of the earth. The force of gravity on earth is always
equal to the weight of the object as found by the equation:
FG = mg
Even when your feet leave the earth and you are no longer in physical
contact with the earth, there is a gravitational pull between you and the Earth.3.8 Unit 3 assessment
1. Fill in the blanks.
(a) Friction opposes the ________ between the surfaces in contact
with each other.
(b) Friction depends on the _________ of surfaces.
(c) Friction produces _________.
(d) Sprinkling of powder on the carrom board ________ friction.
(e) Sliding friction is ________ than the static friction.
2. Four children were asked to arrange forces due to rolling, static
and sliding frictions in a decreasing order. Their arrangements
are given below. Choose the correct arrangement.
(a) rolling, static, sliding
(b) rolling, sliding, static
(c) static, sliding, rolling
(d) sliding, static, rolling3. Alida runs her toy car on a dry marble floor, wet marble floor, a newspaper and a towel spread on the floor. The force of friction acting on the car on different surfaces in increasing order will be;
(a) Wet marble floor, dry marble floor, newspaper and towel.
(b) Newspaper, towel, dry marble floor, wet marble floor.
(c) Towel, newspaper, dry marble floor, wet marble floor.
(d) Wet marble floor, dry marble floor, towel, newspaper.
4. Suppose your writing desk is tilted a little. A book kept on it
starts sliding down. Show the direction of frictional force acting on it.5. You spill a bucket of soapy water on a marble floor accidentally.
Would it make it easier or more difficult for you to walk on the
floor? Why?
6. Explain why athletes use shoes with spikes.
7. Ineza has to push a lighter box and Shema has to push a similar
heavier box on the same floor. Who will have to apply a larger
force and why?
8. Explain why sliding friction is less than static friction.
9. Give examples to show that friction is both a friend and a foe.
10. Explain why objects moving in fluids must have special shapes.
11. A 5kg object is sliding to the right and encountering a friction force which slows it down. The coefficient of friction (μ) between the object and the surface is 0.1. Determine the force of gravity, the normal force, the force of friction. (Neglect air resistance).
12. A rightward force is applied to a 10-kg object to move it across a rough surface at constant velocity. The coefficient of friction between the object and the surface is 0.2. Use the diagram to
determine the gravitational force, normal force, applied force, frictional force, and net force. (Neglect air resistance).
13. A rightward force is applied to a 5kg object to move it across a rough surface with a rightward acceleration of 2m/s2. The coefficient of friction between the object and the surface is 0.1.
Use the diagram to determine the gravitational force, normal force, applied force, frictional force, and net force. (Neglect air resistance).
14. Eduardo applies a 4.25N rightward force to a 0.765kg book to accelerate it across a table top. The coefficient of friction between the book and the table top is 0.410. Determine the acceleration of the book.
Key unit competence
The learner should be able to define pressure and explain factors affecting
it.
My goals
By the end of this unit, I will be able to:
define and explain the pressure as a relationship of force acting
on a surface area.
identify force and area as factors affecting pressure in solids.
give the relationship between force, pressure and area.
explain how pressure varies with force and the area of contact.
describe liquid (mercury) in glass barometer.
explain floating and phenomena.
describe how to measure atmospheric pressure.
carry out calculations using the equation.Pressure = force/area and P = ρgh.
explain the variation in atmospheric pressure with altitude.
explain the change in pressure by reducing or increasing area of contact and vice versa.
measure atmospheric pressure using a barometer, liquid in glass barometer. explain the functioning of aneroid barometer.
describe and explain pressure transmission in hydraulic systems.
explain functioning of a hydraulic press and hydraulic brakes.Key concepts
1. How experimentally can one determine the effect of force exerted on a solid?
2. How can one define pressure in a fluid?
3. How can pressure in liquids be measured?
4. What instrument should be used in measuring pressure?
5. Where can pressure in solids and liquids be applied in real life?Vocabulary
Pressure, atmospheric pressure, fluids, hydrostatic pressure, barometer, manometer.Reading strategy
As you read this unit – mark the paragraphs that contain definitions of key terms. Use all the information to write a definition of each key term in your own words. Perform calculations related to pressure in solids and fluids.4.1 Force exerted by solids
Activity 4.1: Investigation pressure of a solid
Materials:
⚫ One concrete brick
⚫ Balance
⚫ A pile of sand
⚫ A ruler
⚫ A long beam of woodProcedures:
Measure the mass (m) of the brick and calculate its weight (w = mg).
Pour two bucketfuls of sand outside your laboratory such that it forms a pile as shown in (i).
Use the long wooden beam to spread the sand such that you have a
fairly large plain surface on top of the sand pile, as shown in (ii).
⚫ Take measurement of dimensions of one of the large surface side and calculate its area A1.
⚫ Take measurement of the dimensions of the small side and calculate its area A2.
⚫ Gently place the brick in the sand on its big side and let it rest on the sand for 15s. Carefully remove the brick from the sand. Note the depression formed on the sand and carefully measure its depth. Measure the depth at four different places and determine the average depth of the depression on the sand left by the brick. Calculate pressure exerted by the brick using
⚫ Gently place the brick on the sand on its smaller side but at a
point away from the first experiment. Calculate the pressure
exerted by the brick
Compare and discuss the result obtained.
If the force is concentrated on a small area, it will exert higher pressure
than if the same force is distributed over a larger surface area.
As an example of varying pressures,
⚫ A finger can be pressed against a wall without making any lasting impression; however, the same finger pushing a thumbtack can easily penetrate the wall. Although the force applied to the surface
is the same, the thumbtack applies more pressure because the point concentrates that force into a smaller area.
⚫ If we try and cut a fruit with the flat side of the knife it obviously won’t cut. But if we take the sharp side, it will cut smoothly. The reason is, the flat side has a greater surface area(less pressure) and so it does not cut the fruit. When we take the thin side, the surface area is very small and so it cuts the fruit easily and quickly.
⚫ A bus or truck is heavy. It may have large tyres, so that its weight is spread over a large area. This means that the pressure on the ground is reduced; so it is less likely to sink in soft ground. This is one example of a practical application of pressure.4.2 Definition and units of pressure
Pressure (symbol “p”) is the force acting normally per unit area applied in a direction perpendicular to the surface of an object. Gauge pressure is the pressure relative to the local atmospheric or ambient pressure. The pressure is directly proportional of force and proportional of square area. In mathematical terms, pressure can be expressed as:
The pressure within a fluid (gas or liquid) is a scalar quantity—that is, it has magnitude but no particular direction associated with it in space.
Pressure arises from two fundamentally different kinds of sources: ambient and localised.1. Ambient sources of pressure are usually a gas or a liquid in which an entity is immersed, such as a human being on the surface of the earth or a fish in a stream or lake. Life forms are generally insensitive to ambient pressures and become aware of the source of that pressure when currents become strong enough that the fluid exerts a non-uniform localised pressure on the life form, such as when the wind blows.
2. Localised pressure sources are usually discrete objects, such as the finger pressing on the wall, or the tyres of a car pressed against the pavement. A liquid or gas can become the source of a localised pressure if either of them is forced through a narrow opening.
4.2.1 Unit of pressure
The unit is the pascal and is named after Blaise Pascal, the eminent
French mathematician, physicist, and philosopher noted for his
experiments with a barometer, an instrument to measure air pressure.
The name Pascal was adopted for the SI unit Newton per square meter
by the 14th CGPM in 1971.The Pascal (symbol: Pa) is the SI derived unit of pressure. It is a measure of force per unit area, equivalent to l Pa ≡ 1 N/m2 ≡ 1 kg/(m·s2).
The Pascal is perhaps best known from meteorological barometric
pressure reports, where it occurs in the form of hectopascals or millibar
(1hPa ≡ 100 Pa ≡ 1 mbar).
barye ≡ 1 μbar = 1 dyn/cm2The standard atmosphere (atm) of pressure is approximately equal to air
pressure on earth above mean sea level and is defined as:
In 1985, the International Union of Pure and Applied Chemistry (IUPAC)
recommended that standard atmospheric pressure should be harmonised
to 100,000 Pa = 1 bar = 750 Torr.
Another unit for pressure measurement is millimeters of mercury
(.1 mmHg = 9.80669Pa)
We use a manometer to measure pressure in liquids and a barometer to
measure air pressure.Activity 4.2: Using a Manometer
⚫ Manometer
⚫ Water
⚫ Beaker
⚫ A ruler (Optional)
Procedures:
Try to refer to the Fig. 4.2 and do the following:
a) Pour water into the beaker.
b) Note the level of the manometer liquid.
c) Lower the manometer nozzle in water.
d) Note the change in the level of the manometer liquid.
e) Lower it deeper than before and note the new changes.Questions:
1. What changes are you observing?
2. Discuss the meaning and the cause of that change.
3. Prove that liquids exert pressure on a body submerged in.
The figure above is a simple pressure gauge and it measures differences in
pressure exerted at the two ends of the apparatus. It is called a Manometer.
The mouth of a thistle funnel is tightly covered with a thin plastic sheet.
The thistle funnel is connected to a U-tube manometer containing water, by rubber tubing.Now, lower the mouth of the funnel into a glass vessel containing water.
You will notice that the deeper it goes; greater is the difference in the levels
of the water in the manometer. This indicates that the pressure in a liquid increases with depth.Repeat the experiment by turning the thistle funnel in different directions
keeping the depth constant. You will observe that as long as the depth
remains the same, there is no change in the level of the water in the
manometer. Thus, the pressure exerted by a liquid at a given depth is the same in all directions.Now, lower the thistle funnel to the same depth in a number of liquids
having different densities. You will notice that in liquids having greater
density, the pressure at the same depth is greater. This indicates that
the greater the density of the liquid, the greater the pressure at the same depth.4.3 Static fluid pressure
The pressure exerted by a static fluid depends only upon the depth of the
fluid, the density of the fluid, and the acceleration of gravity.
The pressure in a static fluid arises from the weight of the fluid and is given by the expression:Pstaticfluid = ρf gh
Where ρ is the density of fluid; g is acceleration of gravity; and h is depth
of fluid.
The pressure from the weight of a column of liquid of area A and heighth is:
Fig. 4.3: Pressure in liquid
The most remarkable thing about this expression is what it does not
include. The fluid pressure at a given depth does not depend upon the total mass or total volume of the liquid. The above pressure expression is easy to understand for the straight, unobstructed column, but not obvious for the cases of different geometry which are shown.Because of the ease of visualising a column height of a known liquid, it has become common practice to state all kinds of pressures in column height units, like mmHg. Pressures are often measured by manometers in terms of a liquid column height.
4.3.1 Fluid pressure calculation
Fig. 4.4: Pressure at a depth h P = ρgh
4.4 Atmospheric pressure
4.4.1 Measuring atmospheric pressure
Fig. 4.5: Mercury barometer
When the tube is inverted over the pool, mercury flows out of the tube, creating a vacuum in the head space, and stabilises at an equilibrium height, h over the surface of the pool. This equilibrium requires that the pressure exerted on the mercury at two points on the horizontal surface of the pool, (inside the tube) and (outside the tube), be equal. The pressure at the point inside the tube is that of the mercury column overhead, while the pressure at that point is that of the atmosphere overhead. We obtain,
from measurement of h,
where ρHg = 13.6 gcm–3 is the density of mercury and g = 9.8ms–2 is the
acceleration of gravity. The mean value of h measured at sea level is 76.0
cm, and the corresponding atmospheric pressure is 1.013 × 105 kg m–1
S–2 in SI units. The most commonly used pressure unit is the atmosphere
(atm) (1atm = 1.013 × 105 Pa) the bar (b)
(1 b = 1 × 105 Pa), the millibar (mb) (1 mb = 100 Pa) and the torr.
(1 torr = 1 mm Hg = 134 Pa) The use of millibars is slowly giving way to
the equivalent SI unit of hectoPascals. The mean atmospheric pressure at sea level is given equivalently as;P = 1.013 × 105 Pa = 1013hPa = 1013mb = 1 atm = 760torr.
R is the mean height of the atmosphere from the surface of earth.4.4.2 Mass of the atmosphere
The global mean pressure at the surface of the Earth is slightly less than
the mean sea-level pressure because of the elevation of land. We deduce
the total mass of the atmosphere ma:
4.4.3 Torricelli’s experiment
Fig. 4.6: Torricelli experiment of a simple mercury barometer
Torricelli’s experiment was a curious project made in 1643 by the Italian
physicist and mathematician Evangelista Torricelli (1608-1647) in a
laboratory that attained to measure the atmospheric pressure for the first
time.
Process:
Torricelli filled a 1 metre long tube with mercury, (closed at one end) and
inverted it on a tray full of mercury. Immediately the column of mercury
went down several centimetres, remaining static at some 76cm (760mm) of height.
As it was observed that the pressure was directly proportional to the height
of the mercury column (Hg), the millimetre of mercury was adopted as a
measurement of pressure.
That way, the pressure corresponded to a column of 760mm.Conclusion:
The column of mercury did not fall due to the fact that the atmospheric
pressure exerted on the surface of the mercury was able to balance the
pressure exerted by his weight.
760 mmHg = 1 atm
1 atm = 1.013 mbar or hPa
1 mbar or hPa = 0,7502467 mmHg4.5 Gas pressure
Activity 4.3: Investigating gas pressure
Imagine the case of pumping air in the ball (e.g Football):
Materials:
⚫ Bicycle tube
⚫ Brick
⚫ Plank of wood
⚫ Bicycle pump
Procedures:
⚫ Take an empty bicycle tube.
⚫ Lay a plank of wood across it.
⚫ Place the brick at the centre of the plank.
⚫ Take the pump and start to pump the gas into the bicycle tube.
⚫ Note the changes in the position of the brick as one continues to pump.Questions:
1. Explain why pumping air into the bicycle tube results in rise of the brick placed on the wooden plank.
2. Explain what would happen, if one continues to pump in more and more gas.Pressure is determined by the flow of mass from a high pressure region to a low pressure region. Air exerts a pressure which we are so accustomed to that we ignore it. The pressure of water on a swimmer is more noticeable. You may be aware of pressure measurements in relation to your bicycle tyres.
Atmospheric pressure varies with height just as water pressure varies with depth. As a swimmer dives deeper, the water pressure on his/her increases. As a mountain climber ascends to higher altitudes, the atmospheric pressure on him/her decreases. His body is compressed by a smaller amount of air above it. The atmospheric pressure at 6100m is only one-half of that at level because about half of the entire atmosphere is below this elevation.
4.6 Simple pressure related applications
4.6.1 Drinking straw
Activity 4.4: Investigation of atmospheric
pressure in using drinking straws
Materials:
⚫ Drinking straws
⚫ Very clean bottles of mineral water
⚫ Safe drinking water
⚫ 50mm beakersProcedures:
⚫ Make a hole on the cover of the mineral water bottle that exactly fits the straw tube.
⚫ Insert the straw tube through the hole such that its bottom is about 5mm from the bottom of the plastic water bottle.
⚫ Fill the water bottle with clean and safe water and close.
⚫ Make sure that the contacts between the tube and cover are airtight.
⚫ Fill your beaker with safe drinking water and suck it using a straw.
⚫ Take water/juice in the bottle (closed such that no air can get inside), but with an opening of the straw only and suck.Questions:
1. What have you noticed when drinking from the glass?
2. What have you noticed when drinking from the bottle?
3. Discuss and explain the causes of your observations.
Fig. 4.7: Drinking straw
A drinking straw is used to create suction with your mouth. This causes a
decrease in air pressure inside the straw. Since the atmospheric pressure
is greater on the outside of the straw, liquid is forced into and up the straw.4.6.2 Siphon
Activity 4.5: Investigation of the siphon
Materials:
⚫ A jerrycan
⚫ Bucket
⚫ Water
⚫ A long flexible plastic pipe
⚫ TableProcedures:
⚫ Fill a jerrycan of water on the table.
⚫ Place the bucket down on the lower level of the table.
⚫ Lower one end of the plastic pipe in the jerrycan.
⚫ Let the other end of the plastic pipe be at a lower level than of the water in the jerrycan.
⚫ Suck water from the jerrycan, and release after the water has come to your mouth.
⚫ Let water flow from the jerrycan to the bucket freely.
Questions:
1. What causes the water to flow from the jerrycan to the bucket?
2. Why does the water continue to flow without sucking again?
3. Discuss and explain where this can be applied.
Fig. 4.8: The Siphon
With a siphon water can be made to flow “uphill”. A siphon can be
started by filling the tube with water (perhaps by suction). Once started,
atmospheric pressure upon the surface of the upper container forces water
up the short tube to replace water flowing out of the long tube.4.7 Pressure with altitudes
Pressure decreases with increasing altitude
The pressure at any level in the atmosphere may be interpreted as the total
weight of the air above a unit area at any elevation. At higher elevations,
there are fewer air molecules above a given surface than a similar surface
at lower levels. For example, there are fewer molecules above the 50km
surface than are found above the 12km surface, that is why the pressure
is less at 50km as shown in the Fig. 4.10 below.
Fig. 4.19: Pressure variation with altitude
Fig. 4.10: Pressure variations with altitude
Since more than half of the atmosphere’s molecules are located below
an altitude of 5.5km, atmospheric pressure decreases roughly 50% (to
around 500mb) within the lowest 5.5km (Fig 4.10). Above 5.5km, the
pressure continues to decrease but at an increasingly slower rate.4.8 Common observations of pressure
Activity 4.6: Investigation of application of pressure in real life
Study the following questions. Discuss them and give other more
examples in each case.
1. Why do heavy lorries have many tyres?
2. Why do tractors (excavators, loaders, etc) have wide tyres?
3. Discuss and explain why ducks, geese have webbed feet?
4. Why do the feet of camels and elephants have wide pads?4.8.1 Ducks, geese, and swans all have webbed feet. The primary use for webbed feet is paddling through water
Fig. 4.11: Duck webbed feet
Here’s how it works: as the bird pulls its foot backwards through the water, the toes spread apart,
causing the webs to spread out. The webs push more water than just a bird foot with spread-out toes would push. (It would be like trying to swim with your fingers spread apart.) The webbed feet propel the bird through the water.When the bird pulls its foot forward for the next push, the toes come together, folding up the webs. The foot is instantly less resistant, moving through the water easily to get into place for the next stroke without
pushing the bird backwards. Webbed feet are useful on land as well as on water because they allow birds to walk more easily on mud. Most swimming or paddling birds have their legs and feet located at the rear of their body. This adaptation is an advantage on the water as it helps to propel the birds along.
But what’s good on the water isn’t necessarily good on land. Having their legs and feet located at the rear of their body makes walking more difficult for these birds.4.8.2 Camel or elephant wide pads
Fig. 4.12: Wide pads of a Camel and an Elephant
Why do camels or elephants have wide and large feet? Those features
help them walk across desert sands. So that they are able to walk across
sand without sinking in. To walk on sand so they have a bigger surface area
to handle their weight and the objects that are put on it to carry - so they
don’t sink into the sand to spread their weight out over the sand, which
helps to prevent them from sinking into it. Camels are adapted to walk
long distances in deserts, hence, they have evolved to form large, broad,
flat feet. More surface area means less pressure exerted on that surface,
and vice-versa as the pressure is distributed on a large area, because it
would give less pressure on the sand which prevents it from sinking.4.8.3 Lorries with many tyres
Why do trucks that carry heavy loads have so many wheels often eighteen?To distribute the load over a greater area.
Fig. 4.13: Lorries with many tyres
Pavement strength is all about pressure (or stress more correctly). This is a force divided by area. If you increase the area (number of tyres) that the load is distributed over, there will be less pressure (stress) on the pavement. Think about an extreme example: Say a really heavy dump truck has only four tyres, then there are only four places for the load to go to. Further still, think about a dump truck with only one tyre! all the weight of the truck will be put onto that one tyre!
This is the same principle by which knives work, you are applying a small force, but over a very small area, which makes the thing you are cutting, cut. If you have a dull knife, it will be harder to cut because the force is distributed over a larger area.4.9 Unit 4 assessment
3. In the equation for Pressure p=p×g×h, the units for g (SI
system) are:
a) kg/m 3
b) m/s
c) kg-m/s
d) m/s 2
4. What is the pressure at the bottom of a swimming pool that is 3
meters in depth?
a) (1.01 x 105) + (1.09 x 105) Pa = 2.10 x 105 Pa
b) (1.01 x 105) + (3.0 x 104) Pa = 1.31 x 105 Pa
c) (1.01 x 105) + (7 x 104) Pa = 1.71 x 105 Pa
5. A substance has mass of 3kg submitted by acceleration of
9m/s2.
a) Find the force in Newton.
b) What is the pressure of it on a square of 4m for a side?7. (a) Define pressure and state its SI unit.
(b) Find the pressure in Pa of force, F = 45N and applied on a
triangle of base of 5m and height of 3m.
8. Calculate the Pressure on the surface when a force of 30N acts
on area of 0.2m2.
9. a) Define pressure and state the S.I unit in which pressure
can be expressed.
b) A brick of mass 3kg measures 6cm by 4cm by 3cm.
(i) What is the greatest pressure it can exert when placed on
a flat surface.
(ii) What is the least pressure it can exert?
10. Which of the shoes shown below causes most damage?
Fig 4.15
11. Copy and fill in the blanks the missing words:
Pressure tells us how concentrated a ------------ is. It is measured in
------------- or ------------, and is calculated using the equation: p = ---
---------. A force of 12N acting over an area of 2m2 causes a pressure
of ------------. If the area were less, the pressure would be ------------
----------. The dimensions of velocity are ------------. The dimensions
of pressure are ------------.
12. A book of mass 500g is lying on a table. Its cover measures
25cm by 29cm. What pressure does it exert on the table?- Key unit competenceThe learner should be able to explain the working principle of a manometer used to measure the pressure in liquids.My goalsBy the end of this unit, I will be able to: describe a manometer.
explain the principle of a manometer.
explain hydrostatic pressure and atmospheric pressure and their measurement.
explain equilibrium of a liquid at rest in a vessel and
communicating container.
explain why a liquid surface is an isobar and describe its application.
analyse the equilibrium of non-miscible liquids in a container and in communicating container.
solve problems on a manometer.
recognise the application of the same level of liquid in communicating vessels.
appreciate the results of measurement of liquid pressure using a manometer.
identify the use of pressure in everyday activities (aviation,, sports).Key concepts
1. What is needed to describe pressure in liquids?
2. Which factors affect the variation of pressure in liquids?
3. Discuss the pressure effect in non-miscible liquids?
4. Discuss different applications of pressure in liquids?
Vocabulary
U-tube, miscible and non miscible liquids, hydrostatic pressure, isobar, fluids in equilibrium.Reading strategy
As you read each section of this unit, put emphasis on paragraphs that contain definitions of key terms. Use this information to write the meaning of each key term in your own words.5.1 Pressure in liquids in equilibriumActivity 5.1:Investigating of pressure in liquidsMaterials:
⚫ Large tin can or plastic bottle.
⚫ Water
⚫ Hammer and nail
⚫ Ruler
Fig. 5.1: Water squirts further at greater depthsProcedure:
1. Punch 5 holes in the sides of the container, one below the other at 4cm intervals.
2. Fill the container with water.
3. Measure the distance from the bottom of the container to the point that the water squirts on the ground from each hole.
4. Plot a graph of depth (distance of the hole from the top of the water level) versus the distance water squirts on the ground.
Questions:
1. What is pushing water to squirt out from the container?
2. Why is water falling at different distances?
3. Discuss and explain the situation.Table 5.1: Hydrostatic Pressure in a Liquid
Activity 5.2: Water pressure experiments
Materials:
Plastic bottles can be used to make apparatus to investigate water
pressure in containers of water. Figure 5.3 shows three sizes of plastic
bottles: 1L, 2L and 4L.
One hole, approximately 5 mm in diameters (use a pin hole), has been
drilled through the side of each bottle, at a height of 5cm from the
bottom.
Fig. 5.3: Investigating water pressure
Procedure:
1. Using a felt pen, draw a line circling the bottle at the height
of the hole (5 cm from the bottom of the bottle) and another
line circling the bottle at a height of 20cm from the bottom of the bottle.
2. Place a piece of masking tape over the holes in each bottle.
Leave a loose tab so that you can use it to pull off the tape
quickly when you want to let water run out of the hole.
Questions:
1. Predict what will happen in this situation:
2. Out of which bottle will the water travel furthest horizontally?
3. Try the experiment and test whether your prediction was correct.
Suggestion! Do this experiment outdoors on a paved area where there is no traffic.Table 5.2: Investigating pressure in liquids
If the pressure is measured above atmospheric pressure then the pressure
is called the gauge pressure (hydrostatic pressure):Pg
The air pressure will decrease at altitudes above sea level (and increase
below sea level).
The liquid pressure at different depths based on gauge pressure is rewritten
as pg = ρgh
The pressure in a liquid increases linearly with depth from its value P0 at
the surface that is open to the atmosphere or from some other reference point.5.1.1 Pressure in relation to diving and aviation
Pressure increases with depth. It is dangerous to stay longer at a depth of
45m, since as result of high pressure, an excess of nitrogen dissolves in
the blood and on return to the surface nitrogen bubbles form in the blood
in the same way that bubbles form in a bottle of soda water when the cap
is removed. Such a condition causes severe pain or even death. And in
cases of emergency the diver is immediately placed in a decompression
chamber. This is a steel tank full of compressed air and, by slowly reducingthe pressure over a long period, the nitrogen becomes gradually eliminated from the blood without forming bubbles.
In contrast with the problem encountered by a diver, the crew and passengers in aircraft flying at high altitude would experience difficulty in breathing and consequent danger owing to low pressure. The problem is overcome by pressurising the aircraft. All openings are sealed, and a normal atmospheric pressure is maintained inside by the use of air pump.5.2 Equilibrium of a liquid at rest
5.2.1 Equilibrium of a liquid at rest in a container
Fig. 5.6: Isobar and free surface of a liquid at equilibriumWhen water or any other liquid is poured into the communicating tubes, it stands at the same level in each tube which means that water finds its own level.
“Iso” means “same” and “bar” means “pressure”, so an isobar is a surface of constant pressure. In hydrostatics, isobars are horizontal surfaces, since pressure does not change horizontally through the same fluid.Recall that a free surface exposed to atmospheric pressure always has a pressure equal to the local atmospheric pressure. Thus, a free surface is always an isobar.
Fig. 5.7: Pressure difference between two pointsActivity 5.3: Pressure difference between any two points in the same liquid
Materials:
⚫ Water
⚫ Bucket
⚫ Manometer
⚫ Ruler
Procedure:
⚫ Pour water in the bucket
⚫ Lower the manometer at a given height h1, and calculate the pressure p1.
⚫ Increase the depth at h2, and calculate the pressure p2.Questions:
Find the pressure difference.
Application: the level indicator, distribution of water in cities.
Fig. 5.8: Communicating vessel or Pascals vessel5.2.2 Equilibrium of a liquid in a communicating containerActivity 5.4: Investigating pressure in liquids using a communicating vessel
Materials:
⚫ Communicating vessel (Fig.5.8)
⚫ Coloured waterProcedure:
⚫ Pour water in the big branch of the communicating vessel.
⚫ Open the tap and observe what happens.
Questions:
1. Suggest what will happen after the tap is.
2. Why are the levels of water in all branches like that, after opening the tap?
Fig. 5.9Communicating vessels, sometimes referred to as communicating vases,
is a name given to a set of containers of different shapes and sizes that are
attached to a common tube at different places along the tube. This tube
is scaled at one end and is attached to a larger vessel at the other end.
See figure 5.9(a) or 5.9(b). When a liquid is poured into a communicating
vessel, the liquid balances out to the same level in all of the containers
regardless of the shape and volume of the containers. If additional liquid
is added to one vessel, the liquid will again find a new equal level in all
the connected vessels. This process is part of Stevin’s Law and occurs
because gravity and pressure are constant in each vessel (hydrostatic pressure).
Fig. 5.10: Communicating vessels or Pascal’s vesselThe fluid levels are the same in each tube irrespective of their shape and size.
Pascal’s vases is used for demonstrating that pressure in a liquid is a function of depth only.The pressure at any point in a liquid at rest depends only on the depth and on the density of the liquid but not
the shape of the vessel.The apparatus consists of a group of glass flasks of assorted shape (see Fig. 5.10) linked at their base by a communal reservoir. With the pressure being dependent on the depth of the liquid only, an equilibrium situation must have the surface level in each vase equal. When the liquid is at rest
in the vessel the pressure must be the same at all points along the same horizontal level, otherwise the liquid would move until the pressures were equalised. The pressure at the bottom of the fluid at rest depends upon the depth and on the density of the fluid. It is independent on the shape of the container
or the amount of liquid in the vessel5.2.3 Equilibrium of several non-miscible liquidsActivity 5.5 Investigating pressures in non– miscible liquidsMaterials:
⚫ A tall glass jar or a tall transparent plastic container
⚫ Water
⚫ Cooking oil
⚫ Glycerin
⚫ Engine oil
Procedure:
⚫ Without any preferantial order, pour each of the liquids, one by one and carefully into the same glass or plastic jar. Determine the density of each of the liquids that you are going to use in
the exercise.
⚫ Shake the vessel or stir the liquids you have poured in and allow them to settle for a period of 20 minutes.
⚫ After twenty(20) minutes carefully observe and note the different layers of liquid in the jar and record the order from top to bottom.
Questions:
1. List the order of liquids from the top to the bottom.
2. Discuss and explain why the situation is as you have observed.
Fig. 5.11: Non-miscible liquids in a containerIn a container, immiscible fluids are superposed according
to their decreasing relative densities from the lightest to the
heaviest.Where RD denote the Relative density.The area of separation between any two immiscible liquids is flat and
horizontal. We use the funnel method or decanting to separate immiscible
liquid e.g. mixture of water and paraffin.
Fig. 5.12: Two non-miscible liquids in a communicating container.5.2.4 The manometer density measurement methods
The method can only be used with non-miscible liquids. The two liquids
are separated by air and never mix (if you are careful!).
The formula relating the densities and height is:
5.3 Applications of hydrostatics
5.3.1 Pressure measurement with hydrostatics
A standard mercury barometer has a mercury column of about 76 cm in height, in a glass or plastic jar or tube closed at one end, with an openmercury-filled reservoir at the base (Fig. 5.13). The first barometer of this
type was devised in 1643 by Evangelista Torricelli.
Torricelli had set out to create an instrument to measure the weight of air, or air pressure, and to study the nature of vacuums. If the indicator scale is calibrated to give altitude instead of air pressure,
the device is called an “altimeter”.
Fig. 5.13: Mercury BarometerThis gadget is formed by inverting a glass tube filled with mercury into a mercury bath.
At the top of the mercury column in the tube (point 3 in the sketch), the pressure is
nearly a total vacuum. The pressure at point
1 is atmospheric, and this pressure holds the mercury column at some height h, as measured by a ruler.
For e.g., the weatherman may say “…the barometer reads 76cm of
mercury.” This means that h = 76cm of mercury column in the barometer.
Using the above equation, one can calculate atmospheric pressure in more
standard pressure units, such as kPa:
Atmospheric pressure can support a column of water 10.3m high, or a
column of mercury of length 76cm.Activity 5.6: Investigating pressure using a u-tube manometerTake the case of U-Tube manometer in Fig.5.15; discuss and
explain the steps in a (Fig.5.15),
b (fig. 5.17), c (fig. 5.18) and d (fig. 5.19) using the rules of
the U-tube manometer.
Questions:
1. What does the pressure of liquid in a manometer tube depend on?
2. Discuss other areas where a manometer tube can be used in real life.The U-tube Manometer case
This device consists of a glass tube bent into the shape of a “U”, and is
used to measure some unknown pressure.
For example, consider the sketch below (Fig. 5.14), where a U-tube
manometer is used to measure pressure PA in some kind of tank or machine.
Fig. 5.14: The U-Tube manometerConsider the right side and the left side of the manometer separately. The
equation for hydrostatics gives:
Fig. 5. 15: The Manometer and its diameter
b) Manometer height difference does not depend on tube length
(provided, of course, that the length is enough to handle the height difference).
Fig. 5.16: The Manometer and its tube lengthc) Manometer height difference does not depend on tube shape
(except, of course, if the tube is of very small diameter, and surface tension effects are significant).
Fig. 5.17: The Manometer and its shape
Recall that the shape of a container does not matter in hydrostatics. This
implies that a U-tube manometer does not have to be in a perfect U shape.
Although the column height difference between the two sides does not
change, an inclined manometer has better resolution than does a standard
vertical manometer because of the inclined right side. Specifically, for a
given ruler resolution, one “tick” mark on the ruler corresponds to a finer gradation of pressure for the inclined case.
d) Manometer height difference does depend on the fluid used in the manometer.
Fig. 5.18: The Manometer and manometer fluid
For the same pressure difference, a dense manometer liquid will have a
smaller difference in column height than a less dense manometer liquid.
This too can be used advantageously. If a small pressure difference is
being measured, it is better to use a light fluid, since the resolution and accuracy are improved.Why not always use a very light liquid in a manometer then? Well, for measurement of large pressure differences, the manometer may have to be too high to be practical. In such cases, a very dense liquid, such as mercury, should be used. Furthermore, the manometer liquid must be more dense than the fluid in which the pressure is being measured, for obvious reasons.5.3.2 Pressure measurement with bourdon gauges
An aneroid barometer (Bourdon gauges) uses a small, flexible metal box
called an aneroid cell. This aneroid capsule (cell) is made from an alloy
of beryllium and copper. The box is tightly sealed after some of the air
is removed, so that small changes in external air pressure cause the cell
to expand or contract. This expansion and contraction drives mechanical
levers and other devices which are displayed on the face of the aneroid barometer.
Fig. 5.19: Bourdon GaugesA mercury barometer is long and inconvenient, heavy; and contains a
liquid that is hazardous and easily split; therefore, an aneroid barometer is
commonly used. Aneroid means without liquid. It is compact and portable;
it has no liquid to spill and no problem with vapour or air getting in.5.3.3 Sphygmomanometer
Blood pressure is measured in millimeters of mercury (mm Hg). A typical
blood pressure is 120/80mm Hg, or "120 over 80." The first number
represents the pressure when the heart contracts and is called the systolic
blood pressure. The second number represents the pressure when the heart relaxes and is called the diastolic blood pressure.Activity 5.7: Measuring blood pressure
Materials:
A Sphygmomanometer (blood pressure cuff)
1. Deflate the air bladder of the cuff and place it around the upper arm so it fits snugly. If you’re right handed, you should hold the bulb/pump in your left hand to inflate the cuff. Hold it in
the palm so your fingers can easily reach the valve at the top to open and close the outlet to the air bladder.
2. Put the head of the stethoscope just under the edge of the cuff, a little above the crease of the person’s elbow.
3. Inflate the cuff with brisk squeezes of the bulb. Watch the pressure gauge as you do it, you should go to around 150 mmHg or until the pulse is no longer heard. At this point blood flow in the underlying blood vessel is cut off by pressure in the cuff.
Fig. 5.20: The Sphygmomanometer4. At around 150, slightly open the valve on the air pump (held in your left hand). This
part takes practice, it’s important that you don’t let the air out too suddenly.
5. Now, pay attention to what you hear through the stethoscope as the needle
on the pressure gauge falls. You will be listening for a slight “blrrp” or something
that sounds like a “prrpshh”. The first time you hear this sound; note the reading on
the gauge. This value is the systolic blood pressure.
6. The sounds should continue and become louder in intensity. Note the reading when
you hear the sound for the last time. This is the diastolic blood pressure.Questions:
1. In your own words, describe how to use a blood pressure cuff (sphygomomanometer).5.4 Hare’s apparatus
Activity 5.8: Calculating relative density of a liquidObserve at the fig. 5.21 below and try to answer the following
questions:
Fig. 5. 21. Hare’s apparatusQuestions:
1. Calculate pressure at B and B’ due to h1 and h2 respectively.
2. If Pressure at A and A’ are equal, find the relative
density of the liquids. (take density of water to be 1000kg/m3)Finding the Relative Density of Two Liquids Using Hare’s Apparatus
Fig. 5.22: Hare’s apparatusA vacuum pump is connected to the T- clip and some air is sucked out.
The clip is then closed. The pressure inside the glass tubes falls and theliquids rise up the tubes. The liquids rise until the pressure due to each
liquid column is equal to the difference between atmospheric pressure
and the pressure P inside the glass tube. To measure the heights of the
liquids h1 and h2 accurately, bend a paperclip into an asymmetric U
shape and attach it to the bottom of the ruler. Measure the readings on the ruler respectively.
5.5 Unit 5 assessment
1. Find the pressure of that gas sample in Fig 5.17Air pressure, 101.3 kPa
Fig. 5.23: The Opened and closed tube manometer measures pressure2. Mercury has a density that is about 14 times greater than that of
water. If you were to build a barometer that uses water instead of mercury, how would the height of the column of water needed compare to that of the mercury?a) higher than c) equal to
b) lower than d) can’t tell3. Another barometer is placed next to the one discussed previously.
The second barometer uses a wider glass tube but is otherwise the same. The top of the column in the second barometer is the height in the narrow barometer.a) higher than c) at the same height as
b) lower than d) don’t know.”4. A nurse administers medication in a saline solution to a patient
by infusion into a vein in the patient’s arm. The density of the
solution is 1.0 × 103kg/m3 and the gauge pressure inside the
vein is 2.5 × 103. How high above the insertion point must the
container be hung so that there is sufficient pressure to force the fluid into the patient?
5. A tube in a form of U with uniform section contains mercury. In
one of the branches, they pour successively 8cm of water and
6cm of ether. Determine the difference in height between the
two free levels; the volume weight of ether is 7115N/m3, that of mercury is 1333 × 102N/m3.6. In a tube in a form of U, Fig 5.24, they pour mercury. Then in one branch
they pour 20cm of water and 20cm of naphtha in the other branch. Calculate
the difference in height from the surfaces of separation. Volume weight of naphtha is 6157N/m3.
Fig. 5.24: U-tube manometer 17. Determine the pressure difference between points A and B, for
the set-up shown in Fig 5.25. Volume weight unit of water is taken as ω = ρg = 9790N/m3.
Fig. 5.25: U-Tube manometer 28. In a tube in a form of U, they pour mercury and then water in
the other branch. The height of water column is 10cm. What
could be the height of oil column that could bring the two levels
of mercury into the same horizontal plan? The oil volume weight
is .7752 kg/m3
9. Calculate the pressure at a depth of 2m in swimming pool filled
with water.
10. Water and oil are poured into the U-shaped tube, fig 5.26, open
at both ends, and do not mix. They come to equilibrium as shown
in the fig. below. What is the density of the oil?
Fig. 5.26: The U-tube manometer 3 Key unit competence
The learner should be able to explain transmission of pressure in fluids at
rest and describe its applications.My goals
By the end of this unit, I will be able to: explain static pressure of fluids at rest.
describe transmission of pressure in static fluids.
explain Pascal’s principle.
describe applications of Pascal’s principle (Hydraulic press, Hydraulic brake, Water Towers, Hydraulic jack).
Illustrate Pascal’s principle.
explain the functioning of hydraulic jack, lift and dump truck, and car brakes.
learn that pressure exerted on an enclosed fluid is equally transmitted in all directions.
understand how pressure transmitted in a fluid produces a large force when a small force is applied to it.Key concepts
1. What do you understand on Pascal’s Principle?
2. Which factors affect the application of Pascal’s principle?
3. Discuss transmission of pressure in a hydraulic press.
4. Discuss different applications of Pascal’s principle in real life.Vocabulary
Hydraulic press, water tower, cylinder, piston,, liquids in equilibrium, cross section area.Reading strategy
When you are reading this section, take time to understand what you are reading especially the meaning of key words. This will help you to express them in your own words. You will be able to express your calculations and draw your own experiments about Pascal principle.6.1 Fluid static and pressure in fluids at rest
Activity 6.1: Investigating the variation of pressure with depth
Materials:
⚫ Bath
⚫ Water
Procedures:
⚫ Pour water in the bath.
⚫ Let it be in equilibrium for like 5 minutes.
Questions:
1. Is that water flowing?
2. What is the state of motion that water has?
3. How would you find the pressure at the bottom of the bath?Fluid static (also called hydrostatics) is the science of fluids at rest, and is
a sub-field within fluid mechanics. It embraces the study of the conditions
under which fluids are at rest in stable equilibrium. The use of fluid to
do work is called hydraulics, and the science of fluids in motion is fluid
dynamics.
A fluid is defined as a substance that continually deforms (flows) under
an applied shear stress (deformation). All gases and liquids are fluids.
Fluids are a subset of the phases of matter and include liquids, gases
and plasmas.Several properties of fluids need to be established before we concentrate
our discussion on Archimedes’ Principle and Pascal’s Principle.
Fluids display such properties as Density which is defined as the ratio of
mass per unit volume. It is generally represented by the Greek letter
and measured in terms of kilograms/cubic meter, or kg/m3.
Since the volume of a fluid expands and contracts, the density of fluids
varies with temperature. The most common fluid, water, has maximum
density of 1000kg/m3 at 4ºC. Air, a mixture composed principally of the
gases nitrogen (78%) and oxygen (21%), has a density of 1.29kg/m3 at
0ºC and 1.20kg/m3 at 20ºC.How a liquid’s density compares to that of water at 4ºC is called its specific
gravity. If a liquid has a specific gravity of 0.9, then its density is 0.9 times
that of water, or 0.9 x 1000kg/m3 = 900kg/m3.6.2 Pascal’s principle and its application
In the physical sciences, Pascal’s law or Pascal’s principle (1647) states
that: “pressure applied to an enclosed fluid, is transmitted equally to every part of the fluid.”Activity 6.2: Investigating the variation of pressure with depth
Materials:
⚫ A clean water bottle (eg: Inyange mineral water bottle)
⚫ Water
⚫ A pin
Procedure:
⚫ Take water in the bottle.
⚫ Using a pin to make holes on different sides.
⚫ Pump air in the bottle from its opening.Questions:
1. Does water fall at the same distances as before?
2. Compare the distance of water from the lower part and that
of the upper part of the bottle.
3. What causes water to appear to fall at the same distances while it is being pressed?Table 6.1: Transmission of pressure in fluids
Fig. 6.1: Transmission of pressure in liquid
This can be demonstrated using a glass vessel as shown in Fig.6.1.
When force is applied to the piston the pressure exerted on the water is
transmitted equally throughout the water so that water comes out of all
the holes with equal force.
When pressure is applied at a point in a confined fluid, it is transmitted
equally in all directions.
Fig. 6.2: Transmission of pressure in fluid
“Any external pressure applied to a fluid is transmitted undiminished
throughout the liquid and onto the walls of the containing vessel”
Hydraulic devices like the hydraulic press and car brakes are based on the
above principle.“For all points at the same absolute height in a connected body of an incompressible fluid at rest, the fluid pressure is the same. The difference of pressure due to a difference in elevation within a fluid
column is given by:
Where, using SI units,
⚫ ΔP is the hydrostatic pressure (in pascals Pa), or the difference in
pressure at two points within a fluid column, due to the weight of the fluid;
⚫ ρ is the fluid density (in kilograms per cubic meter kg/m3);
⚫ g is sea level acceleration due to Earth’s gravity (in meters per second squared m/s2);
⚫ Δh is the height of fluid above (in meters m), or the difference in elevation between the two points within the fluid column.
The intuitive explanation of this formula is that the change in pressure
between two elevations is due to the weight of the fluid between the elevations.Note that the variation with height does not depend on any additional
pressures. Therefore Pascal’s law can be interpreted as saying that any change in pressure applied at any given point of the fluid is transmitted undiminished throughout the fluid.6.2.2 Pascal’s Principle calculation
A hydraulic pump is used to lift a car; When a small force F is applied
to a small area A of a movable piston it creates a pressure p = F/A. This
pressure is transmitted to and acts on a larger movable piston of area A’ which is then used to lift a car.
Fig. 6.3: The Hydraulic pump
When a force is applied at one end it is transferred to the other end. From
the relation: F = p × A
We see that a small force (thrust) applied at the end with a small area will
produce a larger force at the end with the larger area: F '= p × A'Since pressure equals force per unit area, then it follows that
In both cases the volume of the fluid remains the same. So if the small
piston moves through a distance h and as a result the large piston moves through a distance h’, then the two volumes replaced by the piston and the fluid must be equal i.e. volume displaced by small piston = volume occupied when large piston moves or V = h × A = h' × A'. This gives the
velocity ratio:
So if the area of the large piston is πR2 and that of the small piston is
,
then
This same principle is also used to produce a great force in a car lift and
car transmission systems as well as the hydraulic brake.
For example, if the ratio of the areas is 5, a force of 100N on the small piston will produce a force of 500N on the large piston, and the small piston must be pushed 50cm to get the large piston to rise 10cm. This is how energy, in the form of work in this case, is conserved and the Law of Conservation of Energy is satisfied. Work is force times distance, and since the force is increased on the larger piston, the distance the force is applied over must be decreased. The work of the small piston, 100N multiplied by
0.5m (50cm) is 50 Joules (J), which is the same as the work of the large piston, 500N multiplied by 0.1 m (10cm).6.2.3 Application
6.2.2.1 Hydraulic press
A hydraulic press is a hydraulic mechanism for applying a large
compressive force. It is the hydraulic equivalent of a mechanical lever, and is also known as a Bramah press after the inventor, Joseph Bramah, of England in 1795. Hydraulic presses are the most commonly-used and efficient form of modern press. A fluid, such as oil, is displaced when either piston is pushed inward.
Pressure applied to small piston
Pressure transmitted by fluidFig. 6.4: The Hydraulic press
6.2.2.2 Hydraulic brakeThe hydraulic brake is a braking mechanism which uses brake fluid,
typically containing ethylene glycol, to transfer pressure from the
controlling unit, which is usually near the driver of the vehicle, to the brake
cyclinder, which is usually at or near the wheel of the vehicle.
Fig. 6.5: The Hydraulic brake
The most common arrangement of hydraulic brakes for vehicles consists of a brake pedal, a vacuum assist module, a master cylinder, hydraulic lines, and a brake rotor and/or brake drum.
At one time, passenger vehicles commonly employed disc brakes on the front wheels and drum brakes on the rear wheels. However, four wheel disc brakes have become more popular, replacing drums on all but the most basic vehicles. As the brake pedal is pressed, leverage multiplies the force applied from the pedal to a vacuum booster. The booster multiplies the force again and acts upon a piston in the master cylinder.
As force is applied to this piston, pressure in the hydraulic system increases forcing fluid through the lines to the slave cylinders. The slave cylinders for a drum brake are a pair of opposed pistons which are forced apart by the fluid pressure, while for a disc brake a single piston is forced out of its housing.
The slave cylinder pistons then apply force to the brake linings, which are referred to as shoes in the case of drum brakes or as pads in the case of disc brakes. The forces applied to the linings cause them to be pushed against the rotating metal of the drum or rotor. The friction between the linings and the metal causes a braking torque to be generated, slowing the vehicle.
Fig. 6.6: The hydraulic brake
The fluid pressure from the master cylinder is transferred equally to all the brake shoes.
6.2.2.3 Water Towers
A water tower or elevated water tower is a large elevated water storage
container constructed for the purpose of holding a water supply at a height sufficient to pressurise a water distribution system.
Pressurisation occurs through the elevation of water; for every 10.20cm of elevation, it produces 1 kilopascal of pressure. 30m of elevation produces roughly 300 kPa, which is enough pressure to operate and provide for most domestic water pressure and distribution system requirement.6.2.2.4 Hydraulic lift car
Fig. 6.7: The hydraulic lift.
Because the increase in pressure is the same on the two sides, a small
force Fl at the left produces a much greater force F2 at the right. A vehicle undergoing repair is supported by a hydraulic lift in a garage.6.3 Hydrostatic paradox
Activity 6.3: Investigating the hydraulic pressure
Take the case of Fig. 6.8; read and then discuss and explain what you
have realised from case A (Fig. 6.9) up to case E (Fig 6.13).If the height of the fluid’s surface above the bottom of the five vessels is
the same, in which vessel is the pressure of the fluid on the bottom of the
vessel the greatest? The amount of liquid in each vessel is not necessarily the same.
Fig. 6.8: The Hydrostatic Paradox: The pressure, p is the same on the bottom of each vessel.
Why the pressure does not depend upon the shape of the vessel or the
amount of fluid in the vessel but rests upon three things below:
1. Pressure is force per unit area and this is not the same as the total weight of the liquid in a vessel.
2. A fluid cannot support itself without a container. Thus the walls of the container exert a pressure on the fluid equal to the pressure of the fluid at that depth.
3. The pressure at a given level is transmitted equally throughout the fluid to be the same value at that level.Explanations
Table 6.2: Explanation of the Hydrostatic ParadoxVessel A: No matter how wide the vessel, the
pressure is just the weight of the fluid above
unit area on the bottom. Even if you take the
whole weight of the fluid in the container mg
and divide by the area of the bottom A, you
still get the same results since the vessel is
equivalent to a column of water.
Vessel B: Vessel B could be divided into three
parts. The fluid in parts 1 and 3 is supported
by upward force of the vessel on the fluid.
Part 2 could be thought of a vertical column
of liquid similar to vessel A.
One could ask; why doesn’t the fluid in parts
1 and 3 (which are much bigger) not squeeze
column 2 where they meet (along the dashed
line) and thereby increase the pressure on the
bottom of the column?
The answer is that the fluid in column 2
exerts pressure and is equal but opposite
pressure outwards on the two other liquids
to support itself. From the point of view of
column 2, the water outside the dashed lines
in sections 1 & 3 could be replaced by solid
vertical walls (along the dashed lines) and
column 2 would still be in equilibrium.Vessel C: Again we could divide the water into
three sections. The middle section is similar
to that of vessel A or B. Since the height of
the fluid in section 1 or 3 is not high enough
to produce the same pressure as the height of
the fluid in 2, how does the pressure on the
bottoms of section 1 and 3 get to be the same
as that of 2?
The answer is that top of the container’s walls
in sections 1 and 3 produce a downward
pressure that is equal to the fluid pressure
in the middle section at the same level. If
you poked a hole in the top of the container
in sections 1 or 3, water would fountain
upwards from the hole under pressure. From
Pascal’s principle, this pressure has to be that
of the fluid in the middle section at the same level.Description of hydrostatic paradox
Vessel D: Again the center column is similar to vessel A.
The pressure of the vessel’s wall creates a
pressure that vertically supports the fluid
in sections 1 and 3. At the same time the
pressure of the walls create a horizontal
component of pressure that sustains the fluid
in the center column.Vessel E: While one cannot offer simple
arguments like those for the other vessels,
the pressure on the bottom is still the same
basically because of Pascal’s principle.
You go down from the surface to some
depth, and then move sideways until you
can go down again. Repeat the process until
you reach the bottom. Since the pressure
at the same depth is the same, moving
sideways does not change the pressure. Only
downwards motion increases the pressure.At a depth of 10 meters under water, pressure is twice the atmospheric
pressure at sea level, and increases by about 105 kPa for each increase of 10m depth.6.4 Unit 6 assessment
1. The maximum gauge pressure in a hydraulic lift is 18 atm. What largest mass vehicle can it lift if the diameter of the output line is 22cm?
2. Pascal placed a long tube with a radius of 0.30cm into a barrel (Fig 6.8) with a 20.0cm radius top. When the water was filled to the 12.0m height the barrel burst.
a) Calculate the mass of water in the tube.
b) Calculate the net force on the lid.
3. What is the total pressure on
a diver 45.0m below the sea?Fig. 6.14: Barrel
4. The master cylinder of a brake system has a radius of 0.100cm
and the cylinders at the brake pads have radii of 4.00cm. If
Debi can apply a force of 150N on the brake pedal, what is the
force applied to slow down her car?
5. Find the pressure needed to push water to the top of the Tower,
a height of 8.0m (use mass units), the density of water is 1.0 g/cm3.
6. Find the total force on the filled school dam whose dimensions
are 5.0m by 2.0m. The average depth is 1.0m.
7. Calculate the pressure and density of air at an elevation of
5000m above mean sea level. The atmospheric pressure and
temperature at sea level are 101.35 KPa and 15 oC respectively.
The temperature lapse rate is 0.007 K/m. The density of air at sea level is 1.2255 kg/m3.8. Rank the pressures at the three points of the figure below:
9. Three containers of different shape, but with bases of equal area (Fig 6.9) are filled with water to the same height.
Fig. 6.15: Containers
a) The weight of the water is the greatest in container…
(i) A
(ii) B
(iii) C
(iv) The weight of the water is the same in all the three containers.
b) The pressure at the bottom of the container is the greatest in container…
(i) A
(ii) B
(iii) C
(iv) The pressure at the bottom is the same in all the three containers10. Calculate the pressure due to water column of height 100m
(Take g = 10 m s-2 and density of water = 103 kg m-3). What
height of mercury column will exert the same pressure? (Density of mercury = 13.6 x 103 kg m-3)
11. What is the pressure due to water pressure 100m below the surface of a lake?
12. Figure 6.16 shows a hydraulic weight bridge which works on the principle of Pascal’s law.
Fig. 6.16: The Hydraulic press
a) What is the pressure at B?
b) What is the pressure at A?
c) What is the weight of the vegetable on the large piston A if the weight bridge is in equilibrium?
13. A regularly shaped object is immersed in water of density 1000 kgm-3 (Fig 6.17)
Fig. 6.17: Pressure exerted on a body in water
a) Calculate the water pressure at the top and the bottom of the object.
b) What is the resultant pressure on the object?My goals
By the end of this unit, I will be able to:
explain atmospheric pressure and state its units.
explain applications of atmospheric pressure.
illustrate Archimedes principle in air.
explain buoyant/up thrust force and Archimedes principle in liquid.
explain the existence of atmospheric pressure.
state the S.I. units of atmospheric pressure.
identify and name the instruments for measuring atmospheric pressure.
mention and explain the applications of atmospheric pressure.
explain the Archimedes principle in fluids: up thrust, factors affecting up thrust, state the principle and formula.
apply the Archimedes principle: floating and sinking.
explain the applications of Archimedes principle in air. (Aerostat, Baroscope)Key concepts
1. How to detect the existence of atmospheric pressure in our environment?
2. What is atmospheric pressure?
3. What do we use to measure atmospheric pressure?
4. Explain the key concept of buoyancy and up-thrust.
5. Discuss Archimedes’ principle in fluids (liquids and gases).
6. Discuss different applications of Archimedes principle in fluids in real life.Vocabulary
Barometer, up-thrust force, floating, sinking, aerostat, atmospheric
pressure, buoyancy.Reading strategy
When reading this unit, emphasise the paragraphs that contain definitions of key terms. Use all the information you have learnt to define each key term in your own words, describe and discuss Archimedes principle and its application in real life and other related calculations.7.1 Atmospheric pressure
7.1.1 Existence of atmospheric pressureThe existence of the atmospheric pressure can be proved by the following experiments.
1. Crushing can experiment.
2. Overturned glass full of water
3. Magdeburg Hemisphere.Activity 7.1: Crushing can experiment
With reference to the Fig. 7.1, and the provided materials, do the following activity and answer the questions.Materials:
⚫ Bunsen burner
⚫ A metal can with a lid
⚫ Water
⚫ Tripod stand
⚫ Wire gauze
Procedure:
⚫ Pour water into the metal can.
⚫ Pour the water till it fills about 1/3 of the volume of the can.
⚫ Heat the water in the can on the lighted Bunsen burner until it boils.
⚫ Remove the can from the burner and close its lid.
⚫ Allow the can to cool and carefully observe.
Questions:
1. What do you think may have caused the changes observed?
2. Discuss and explain your observations.
3. Comment on the situation relating it with atmospheric pressure.
Water in a can is heated The can is closed and is cooled
down rapidly by pouring cold
water on it, it crushes instantly,
due to the high atmospheric
pressure from the sorroundingFig. 7.1: The collapsing or crushing can
When a can filled with hot water is closed and is cooled down it will crush instantly. This experiment proves that atmospheric pressure acts on everything on the surface of the earth.
Activity 7.2: Overturned glass full of water
Materials:
⚫ Glass
⚫ Water
⚫ Stiff paper or cardboard
Procedures:
⚫ Pour water in the glass and make it full.
⚫ Make sure there is no air bubble inside.
⚫ Place the paper on the glass.
⚫ Put your hand on the paper and the other hand holds the glass, at the bottom.
⚫ Very quickly, turn the glass upside down with your hand still on the paper.
⚫ Then remove the hand holding the paper as in Fig. 7.2.
Questions:
1. What have you observed?
2. Discuss and explain why the water has not poured out from the glass.
3. Comment on the observation relating it to the atmospheric pressure.
Fig. 7.2: Inverted glass full of water
The cardboard does not fall and the water remains in the glass even
though it’s not supported by anything. This is because the force due to the atmospheric pressure acting on the surface of the cardboard is greater than the weight of the water in the glass. This experiment proves that atmospheric pressure is present on the surface of the earth.Magdeburg Hemisphere
Fig. 7.3: Magdeburg hemisphere
When the air inside the hemisphere is pumped out so that the hemisphere
becomes a vacuum, the hemisphere cannot be separated even by a very
great force. This is because when the air is pumped out, the pressure
inside the hemisphere becomes very low. The atmospheric pressure
exerts a strong force on the outer surface of the hemisphere, holding the hemisphere tightly together.7.1.2 Atmospheric pressure units
The standard atmosphere (symbol: atm) is a unit of pressure defined
as 101325Pa (1.01325 bar). It is sometimes used as a reference or
standard pressure.
In 1954 the 10th Conférence Générale des Poids et Mesures (CGPM)
adopted standard atmosphere for general use and affirmed its definition
of being precisely equal to 101325Pa. This value is the mean atmospheric
pressure at mean sea level at the latitude of Paris, France.
In chemistry and in various industries, the reference pressure referred
to in “Standard Temperature and Pressure” (STP) was commonly1 atm (101.325kPa) but standards have since diverged; in 1982, the
International Union of Pure and Applied Chemistry (IUPAC) recommended
that for the purposes of specifying the physical properties of substances,
“standard pressure” should be precisely 100kPa (1 bar).Pressure units and equivalencies
Table 7.1: Atmospheric pressure unit conversion
A pressure of 1 atm can also be stated as:
≡ 1.01325 bar
≡ 101325 pascal (Pa) or 101.325 kilopascal (kPa)
≡ 1013.25 millibars (mbar, also mb)
≡ 760 torr[
≈ 760.001 mm-Hg, 0°C, subject to revision as more precise measurements
of mercury’s density become available
≈ 1033.227 452 799 886 cm–H2O, 4°C7.1.3 Instruments for measuring atmospheric pressure
1. Mercury Barometer
A mercury barometer consists of a thick-walled glass tube, which is closed at one end.
Fig. 7.4: The Mercury barometer
The tube is completely filled with mercury and inverted several times to remove air bubbles. The tube is then completely filled again with mercury.
After all the air has been removed, the open end of the glass tube is inverted into a container of mercury. The mercury column drops until it reaches a height about 76cm above the lower surface. The space between the top of the mercury and the end of the tube contains no air; it is a complete vacuum. The column of mercury in the tube is supported by the atmospheric pressure and its height depends on the magnitude of the atmospheric pressure.2. Fortin Barometer
A fortin barometer is a type of mercury barometer which has a higher accuracy. This barometer has a vernier scale which gives a more accurate reading of the atmospheric pressure. The mercury level in
the container can be adjusted by a screw until the pointer touches the surface of the mercury. This eliminates the zero error.
Fig. 7.5: The Fortin Barometer
3. Aneroid Barometer
An aneroid barometer does not use any liquid. It consists of a sealed
metal chamber in the form of a flat cylinder with flexible walls. The chamber is partially evacuated and a spring helps prevent it from collapsing.
Fig. 7.6: The Aneroid barometer
The chamber expands and contracts in response to changes
in atmospheric pressure. The movement of the chamber walls is transmitted by a mechanical
lever system which moves a pointer over a calibrated scale.The Aneroid Barometer can be used as an altimeter (to determine
altitude) by mountaineers or pilots to determine an airplane’s altitude. The scale can be calibrated to give readings of altitude equivalent to a range of values of atmospheric pressure.
An aneroid barometer is also used as a weather glass to forecast the weather. Rain clouds form in large areas of lower pressure air, so a fall in the barometer reading often means that bad weather is coming.7.1.4 Application of atmospheric pressure
Drinking StrawActivity 7.3: Effect of atmospheric pressure
when using a drinking straw
With reference to activity 4.4 in unit 4, discuss and explain how
atmospheric pressure is applied when drinking using a straw.
When drinking with a straw, one has to suck the straw. This causes the
pressure in the straw to decrease. The external atmospheric pressure,
which is greater, will then act on the surface of the water in the glass,
causing it to rise through the straw.Rubber Sucker
Activity 7.4: Investigating the atmospheric
pressure with a rubber sucker
Materials:
⚫ Rubber sucker
⚫ Window glass or a telephone
Procedure:
⚫ Lay the rubber sucker on the window glass and push slightly.
⚫ Lay the rubber sucker on the screen or the back part of the
telephone as shown in Fig 7.7 and try to pull it back.
Question:
Discuss and explain why the rubber sucker is sticking on the window
glass or on the telephone’s screen.
Fig. 7.7: The Rubber sucker used in holding phones
When the rubber sucker is put onto a smooth surface, usually a glass
or tiled surface, the air in the rubber sucker is forced out. This causes
the space between the surface and the sucker to have low pressure. The
contact between the rubber sucker and the smooth surface is airtight. The external atmospheric pressure, which is much higher, acts on the rubber sucker, pressing it securely against the wall.Activity 7.5: Investigating atmospheric pressure when using siphon
With reference to activity 4.5 in Unit 4, discuss and explain the role of
atmospheric pressure in pushing out the liquid from the can as shown in Fig. 7.9.A rubber tube can be used to siphon liquid from a container at a higher
level to another at a lower level. For example, we can remove petrol from the petrol tank of a vehicle or dirty water from an aquarium. The tube is first filled with the liquid and one end is placed in the liquid in the container A. The other end is placed at a level which must be lower than the surface of the liquid in container A.
Fig. 7.8: Siphoning water from a jerrycanThe pressure in the rubber at the lower end is equal to atmospheric
pressure plus the pressure due to h cm column of liquid. As the pressure
at the lower end is greater than the atmospheric pressure, the liquid flows out.
Fig. 7.9: A Vacuum cleaner A Lab lift pumpActivity 7.6: Try this in groups. Application of pressure of liquid in using a lift pump
Materials:
⚫ Lab lift pump
⚫ Bucket
⚫ Water
Procedure:
⚫ Take water in the bucket.
⚫ Try to fetch water from the bucket using the lab lift pump in Fig. 7.10
Questions:
1. Discuss and explain the principle function of the lift pump
used above.
2. Where can this be applied to help the society?
Fig. 7.10: A Lift pump sucking water from a wellA lift pump is used to pump water out of a well or to a higher level.
The greatest height to which the water can be pumped is 10m. This is equivalent to the atmospheric pressure. When the plunger is lifted, the upper valve closes and the lower valve opens. The atmospheric pressure, acting on the surface of the water, causes water to flow past valve B into
the cylinder. When the plunger is pushed down, the lower valve closes and the upper valve opens. Water flows above the plunger. When the plunger is next lifted, the upper valve closes again and the lower valve opens once more. The atmospheric pressure, acting on the surface of the water, forces
water past the lower valve into the cylinder. Simultaneously, the water above the plunger is lifted and flows out through the spout. This process is repeated until sufficient water is obtained.Quick exercise
1. Explain how rubber suckers could be used to help lift panes
of glass safely.
2. If one of the windows in a plane flying at high altitude breaks,
what do you think will happen?
3. Water has a lower density than mercury. Is the column of
liquid in a mercury barometer taller or shorter than one in a
barometer using water?
4. If the density of water is 1000 kg/m3 and atmospheric
pressure is 100,000 Pa; how high will the column of water
in a perfect water barometer be? (Take g for the Earth as 10 N/kg).
5. What would happen to the mercury level in a mercury
barometer if:
(a) The atmospheric pressure went down
(b) The atmospheric pressure went up
(c) A little air leaked into the tube above the mercury
(d) The barometer was taken up a high mountain
(e) The temperature of the room where the barometer was, got higher.[Remember that mercury is a dangerous substance. It should
not be used by learners and you should certainly not heat it].
6. Find out what is meant by:
(a) a millibar
(b) an isobar
(c) an anticyclone7. (a) What is the atmospheric pressure shown by the mercury
barometer in diagram 1 (Fig. 7.11)?
(b) Some air now leaks into the barometer mentioned in
question 10(a). The result is shown in diagram 2. What is
the pressure of the air in the top of the barometer tube?
Fig. 7.11: The Mercury barometer, diagram 1 and 27.2 The Principle of buoyancy and factors affecting upthrustActivity 7.7: Investigating the upthrust (buoyancy) of waterWith reference to the Table 7.2, do this experiment and answer the questions.
Materials:
⚫ A stone of less than 1 kg.
⚫ Water
⚫ Eureka can
⚫ Sewing thread
⚫ Dynamometer
⚫ Measuring cylinder
Procedure:
1. Pour water in the Eureka can and make it full.
2. Tie the thread on the stone3. Measure its weight in air using the dynamometer and record
it to be w1
4. Submerge the stone in water still on the dynamometer and
record the new weight w2
5. Measure the weight of water overflown in the measuring
cylinder and record it as w3
6. Find the difference w' = w1 – w2
Questions:
1. Compare the results obtained from step 5 and 6.
2. What is the volume of the stone?
3. Discuss and explain your findings in question 1.If you walk in water you feel water pushing your legs up. If you lift a bucket
of water from a tank, the bucket appears to be lighter inside the water and
suddenly heavy when it comes out of water. A weather forecasting balloon
floats in air with its heavy equipment! A ship made of steel floats while
a steel pin sinks in water! These experiences are based on Archimedes’ Principle.
An object weighs less in water than it does in the air. This loss of weight is
due to the upthrust of the water acting upon it and is equal to the weight of the liquid displaced.
Because salt water is denser than pure water the object displaces a greater
weight of salt water and, therefore, weighs less.
The denser the liquid, the easier it is to float in it making it easier to swim
in the ocean or a chemical filled pool than a mountain streamTable 7.2: Archimedes principle of buoyancy
Buoyancy reduces the apparent weight of objects that have sunk completely
to the sea floor. It is generally easier to lift an object up through the water
than it is to pull it out of the water. In the case of a submerged body, the
apparent weight of the body is equal to its weight in air less the weight of
an equal volume of fluid and the object that floats will displace a volume
of water equal to its weight:
According to Archimedes’ principle,
“Any object that is completely or partially submerged in a fluid at rest is
acted on by an upward (or buoyant) force. The magnitude of this force is
equal to the weight of the fluid displaced by the object and the volume
of fluid displaced is equal to the volume of the portion of the object submerged.”
Fig. 7.12: Force acting on a submerged bodyThus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy. The horizontal forces on its vertical sides cancel and are removed from consideration.
The upward forces against the bottom surface of the object are greater than the downward forces against its top surface. This net force is called the buoyant force. In other words the “buoyant force” on a submerged body is directed in the opposite direction to gravity and is equal in magnitude.
The net force on the object is thus the sum of the buoyant force and the
object’s weight:
⚫ If the buoyancy of an object exceeds its weight, it tends to rise.
F = B – W
⚫ An object whose weight exceeds its buoyancy tends to sink.
F = W – B⚫ Commonly, the object in question is floating in equilibrium and the
sum of the forces on the object is zero, therefore; F = W – B = 0
Fig. 7.13: Forces on a suspended body. “Loss of weight” in a liquid
You probably noticed that all of the methods employing Archimedes’
principle with a balance scale required the subtraction of two scale
readings. Balances compare forces (weights), but their scales read mass
units (grams). From Fig 7.13 we find that:
7.2.2 Application
7.2.2.1 Archimedes’ principle and densityActivity 7.8: Investigating the floating condition
Materials:
Observe Fig.7.14 and then try to answer the questions below:
1. Compare the three cases and describe each case.
2. Comment on the situation relating with the pressure in liquids.We use density in the determination of whether a substance will float or
sink in another substance usually a liquid or gas. If the buoyancy of an
object exceeds its weight, it tends to rise. An object whose weight exceeds
its buoyancy tends to sink. Commonly, the object in question is floating in
equilibrium and the sum of the forces on the object is zero. When a nonporous
object is placed in a fluid, there are three possible outcomes:
Fig. 7.14: The Density and Floating condition
Case 1: Neutral buoyancy
A material or object having the same density as a liquid will remain
submerged in the liquid, neither floating to the top nor sinking to the bottom.
Case 2: Sinker object
The object sinks when the density of the object is greater than the density
of the displaced fluid. For an object that sinks, the volume of displaced
water equals the volume of the object i.e. if ρb > ρf then
7.2.3 Relative density of liquidsUsing the same procedure as in the previous experiment, a sinker is
weighted first in air, then in liquid, and finally in water. The sinker is any
convenient solid body. Since the same sinker is used in both liquids, the
two apparent losses in weight will be the weight of equal volumes of liquid and water respectively.
Therefore, the density of the sinker:
Case 3: Objects that float
Fig. 7.15: Block of wood immersed in fluid by sinker M1
Density of the solid is less than the density of the liquid and hence the
object will float or, in the case of a balloon, it will rise i.e. If ρb < ρf then
F = W – B where W is the weight of the body and B is the upthrust.
F = ρb gV – ρf gV'
This equation tells us that the fraction of the volume of a floating object
that is below the fluid surface is equal to the ratio of the density of the object to that of the fluid.
Under normal conditions, the weight of a fish is slightly greater than
the buoyant force due to water. It follows that the fish would sink if it
did not have some mechanism for adjusting the buoyant force. The fish
accomplishes this by internally regulating the size of its air-filled swim
bladder to increase its volume and the magnitude of the buoyant force
acting on it. In this manner, fish are able to swim to various depths.
For a floating object the mass of displaced water equals the mass of the object. Law of flotation⚫ A body will rise if V = V’
If the body whose density is to be measured is less dense than the liquid,
it is necessary to fasten the body to a sinker so that the two together
sink in the liquid. Let M1 be the mass in air of the body, say a block of
wood, whose density ρ1 is less than that of the liquid whose density is ρf.
Suppose the body of mass M1, density ρ1, referred to above, is such that
when fastened to the sinker the two together sink in the liquid. Let the
weight of the two together when completely immersed in the liquid be
M'L g. Then from Fig. 7.15 it is readily seen that:
The upthrust in a fluid is given by:
A liquid will float on another liquid, and a solid will float in a liquid, if it is less dense than the liquid.
Example: A ship of mass 1200t floats in sea-water. What volume of sea-water does it displace? If the ship enters fresh water, what mass of cargo must be unloaded so that the same volume of water is displaced as before? (Density of fresh water = 1000kg/m3, relative density of sea-water = 1.03 and 1 t = 1000kg).
Answer:
The ship displaces a weight of sea-water equal to its own weight and therefore a mass of sea-water is equal to its own mass. Mass of sea-water displaced: md = 1200t = 1,200,000kg
7.2.2.2. Submarines
Fig. 7.16: A SubmarineA submarine (the word submarine was originally an adjective meaning
“under the sea”) is made to float or sink by altering the average density.
By average density is meant the value obtained by dividing the total mass
of the submarine (including air in it, the crew, and so on) by its volume.
When submerged, the submarine has an average density equal to that of
the water around it. In order to bring it to the surface, its mass must be
made less and this is done by expelling water from the tanks (this is called
ballast) situated along the sides of the submarine, replacing the water by compressed air.The boat is provided with large ballast tanks which can be filled with
water. This increases the weight of the submarine, so that it sinks lower into the sea.When the submarine is ready to surface, the rudders are moved to drive
the boat upwards, and compressed air is forced into the ballast tanks to drive the water out so that the submarine can rise.Most of the marine animals also use this principle to remain at a selected level in the sea. For example, fish has an air sac, called a swim bladder in its body. This is filled with air and usually occupies about 5% of its total body volume. Its size is adjusted so that the fish is posed at the depth at which they usually live and feed. At that level, the condition of its weight is exactly balanced by the upthrust it experiences.7.2.2.3 Ships
In the case of a ship, its weight is balanced by a buoyant force from the
displaced water, allowing it to float. If more cargo is loaded onto the ship,
it would sink more into the water - displacing more water and thus receive
a higher buoyant force to balance the increased weight.
Fig. 7.17: A Boat on Lake MuhaziSo, how can objects made of aluminium or iron float? The secret lies in
increasing the volume of the displaced water. Although a small cube of
iron will immediately sink when placed in water, a large boat can float by
adjusting the amount of water it displaces. A sheet of aluminium foil can float when formed into a “barge” with a large surface area whereas the same size sheet will immediately sink when crushed. In all cases, you are reducing the density by increasing the volume.Did you know?
⚫ Boats float higher in salt water than they do in fresh water.
⚫ Some liquids sink in other liquids.7.2.2.4 Densimeter (Hydrometer bottom)
Fig. 7.18: Force acting on a submerged bodyAn hydrometer (Fig. 7.18), is a scientific instrument used to measure
the weight and specific gravity of a gas or liquid in which it floats. It is a
hollow tube, widened at the bottom where a weight is placed (B). A scale is present on the upper part of the rod. The hydrometer is placed in the liquid needing to be tested. The scale (A) will be held upright by the weight in the lower part (B). The specific gravity of the liquid is read where the scale penetrates the surface of the liquid.Lactometer
It is a special type of hydrometer used for testing the purity of milk or to
check the richness of milk. It has a range of relative density 1.105 to 1.045.
Battery Hydrometer
It is used for measuring the relative density of accumulator acid. It is kept
inside a glass tube fitted with a rubber bulb at the upper end. The lower
end of the glass tube is connected with a narrow tube which is made of acid resistant material. When in use, this end is submerged in the acid in the accumulator. The acid in a fully charged cell should have a
relative density of 1.25 to 1.30. A reading of less than 1.18 indicates that recharging is necessary.Project 7.9: Making your own hydrometer
Snip off a section of one of those fat drinking straws to about 6cm or so and put a blob of plasticine on one end making sure that the end is fully sealed.
Some interesting experiments with your hydrometer
One thing that the designer of a ship must consider is ‘how high will it float’.
Try your hydrometer in different types of liquid, as below, and note
which mark the water comes up to.
⚫ Filtered water (cold and warm),
⚫ Tap water (cold and warm),
⚫ Rain water (cold and warm),
⚫ The swimming pool,
⚫ Tap water with salt dissolved in it (cold and warm).
Fig. 7.19: My HydrometerYou should be able to tell which of your waters (filtered, tap and rainwater) contain the most minerals. Try your hydrometer on a variety of other fluids including petrol, oil etc.Airship aerostat and hot air balloon
Activity 7.10: Investigating air pressure effect on a balloon
Materials:
⚫ Balloon(s)
⚫ Pump
Procedure:
⚫ Pump air in the balloon
⚫ Let it fly in air and write down the observation made.
Questions:
1. What is pushing up the balloon?
2. Why is the balloon not falling down easily like other bodies
do (like stones)?
3. Comment on common observations you have made in real life where this case may be observed.Rigid Airships
Semi-rigid airships were more popular earlier this century. They usually comprise a rigid lower keel construction and a pressurised envelope above that. The rigid keel can be attached directly to the envelope or hung underneath it. The airships of Brazilian aeronaut Alberto Santos-Dumont
were of this type. One of the most famous airships of this type was Italia, used by General Umberto Nobile in his attempt to reach the North Pole.
Fig. 7.20: Semi–rigid airshipNon-rigid Airships
Non-rigid airships, also known as Blimps, are the most common type
nowadays. They are large gas balloons whose shape is maintained only by
their internal overpressure. The only solid parts are the passenger car and
the tail fins. All the airships currently flying for advertisement purposes are
of this type; the Goodyear Blimps, the Budweiser and the Metlife Blimps
in the USA, and the Fuji Blimp in Europe.
Fig. 7.21: Non rigid airship
Fig. 7.22: Hot Air Airships
Hot air airships, also known as thermal airships, are counted as a fourth
kind although they are technically part of the non-rigid category. Hot
air airships are derived from traditional hot air balloons. Early models
were almost like balloons with an engine and tail fins added. Later, the
envelopes were lengthened and the tail fins and rudder were pressurised
by air from the wash of the propeller. Newer hot air airships maintain their
shape with internal overpressure in the whole envelope, a feature which
older models did not have.7.2.1 Principle of Archimedes
It states that: “when a body is totally or partially immersed in a fluid it
experiences an upthrust equal to the weight of the fluid displaced.”
Fig. 7.24: The stone weighed 0.67N in air and 0.40N when
immersed in water. The displaced water weighed 0.27N (= 0.67N-0.40N).Experimental Verification of Archimedes principle
⚫ Place a Eureka can (over-flow vessel) on a table and place a beaker under its spout as shown in figure below.
⚫ Pour water into the can till the water starts overflowing through the spout.
⚫ When the water stops dripping replace the beaker by another one of known weight.
⚫ Suspend a stone with the help of a string from the hook of a spring balance and record the weight of the stone.
⚫ Now, gradually lower the body into the Eureka can containing water and record its new weight in water when it is fully immersed in water.
⚫ When no more water drips from the spout, weigh the beaker containing water.
⚫ Write down the results of the experiment as follows: Weight of the stone in air Wa = 0.67N
Weight of the stone in water Wf = 0.40NWeight of the empty beaker = aN
Weight of the beaker + water displaced = bN
Apparent loss of weight of the stone = Wa–Wf = 067N – 0.40N = 0.27N
Weight of water displaced = (b - a) N
You will notice that Wa – Wf = b – a
Thus, the apparent loss of weight of the body, or the upthrust on the
body equals the weight of the water displaced.
B = ρfgV = weight of displaced fluid
B = weight in air – weight in fluid = 0.67N – 0.40N = 0.27NQuick exercises
1. A body weighs 450g in air and 310g when completely
immersed in water. Find:
a) the loss in weight of the body
b) the upthrust on the body
c) the volume of the body
d) the relative density of the solid.
2. A piece of aluminium of volume 200cm3 and density 2.7g
cm-3 is completely immersed in kerosene.
a) Determine upthrust exerted on the piece of aluminium.
b) Determine how much it will weigh in kerosene (density of kerosene = 0.8g cm-3).
3. Equal volumes of lead and aluminum are submerged in
water. Which feels the greatest buoyant force? Explain.
4. When placed in a pycnometer, 20g of salt displaces 7.6g of coal oil. If the density of coal oil is 0.83g/cm3, find the volume and density of the salt.Principle of Archimedes and density
Calculation of the Relative Density of a Solid
When a body is immersed in water, it displaces its own volume of water.
Upthrust on the body is equal to the weight of this displaced volume of
water, which is also equal to the loss of weight of the body. Hence ‘weight
of equal volume of water’ can be replaced by upthrust or loss of weight inwater. Find the weight (W1) of a solid in air using a hydrostatic balance as
shown in the figure below.
Fig. 7.25: Measuring relative density of a substance⚫ Tie the solid with a thread and suspend it from the hook as shown in the figure.
⚫ Lower the solid in water as shown and find its weight (W2).
⚫ Weight of the solid in air: W1
⚫ Weight of the solid in water = W2
⚫ Apparent loss of weight of solid = (W1 - W2)
Quick exercises
1. A body weighs 600g in air and 400g in water. Calculate:
a) Upthrust on the body,
b) Volume of the body, and
c) Relative density of the solid.
2. A solid weighs 50g in air and 44g when completely immersed
in water. Calculate:
a) R.D of the solid
b) Upthrust
c) Sensity of the solid in cgs and SI units.Calculation of the Relative Density of a Liquid
When a solid body is immersed in a liquid and then in water, the volume
of displaced liquid is the same as the volume of displaced water which is
equal to the volume of the solid.⚫ Select a solid (sinker) which is insoluble in the given liquid.
⚫ Weigh the sinker in air.
⚫ Weigh the sinker in water and finally weigh the sinker in the given liquid.
Fig. 7.26: Experimental calculation of relative density of liquids
Quick exercises
With reference to the above information provided about Calculation
of Relative Density of a Liquid.
1. A solid weighs 600g in air, 450g in water and 480g in a
liquid. Find the:
a) volume of the solid,
b) R.D. of the solid, and
c) R.D. of the liquid.
A body weighs 20g in air, 18.2g in a liquid and 18.0g in water.
Calculate:
a) the relative density of the body,
b) and relative density of the liquid.
2. A solid weighs 32g in air and 28.8g in water. Find how much
will it weigh in a liquid of R.D 0.9.7.2.2 Relative Density of Solids which Float in Water
The Relative density of solids like wax, cork etc. is determined by the following method.
⚫ Choose a sinker and find its weight in water by suspending it in water as shown in the figure 7.27(a) below.
⚫ Tie the solid cork to the string attached to the sinker and find its weight in air. Put the sinker in the water as shown in figure 7.27(b) below.
⚫ Remove the cork and tie it together with the sinker and suspend it in water as shown in Figure 7.27(c) and find the weight of the cork together with the sinker in water.
Fig. 7.27: Relative density of a floating bodyQuick exercises
1. A wooden block in the form of a cube of side 10cm is floating
in water with 4cm above the surface of water. If the density
of water is 1g cm-3, find the density of wood.
2. Measuring the relative density of a cork;
⚪ Weight of the sinker in water x = 12.6g
⚪ Weight of the sinker in water + cork in air y = 13.7g
⚪ Weight of the sinker and the cork, both in water z = 10.5g
⚪ Find the R.D and density of cork.3. A hollow cylinder closed at one end weighs 85g which floats
vertically in water when 35g of lead shots are added into it.
If the depth of immersion is 10cm, calculate the:
a) upthrust acting on the cylinder,
b) area of cross section of the cylinder,
c) depth of immersion in a liquid of R.D. equal to 12.
4. A 70kg ancient statue lies at the bottom of the sea. Its volume
is 3.0 × 104 cm3. How much force is needed to lift it?Law of Floatation
When a floating body like wood is placed in water, it sinks until the weight
of water displaced by it is just equal to its own weight and then it floats.
This leads us to the principle of floatation: “When a body floats in a fluid,
it displaces an amount of fluid equal to its own weight. The apparent
weight of a floating body is zero.”Quick exercises
Law of Floatation.
1. The mass of a block made of certain material is 13.5kg and
its volume is 15 x 10-3m3. Will the block float or sink in
water? Give a reason for your answer.
2. A slab of ice of volume 800cm3 and of density 0.9g cm-3
floats in water of density 1.1g/cm3. What fraction of ice is above salt water?
3. A block wood of mass 24kg floats in water. The volume of wood is 0.032m3. Find:
a) The volume of the block below the surface of water.
b) The density of wood (Take density of water 1000kg m-3)
4. What volume. V of helium is needed if a balloon is to lift a
load of 800kg (including the weight of the empty balloon)?
5. A weather forecasting a plastic balloon of volume 15m3
contains Hydrogen of density 0.09kg m-3. The volume of the equipment carried by the balloon is negligible compared to its own volume. The mass of the empty balloon alone is 7.15kg.The balloon is floating in air of density 1.3kg m-3. Calculate:
a) The mass of hydrogen in the balloon.
b) The mass of hydrogen and the balloon.
c) The total mass of the hydrogen, the balloon and the
equipment if the mass of the equipment is ‘x’ kg.
d) The mass of air displaced by the balloon.
e) The mass of the equipment using the law of floatation.
6. Hydrometer is a simple instrument used to indicate specific
gravity of a liquid by measuring how deeply it sinks in the
liquid. A particle hydrometer consists of a glass tube weighted
at the bottom, which is 25cm long, 2.00cm2 in a crosssectional
area, and has a mass of 45.0g. How far from the end should the 1.00 mark be placed?7.3 Unit 7 Assessment
1. Discuss other applications of Archimedes principle that are in
use today.
2. Discuss how these methods might be useful in finding mass and
volume rather than by direct measurement.
a) Using Archimedes Principle to Measure Mass of an object
immersed in a fluid of known density (If an object is immersed
completely it will displace its volume.)
b) Archimedes principle allows us to calculate the mass of floating
objects (If an object is floating, the mass of the displaced water
is equal to the mass of the block).
3. A cubical block made of a certain type of plastic has a density
of 0.75g/cm3. The density of water is 1.0g/cm3. If the block is
allowed to float in water, what fraction of the volume of the block
would be below the water level?a) one quarter
b) one half
c) three quarters
d) some other fraction
4. The density of aluminum is about 2.7 times greater than that of
water, so a block of aluminum will sink when placed in water.How is it possible to build a boat in the figure below using only aluminum?
Fig. 7.23: A Sailing boat in lake Kivu5. A glass beaker is filled with water and placed on a balance.A person holds a finger into the water. The reading on the balance will;
a) go up
b) go down
c) stay the same
d) Can’t tell
6. Some common application of pressure in liquids and Archimedes Princilpe.Key unit competence
The learner should be able to relate work, power and energy.
My goals
By the end of this unit, I will be able to:
apply the knowledge on energy, work and power.
explain the terms work, power and energy.
explain the relationship between work, power and energy.
derive the equations relating work and power.
understand the importance of energy and power for efficiency working of machines.
show concern of work as a product of distance and energy.
be aware of the social, economical, environmental and technological implications of studying work energy and power.
develop an analytical mind to critically evaluate work, energy and power.Key concepts
1. How are work, energy and power realised or manifested in our daily life?
2. How can one relate work, energy and power?
3. Discuss the difference between potential energy and Kinetic energy.4. Discuss different areas where Potential energy and Kinetic energy
can be observed.
5. How can you compare the people’s power?Vocabulary
Work, energy, power, Potential energy, Kinetic energy, Mechanical energy,
Energy conservation.Reading strategy
Draw a diagram which can help you to define work and energy in your
own words. After you read each section, compare your definition to the
scientific definition and explain where work and energy may be observed
useful in life. Identify several activities you have learned that are relevant
to your life, explain why they are relevant to you. (8.1). Relating Work,
Energy and Power8.1 Relating Work, Energy and Power
Activity 8.1: Investigating the work done when lifting a box
Materials:
⚫ Box of 2kg mass.
⚫ A table of at least 1.5m high.
Procedure:
⚫ Lift the box and take it on the table.
⚫ Take the box down.
Questions:
1. Compare the effort that you apply to lift up the box and that when you take the box down.
2. In which case do you require more energy?Activity 8.2: Investigating potential energy and Kinetic energy on a swinging pendulum bob
Let us consider the case study of the Fig. 8.1, in pairs and try to answer the question provided.
As the pendulum bob swings to and fro, its height above the tabletop is constantly changing. As the height decreases, potential energy is lost; and simultaneously the kinetic energy is gained. Yet at all times, the sum of the potential and kinetic energies of the bob remains constant. The total mechanical energy is 6J. There is no loss or gain of mechanical energy, only a transformation from kinetic energy to
potential energy (and vice versa). This is depicted in the diagram below.
Fig. 8.1: Energy in a pendulum bob
Question: As the 2.0kg pendulum bob in the above diagram swings
to and fro, its height and speed change. Use energy equations and the
above data to determine the blanks in the above diagram.Energy is difficult to visualise. You cannot pick it up or touch it because it
has no mass; neither does it occupy space. Instead of defining energy in
terms of what it is, energy is defined in terms of what it does or can do.
It is, therefore, important to know what work is if we are to understand what energy is.
Anyone raising a weight does a certain amount of work which is measured
by the product of the weight and the vertical distance through which it is lifted.In this example work has been done against the force of gravity. Work
done, force used and distance moved by an object in the direction of the
force are related as follows:
Work is defined as using a force to move an object through a distance.
W = F . dThe work done by the force is defined to be the product of a component
of the force in the direction of the displacement and the magnitude of this
displacement.
The SI unit of work is the Joule, which is the work done when
a force of 1N acts trough a distance of 1m. Thus: iJ = 1 Nm.
(In honour of British physicist James Prescott Joule: 1818–1889)
Example: Starting from rest, you push your 1000kg car over a 5m distance, on a horizontal ground, when applying horizontal 400N force. What is the work done on the car?Answer:
The work done on the car, W = F × d = 400 N × 5 m = 2000jDifferent forms of Work
⚫ Positive work when the direction of motion and that of the force are
the same. For example when a person is pushing a car, he does a positive work.
⚫ When the direction of motion is opposite to the direction of the force, the work is negative. Examples; when a stone is thrown up vertically, the work of the force of gravity is negative.
⚫ The work is zero when the displacement is zero despite of the action of the force.Example:
When a person tries to move a lorry and remains at rest, that person has
done zero work.
Since energy is the capacity to do work or transfer of heat energy, it has
the same units as work and heat [Joule].
Whenever work is done; energy is transferred or converted from one
form to another. Work is performed not only in motion and displacement
(mechanical work); it is done also by fire flame and electricity in electric
lamps for instance.8.2 Power
Activity 8.3: Investigating power of a cyclist
Fig. 8.2: Tour du Rwanda: Areruya wins ‘Nyungwe Challenge’
Take the case of Tour du Rwanda, on 17th October 2016 in the step of
Rusizi- Huye of a distance equal to 140.7km, in which the winner was
the Rwandan cyclist Areruya Joseph who came first in Huye Town.Questions:
Answer true or false and then, explain your argument.
1. Did Joseph A. have more force than others?
2. Was Joseph’s average kinetic energy more than that of the other?
3. Which physical quantities were very big for Joseph than others?Often it is interesting to know not only the work done on an object, but also
the rate (change per unit time) at which this work is done. For example,
imagine two cars of the same mass but different engines. Both the cars
climb roadway up a hill. But one car takes less time where as another one
takes more time to reach the top. So it is very interesting to know not only
the work done by the vehicles but also the rate at which it is done.
When we speak of power in Physics we refer to the rate of which work is
done or the rate at which energy is used.The rate of doing work is called power or the rate at which work is done or
energy is transferred is called power.
The SI unit of power is the Watt (or J/s).in honor of James Watt (1736- 1819).
Thus 1W = 1l/s
An imperial unit called the horsepower (hp) is sometimes used in
commercial language: 1hp = 736W = 0.736kW
The units of power can be used to define new units of work and energy.
The kilowatt-hour (kWh) is the usual commercial unit of electrical energy.
One kilowatt-hour is the total work done in 1 hour (3600s) when the
power is 1 kilowatt (103 J/s),
So 1kWh = (103 J/s) (3600s) = 3.6MJ
The kilowatt-hour is a unit of work or energy, not power. Our electricity bills
carry the energy consumption in units of kWh.Activity 8.4: Try these exercises and then discuss your results
8.3 Categories of energy in our environment
There are several forms of energy in our environment such as heat energy,
light energy, electric energy, nuclear energy, sound energy, chemical energy
stored in petrol, food and other materials, mechanical energy in moving
matter such as water, wind, falling rocks, etc.
Scientists classify forms of energy into two major categories: Potential energy and Kinetic energy.8.3.1 Potential energy
Activity 8.5: Investigating potential energy in an arc
Carefully study the following photo in Fig. 8.3; in pairs and try to make
an arc. Answer the following questions and explain your answers.
Questions:
⚫ Where does the energy stored in the arc come from?
⚫ Which type of energy is stored there?
⚫ Where is such energy useful?
Fig. 8.3: Potential energy stored in the arrow by the arc
Potential energy may be defined as the energy possessed by objects or bodies due to their
position or state of strain or the position of their parts. Potential energy is energy deriving from
position. Potential energy is referred to as stored energy because it can be looked at as energy
which will be used when time comes for it to be used.
Potential energy is the stored energy in an object due to its position with respect to some reference
(Normally ground).Potential Energy Formula is given by
P. E = m × g × hWhere m is the mass of the body, h is the height attained due to the
body’s displacement and g is the acceleration due to gravity which is constant on earth.
Potential energy formula helps to calculate the mass, height or potential
energy if any of the two quantities are given. It is expressed in Joules.Examples 1.
⚫ A stretched rubber band has elastic potential energy.
⚫ Petrol, coal or food has energy in their chemical bonds, which is called chemical potential energy.This energy is released when the bonds are broken. Chemical potential
energy in petrol is converted to thermal energy when it is burnt in the
engine and this used to move vehicles. The energy which we use to carry
out the daily activities from the food we eat is stored (as chemical energy)
in the molecules of food such as carbohydrates, proteins and fats. During
respiration, some of these molecules are broken down in the cells of the body.Example 2: A ball of mass 2kg is kept on the hill of height 3km. Calculate
the potential energy possessed by the ball.Answer:
Mass of the body (m) = 2kg, Height= 3km,
Potential Energy possessed by the body = m ×g × h
Where g = 9.8 m/s2
∴ Potential Energy = (2kg)× (9.8m/s2)× (3 × 1000m) = 58800J.Activity 8.6: Quick exercise
A girl is carrying a bucket of water of mass 5kg. If she does 500J of work, to what height will she raise it?8.3.2 Kinetic energy
Activity 8.7: Investigating Kinetic energy of a moving body
Study the Fig. 8.4 in order to answer the following questions.
Questions:
1. If the car and the motorcycle have the same velocity; which
one has more energy? Why?
2. Which type of energy do they possess?
3. Discuss other cases where this form of energy is involved.Kinetic energy is the form of energy possessed by moving bodies like in
Fig. 8.4 where the lorry and the motorcycle are moving on a horizontal road. Such bodies have the ability to do work.
Fig. 8.4: The Kinetic energy of a moving body depends on its mass and velocity
Examples:
⚫ A flying bullet can kill an animal.
⚫ Wind (a moving mass of air,
⚫ Flowing streams,
⚫ Falling rocks,
⚫ Electricity (flowing electrons),
⚫ Moving cars,
⚫ Lorries,
⚫ Buses, etc,
All have kinetic energy. Kinetic energy of a body is dependent upon both
the body’s mass and speed.
In mechanics, for a point particle, it is mathematically defined as the amount of work done to accelerate the particle from zero velocity to the given velocity;
In physics, mechanical work is the amount of energy transferred by a force acting through a distance. If a force F is applied to a particle that
achieves a displacement S, the work done by the force is defined as the
product of force and displacement:
W = F.SActivity 8.8: Investigating energy conservation
Take the case in Fig. 8.5 which shows how energy is being stored,
conserved and converted. Discuss and explain your argument from
position A to C.
Questions:
1. What would be the energy that the car possesses at position A?
2. What is the energy the car possesses at position B?
3. What is the energy that the car has at position C?
4. Discuss the conversions of energy that took place in the cases:
a) From A to B
b) From B to C
c) Explain the energy conservation in this car
Fig. 8.5: Conservation of energy
If the mass of the particle is constant, and Wtotal is the total work done on
the particle obtained by summing the work done by each applied force,
from Newton’s second law:
W Total = Ek where, Ek is called the kinetic energy.
Like energy, it is a scalar quantity, with SI units of joules
⚫ If the force and the displacement are parallel and in the same
direction, the mechanical work is positive. W = Fd
⚫ If the force and the displacement are parallel but in opposite
directions (i.e. antiparallel), the mechanical work is negative.⚫ However, if the force and the displacement act perpendicularly to
each other, zero work is done by the force: W = 0
According to the work-energy theorem if an external force acts upon a
rigid object, causing its kinetic energy to change from Ek1 to Ek2, then the
mechanical work (W) is given by:
Where m is the mass of the object and v is the object’s velocity and Δ is
the change in Kinetic energy.
It can be stated in words:The net work done on an object is equal to the change in its kinetic energy.
Example 1 : A 145g baseball is thrown with a speed of 25m/s.
(a) What is its kinetic energy?
(b) How much work was done on the ball to make it reach this speed, if it started from rest?Answer:
Activity 8.9: Quick exercise
1. How much work is required to accelerate a 1000kg car from
20 m/s to 30 m/s?
2. The Moon revolves around the Earth in a circular orbit, kept
there by the gravitational force exerted by the earth. Does gravity do:
a) positive work,
b) negative work, or
c) no work at all on the Moon?8.4 Relation between work, energy and power
Thus the power associated with force F is given by P = F.v where v
is the velocity of the object on which the force acts.Special cases:
8.5 Measure personal power
Activity 8.10: Finding your power
Fig. 8.6: Determining personal power
A flight of stairs, preferably straight, is needed for this experiment. The
time taken to ascend a known height is measured, and calculation
leads to an estimate of human personal power.
1. Measure your own mass using a bathroom scale or any other
suitable scale and calculate your weight.
2. Have a friend use a stopwatch to measure the time you take
to run up a flight of stairs.
3. Count the number of steps, measure the height of each, and
calculate the total height climbed.
4. Calculate the work done in climbing the stairs (work = force × distance).
5. Finally, calculate the work done per second (i.e., the power
at which you were working when climbing the stairs).For example:
Anna and Tom decide they are going to work out; how much energy is
needed to get upstairs. They measure the height of one step and find it is
20cm or 0.2m high. There are 14 stairs altogether so the total height is:
14 × 0.2 = 2.8m
Ann weighs 500N so the total energy needed for her to climb the stairs is:
Energy = 2.8 × 500 = 1400J
So the energy needed is 1400J whether Ann runs or walks up the stairs.
Tom times Ann walking up the stairs. It takes her 10 seconds, so Ann’s
power is:
Power = 1400 / 10 = 140Watts
Ann returns downstairs, then Tom times her running upstairs. This time
she gets there in 3 seconds, so her power is:
Power = 1400 / 3 = 467Watts8.6 Unit 8 assessment
1. A 7.00kg bowling ball moves at 3.00m/s. How much kinetic
energy does the bowling ball have? How fast must a 2.45g tabletennis ball move in order to have the same energy as the bowling ball?
2. A 193kg curtain needs to be raised 7.5m in as close to 5.0s as possible. Three motors are available. The power ratings for thethree motors are listed as 1.0kW, 3.5kW, 5.5kW. Which motor is best for the job?
3. Starting from rest, you push your 1000kg car over a 5m distance,
on an horizontal ground, applying a horizontal 400N force.
a) What is the car kinetic energy change?
b) What is its final velocity at the end of the 5 meters displacement? Disregard any friction force.4. Define (a) Energy, (b) Kinetic energy, (c) Potential energy and (d) Power.
5. a) A lorry tows a trailer of mass 1800kg at a speed of 45
km/h along a straight road. If the tension in the coupling is
900N, find the power expended by the lorry’s engine.
b) If the trailer is pulled along a stretch of 800m at a new speed
of 60km/h, find the new power output required to create a tension of 1200N in the coupling.
6. A 5.2kg object speeds up from 3.1m/s to 4.2m/s. What is the change in Kinetic energy?
7. A 1.2 HP motor (1 HP = 745.7W) is used to raise a 1300kg
Land Rover 5.7m up into a tree. What time will it take?
8. A massless spring with a spring constant of 34N/m is
compressed 5.8cm horizontally and used to shoot an 18 gram
marble across a frictionless table. What is the speed of the marble?
9. A 3.4kg bowling ball hanging from the ceiling on a long string
15cm swings from side to side like a pendulum. When it is
15cm above its lowest point on the left side, I shove it with a
force of 11N for a distance of .35m in the direction it is going.
How high will it swing on the other side? (Neglect friction)
10. A 580kg rollercoaster is going 7.5m/s on the top of a 1.2m
tall hill, how fast is it going on top of a 3.5m tall hill? (Neglect
friction)
11. In the picture given below, F pulls a box having 4kg mass from
point A to B. If the friction constant between the surface and the
box is 0,3; find the work done by F, work done by friction force
and work done by the resultant force.
12. Applied force vs. position graph of an object is given below. Find the work done by the forces on the object.
13. Different forces are applied to three objects having equal masses.
Forces pull objects to height, h. Find the work done by the forces on objects and work done on gravity.
Key unit competence
The learner should be able to apply the principle of conservation of
mechanical energy for isolated system.
My goals
By the end of this unit, I will be able to:⚫ define terms associated with isolated system and open system.
⚫ describe an isolated and open system.
⚫ state different forms of mechanical energy.
⚫ differentiate kinetic from potential energy.
⚫ explain conversion of kinetic energy into potential energy and vice versa.
⚫ state principle of the conservation of energy.
⚫ identify different forms of mechanical energy.
⚫ apply the principle of conservation of mechanical energy in solving problems.
⚫ discuss applications of the principle of conservation of mechanical energy to isolated system.
⚫ understand the application of the principle of conservation of mechanical energy.
⚫ explain that kinetic energy can be converted into potential energy and vice versa.
⚫ predict the consequences of the law of conservation of mechanical energy on an isolated systemKey concepts
1. How one can define a system and energies that are involved in.
2. How energy of a system is conserved.
3. Discuss the mechanical energy conversion process and its conservation.
4. Discuss different application of energy conservation law.Vocabulary
Energy conservation, isolated system, open system, closed system.Reading strategy
When reading this section, take note of the definitions of key terms. Explain
the key terms in your own words and try to perform calculations involving
energy conservation in a system. Exercise doing calculations regularly until they are grasped fully.9.1 Isolated and open systems
Activity 9.1: Investigating the open and closed system
Materials:
⚫ Vacuum flask
⚫ Cooking vessel
⚫ Tripod stand
⚫ Bunsen burner
⚫ Thermometer
⚫ Stop watchProcedure:
⚫ Light the Bunsen burner, and heat water in the cooking vessel on the tripod stand until it is boiling and record the time taken for the water to boil using your stop watch.
⚫ Measure the temperature of the boiling water.
⚫ After the water has boiled, pour part of it into the vacuum flask
and close it, then leave another part in the cooking vessel.
⚫ Remove the cooking vessel and the boiling water from the bunsen burner.
⚫ Leave the water in the flask and that in the cooking vessel for a period of 20 min.⚫ Measure the temperature of the water in the vacuum flask T1 after those 20 min.
⚫ Measure the temperature of water in the cooking vessel T2 after those 20 min.
Questions:
1. Compare the two temperatures t1 and t2 and discuss the results obtained.
2. Why are the results different?
3. What is the difference between the vacuum flask system and the cooking vessel system of keeping the temperature?An isolated system referred to as closed system is a physical system that does not allow certain types of transfers (such as transfer of energy or mass) in or out of the system. The specification of what types of
transfers are excluded varies in the closed systems of physics, chemistry or engineering.
Fig. 9.1: Closed and Open system
In thermodynamics, a closed system can exchange energy (as heat or work) but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter.
In particular, some writers use ‘closed system’ where ‘isolated system’ is here used. For a simple system, with only one type of particle (atom or molecule), a closed system amounts to a constant number of particles.
An open system is a system that has external interactions. Such interactions can take the form of information, energy, or material transfers into or out of the system boundary, depending on the discipline which defines the concept. An open system is contrasted with the concept of an isolatedsystem which exchanges neither energy, matter, nor information with its environment. An open system is also known as a constant volume system or a flow system.
9.2 Kinetic and potential energy of a system
Activity 9.2: Investigating Kinetic energy and Potential energy of a system
With reference to the activity 8.2 and the Fig.8.2; try to answer the
following questions in pairs;
Questions:
1. Name the energies that are involved in the system (Fig. 8.2).
2. Discuss and explain the energy relationship in the system.Kinetic energy is directly proportional to the mass of the object and to the
square of its velocity:
If the mass has units of kilograms and the velocity of meters per second,
the kinetic energy has units of kilograms-meters squared per second
squared. Kinetic energy is usually measured in units of Joules (J); one Joule is equal to 1kg m2/s2.Activity 9.3: Quick exercise
Calculate the kinetic energy in Joules possessed by each of the following
objects. Remember to use the correct number of significant figures in your answer.
Potential energy is the energy an object has because of its position relative
to some other object. When you stand at the top of a stairwell you have
more potential energy than when you are at the bottom, because the earth
can pull you down through the force of gravity, doing work in the process.
The formula for potential energy depends on the force acting on the two
objects. For the gravitational force the formula is:
Where m is the mass in kilograms, g is the acceleration due to gravity
(9.8m/s2 at the surface of the earth) and h is the height in meters. Notice
that gravitational potential energy has the same units as kinetic energy,
kgm2/s2. In fact, all energy has the same units, kg m2/s2, and is measured
using the unit Joule (J).Example
John has an object suspended in the air. It has a mass of 50 kilograms
and is 50 meters above the ground. How much work would the object do
if it was dropped? Show your work. Work is converted in potential energy.Answer:
m = 50kg
g = 9.8m/s2
h = 50m
Where the Work done on the object was converted to Potential
energy.
PE = mgh = 50kg × 9.8m/s2 × 50m = 24500J9.2.1 Kinds of potential energy
a) Chemical potential energy
Activities such as tug of war or riding a bicycle, make us use energy
provided by the food we eat. In cars or motorcycles, petrol is used to
provide energy. Petrol contains energy which makes these vehicles
move. Food and petrol contain energy called chemical potential
energy. It is called chemical energy because it is from the chemical
bonds found in the food or petrol and also called potential energy
because it is potentially available for use when it is needed.
Activity 9.4: Investigating elastic potential energy
Materials:
⚫ Spring or dynamometer
⚫ Rotor stand set
⚫ Slotted mass (50g or 100g)
⚫ Mass hanger (50g or 100g)
Procedure:
⚫ Fix the retort stand
⚫ Hang the spring on the retort stand
⚫ Hang the mass hanger on the spring
⚫ Put different slotted mass on the mass hanger in order of 100g, 200g, 300g etc
Questions:
1. What is happening on the spring as masses are being added
to the mass hanger?
2. Which energy is involved in that system?A catapult is used to hurl a stone at a high speed by stretching its bands
and then releasing them to hurl the stone. The catapult possesses potential
energy when its bands are in the condition of being stretched, which is
then transferred to the stone and makes it move at high speed.
Fig. 9.4: Elastic potential energy
c) Gravitational potential energy
Activity 9.5: Investigating gravitational potential energy
Materials:
A ball
Procedure:
Throw the ball vertically upward such that it reaches a noticeable
height as shown in Fig. 9.5.
Questions:
1. Describe the energy that made the ball to fly upward.
2. Why, does it come back when at the top of its path?
3. Which kind of energy does the ball have at the top of its path?An object raised to a height has energy due to the position it is at. An
object raised to a higher level has more gravitational potential energy. If
an object such as a hammer or a brick which was placed on a table top is
let to fall, it can break something which is placed in its way or it can hurt
someone whose foot is in its way because the potential energy which was
stored in it is changed into motion (kinetic) energy which is used to break
something or hurt someone in its way. More work is done in raising a brick
to a higher level hence more gravitational potential energy is stored in the brick at a higher level.
Fig. 9.5: Potential energy in the ball
The gravitational potential energy of
a mass m, at a height h, is:
PE = mgh
This expression can be derived as shown below: suppose a mass m
(weight = mg) is raised through a vertical height h, the work done is:
W = mgh9.2.3 Conversion of potential energy into kinetic energy
Activity 9.6: Investigating energy conversions on cyclists
Look at this photo (Fig. 9.6) of cyclists in Tour du Rwanda step of
MUSANZE-KIGALI. As the cyclists were going downhill, explain and
describe how the potential energy is being converted to Kinetic energy.
Fig. 9.6: Potential energy stored in a cyclist when going downhill.
Kinetic energy is the energy possessed by a moving object. That is energy
gained by a body due to virtue of its motion. It generally comes into picture
when some work is done onto the body to set it in motion. We calculate it
Potential energy is the energy stored within a physical system as a result of
the position or configuration of the different parts of that system. It has the
potential to be converted into other forms of energy such as kinetic energy
and to do work in the process. PE = mgh (h is the height of the body from
reference point). When no other form of energy is created or lost in motion
of a body, then from the law of Conservation of energy we can say that the
Potential energy of a body converts to Kinetic Energy.
Fig. 9.7: Change of Potential energy into Kinetic energy and
conservation of mechanical energyConsidering the diagram above, at the highest point the block has Potential
energy. If we suppose it was dropped from rest, then its Kinetic energy at
that point was zero. At that instant :
Here m is the mass of the body, g is acceleration due to gravity, h is the height, v is the velocity of the body. When the block comes to the lowest point which is ground, all its Potential energy is converted to Kinetic energy.
That’s the reason why the velocity of falling objects keeps on increasing and hits the ground with great impact.9.3 Mechanical energy
Where g is the gravitational field strength and h is the position in height.
They are actually very closely related. In fact, the potential energy plus
the kinetic energy due to the force is a constant. When potential energydecreases at exactly the same rate, it implies the increases in kinetic energy.
This is the conservation of energy. In fact, since the particles are moving
at finite velocities, this is the much stronger local conservation of energy
for mechanical systems. We may concisely state the following principle,
which applies to closed systems (i.e. when there are no interactions with
things outside the system).9.4 Law of conservation of energy
This law states that: “in all energy conversions or transformations, energy
is neither created nor destroyed, but it may be converted from one form
to another but the total amount remains constant.”This means that energy does not disappear but is either transferred to
another place or transformed (changed) into some other form. This law
tells us when one form of energy is converted to another form during an
energy conversion, energy in put always equals energy out.The law of conservation of energy can also be stated as follows: “during
transformation of energy from one form to another, the total amount of
energy is unchanged i.e. the amount of the new form which appears is
equal to the amount of the old form which disappeared”In all physical processes taking place in closed systems, the amount
of change in kinetic energy is equal to the amount of change in
potential energy. If the kinetic energy increases, the potential energy
decreases, and vice-versa.When we consider open systems (i.e. when there are interactions with
things outside the system), it is possible for energy to be added to the
system (by doing work on it) or taken from the system (by having the
system do work). In this case the following rule applies:The total energy of a system (kinetic plus potential) increases by the
amount of work done on the system, and decreases by the amount of
work the system does.A conservation law, in its most general form, simply states that the total
amount of some quantity within a closed system doesn’t change. For
instance, the conserved quantity would be socks, the system would be the
dryer, and the system is closed as long as nobody puts socks into or takes
socks out of the dryer. If the system is not closed, we can always regarda larger system which is closed and which encompasses the system we
were initially considering (e.g. the house in which the dryer is located),
even though, in extreme cases, this may lead us to consider the amount of
socks (or whatever) in the entire Universe!
Within a closed system, the total amount of energy is always conserved.
This translates as the sum of the n changes in energy totalling to 0.
An example of such a change in energy is dropping a ball from a distance
above the ground. The energy of the ball changes from potential energy
to kinetic energy as it falls.
It is the energy associated with the motion and position of an object. The
principle of conservation of mechanical energy states that in an isolated
system that is only subject to conservative forces, the mechanical
energy is constant. If an object is moved in the opposite direction of a
conservative net force, the potential energy will increase and if the speed
(not the velocity) of the object is changed, the kinetic energy of the object
is changed as well. In all real systems, however, non-conservative forces,
like frictional forces, will be present, but often they are of negligible values
and the mechanical energy’s being constant can therefore be a useful
approximation. In elastic collisions, the mechanical energy is conserved
but in inelastic collisions, some mechanical energy is converted into heat.
The equivalence between lost mechanical energy (dissipation) and an
increase in temperature was discovered by James Prescott Joule.Many modern devices, such as the electric motor or the steam engine,
are used today to convert mechanical energy into other forms of energy,
e.g. electrical energy, or to convert other forms of energy, like heat, into
mechanical energy.Example: A 10kg object falls from a height of 12m. Fill in the potential,
kinetic and total energy of the object at the given points.
9.5 Applications of law of conservation of mechanical energy
In physics, if you know the kinetic and potential energies that act on
an object, then you can calculate the mechanical energy of the object.
Imagine a roller coaster car traveling along a straight stretch of track. The
car has mechanical energy because of its motion: kinetic energy. Imagine
that the track has a hill and that the car has just enough energy to get to
the top before it descends the other side, back down to a straight and level
track (Fig. 9.7). What happens?
Well, at the top of the hill, the car is pretty much stationary, so where
has all its kinetic energy gone? The answer is that it has been converted
to potential energy. As the car begins its descent on the other side of the
hill, the potential energy begins to be converted back to kinetic energy, and
the car gathers speed until it reaches the bottom of the hill. Back at the
bottom, all the potential energy the car had at the top of the hill has been
converted back into kinetic energy.An object’s mechanical potential energy derives from work done by forces,
and a label for a particular potential energy comes from the forces that are
its source. For example, the roller coaster has potential energy because
of the gravitational forces acting on it, so this is often called gravitational
potential energy.
Fig. 9.8: The change of potential energy to kinetic energy, the
conservation of mechanical energyThe roller coaster car’s total mechanical energy (Fig. 9.8), which is the
sum of its kinetic and potential energies, remains constant at all points of
the track (ignoring frictional forces). The combination of the kinetic and
potential energies does vary, however. When the only work done on an
object is performed by conservative forces, its mechanical energy remains
constant, whatever motions it may undergo.
Say, for example, that you see a roller coaster at two different points on a
track Point 1 and Point 2 so that the coaster is at two different heights and
two different speeds at those points. Because mechanical energy is the
sum of the potential energy = mass × gravity × height and kinetic energy
These equations represent the principle of conservation of mechanical
energy. The principle says that if the net work done by non-conservative
forces is zero, the total mechanical energy of an object is conserved;
that is, it doesn’t change. (If, on the other hand, friction or another nonconservative
force is present, the difference between ME2 and ME1 is equal
to ( Wnc) the net work of the non-conservative forces do: ME2 – ME1 =
Wnc.)
Another way of rattling off the principle of conservation of mechanical
energy is that at Point 1 and Point 2, PE1 + KE1 = PE2 + KE2You can simplify that mouthful to the following: ME1 = ME2
Where ME is the total mechanical energy at any one point. In other words,
an object always has the same amount of energy as long as the net work
done by non-conservative forces is zero.9.6 Unit 9 assessment
1. A 580kg rollercoaster is going 7.5m/s on the top of a 1.2m
tall hill, how fast is it going on top of a 3.5m tall hill? (Neglect friction)2. In the picture given below (Fig. 9.9), forces act on objects.
Works done on objects during time t are W1, W2 and W3. Find
the relation of the works. Find the relation of the three works.
Fig. 9.9: Forces acting on objects in different directions
3. A box having 2kg mass, under the effect of forces F1, F2 and F3,
takes distance 5m. Which ones of the forces do work.
Fig. 9.10: A box acted on by three forces
4. Applied force vs. position graph of an object is given below. Find
the kinetic energy gained by the object at a distance 12m.
Fig. 9.11: Force-position graph
5. Three different forces are applied to a box in different intervals.
Graph, given below, shows kinetic energy gained by the box in
three intervals. Find the relation between applied forces.
Fig. 9.12: Friction force versus distance moved
6. A stationary object at t = 0, has an acceleration vs. time graph
given below. If an object has kinetic energy E at t = t, find the
kinetic energy of the object at t = 2t in terms of E.
Fig. 9.13: Acceleration-time graph
7. An object does free fall. The picture given below shows this
motion. Find the ratio of kinetic energy at point C to the total mechanical energy of the object.
Fig. 9.14: Free fall diagram
8. A box is released from point A and it passes point D with a velocity, V.
The work done by gravity is W1 between AB, W2 between BC and W3 between CD. Find the relation between them.
Fig. 9.15: Box moving under gravity
9. An object is projected with an initial velocity V from point A. It
reaches point B and turns back to point A and stops. Find the
relation between the kinetic energy object has at point A and
energy lost on friction.
Fig. 9.16: Energy conservation of a thrown box
10. An object is projected from point A with an initial kinetic energy
E, and it reaches point C. How much energy must be given to the object in order for it to reach point D.
Fig. 9.18: Object moving under gravity
Key unit competence
The learner should be able to describe and analyse gas laws experiments.
My goals
By the end of this unit, I will be able to: state and explain the behaviour and properties of an ideal gas.
discuss the equation of perfect gas. (Ideal gas).
define Boyle’s law, Charles’s law, pressure law, and Dalton`s law.
apply the gas law equations in problem solving.
carry out an experiment to verify Dalton’s law of partial pressure.
carry out an experiment to verify Boyle‘s law, Charles’s law and pressure law.
explain equations of perfect gas. (Ideal gas)
discuss the gas laws.
explain how a change in volume of a fixed mass of gas at constant temperature is caused by a change in pressure applied to the gas.
understand; think logically and systematically when relating gas laws. adopt scientific methods applied in solving gas problems.
adopt scientific methods in analysing, modeling and establishing the dimensions of gas laws.Key concepts
1. What do you understand by the concept of an ideal gas?
2. Describe the laws of gas and discuss their application in real life.
3. How do you mix gases?
4. What do you use to measure gas pressure? Vocabulary
Ideal gas or perfect gas, partial pressure, real gas, gas constant.Reading strategy
After you read each section, re-read the paragraphs that contain definitions
of key terms. Use all the information you have learnt to explain each key
term in your own words and draw your own experiment.
Master the procedures of each experiment so that you will be able to carry
out similar experiments in the future.10.1 Introduction
Activity 10.1: The Balloon and the Bottle Experiment
Materials:
⚫ Water
⚫ Soda bottle (50ml)
⚫ Balloon
Procedure:
Find an empty glass bottle (a soda bottle will work) and fill it with
about 20 ml of boiling water. Stretch a balloon over the mouth of the
bottle. As the bottle cools, the gas will suck the balloon into the bottle
and it will begin to inflate within the bottle.Questions:
1. Why is the balloon being sucked into the bottle as the water cools?
2. What would happen if you heat the water again when the
balloon is still stretched over the bottle’s mouth?
The study of heat and its transformation to mechanical energy is called
thermodynamics (which stems from Greek words meaning movement of
heat). Thermodynamics is the dynamics of heat.Statistical systems are systems with large numbers of particles (atoms
and / or molecules). By large, we mean on the order of 6.022 x 10 23
(“Avogadro’s Number”, designated NA; one mol).
The measurable quantities are called “state variables”. As their name
implies, their values depend only on the current state of the system,
and not on the path taken to that state: they have no memory of their
past values. Three of the most important state variables are temperature,
pressure and volume. Temperature and pressure are “intrinsic” state
variables, since their value does not depend on the “size” of the system.
Volume is an “extrinsic” state variable, since its value does depend on the
size of the system.In thermodynamics, all temperatures are measured in “Kelvin” (K = Celsius
+ 273.15). Zero K is called “absolute zero”, since it is the lowest possible
temperature. A temperature difference of 1K is equal to a temperature
difference of 1C. We will not be concerned with pressure all that much,
since most physiological functions assume a constant pressure equivalent
to that of the atmosphere. We will typically measure volume in liters.Let us find the volume of one mol of air using its density and gram molecular
weight (the mass of one mol of a substance in grams is numerically equal
to its molecular weight). If air is 78.08% N 2 (with molecular weight 28),
20.95% O 2 (32) and.93% Ar (40), the average molecular weight of air is
28.94g/mol. Since the density of air at standard temperature and pressure
is 1.293kg/m 3, one mole of air then occupies 22.4l (0.224m 3).
A basic axiom for us will be the “Equipartition Theorem”.
10.2 Three gas laws
The early gas laws were developed at the end of the eighteenth century,
when scientists began to realise that relationships between the pressure,
volume and temperature of a sample of gas could be obtained which would
hold for all gases. Gases behave in a similar way over a wide variety of
conditions because to a good approximation they all have molecules which
are widely spaced, and nowadays the equation of state for an ideal gas
is derived from kinetic theory. The earlier gas laws are now considered as
special cases of the ideal gas equation, with one or more of the variables held constant.
Fig. 10.1: Boyle’s law and Charles’ law
10.2.1 Boyle’s lawActivity 10.2: Investigating Boyle’s law
Materials:
⚫ A 50 ml syringe
⚫ A small sized balloon
Fig. 10.2: A Syringe
Procedure:
⚫ First, trap a small amount of air in the balloon and tie a knot.
⚫ Place the balloon in the syringe.
⚫ With your finger, close the syringes' nose and press the piston as in Fig. 10.2.
Questions:
1. Why is the balloon decreasing the size as the pressure increases?
2. Why is its volume increasing as the pressure decreases?
3. Discuss and explain where this can be observed in real life.Robert Boyle (1627-1691), the English natural philosopher and one of
the founders of modern chemistry; is best remembered for Boyle’s law
(1662), a physical law that explains how the pressure and volume of a gas
are related. It states that;“The volume of a fixed mass of gas is inversely proportional to the
pressure, provided the temperature remains constant”:
PV = Ct where Ct is a constant.
We can re-write this equation as: P1V1 = P2 V2
This relationship means that pressure increases as volume decreases, and vice versa.The cubic expansivity is always calculated in terms of an original volume
at 00C. The cubic expansivity of a gas at constant pressure is thus defined
as the fraction of its volume at 00C by which the volume of a fixed mass
Activity 10.3: Experiment to verify Boyle’s law
Visit this link to play the experiment game: http://www.uccs.edu/vgcl/
gas-laws/experiment-1-boyles-law.html
In this experiment you will examine the effect of pressure on the
volume of a gas at constant temperature and formulate the relationship
between the two.
Fig. 10.3: Boyle’s law
Procedure:
1. The experimental setup includes a closed cylinder with a
moveable piston containing a sample of gas (air), some 1
kg weights to place on the piston, and a pressure gauge
and ruler to measure the pressure (in torr) and height of the
piston (in cm) in the cylinder. (We will assume that the gas in the cylinder is at room temperature).2. Open the worksheet for Experiment 1. (You may want to enter the data within Excel, so that the graph can be created, then print out the results afterwards).
3. Click anywhere on the apparatus to start the lab. (If you are using Internet Explorer, you will have to click once to activate the control, then click again to start the lab).
4. The piston itself has a mass of 1kg. Click on the pressure gauge to see a close-up of the measuring devices. Record the pressure and height of the piston in the appropriate cells in the worksheet.
5. Click on the stack of weights to move one onto the piston.
6. Record the values in the next row of the worksheet.
7. Repeat steps 5 and 6 to record the pressure and height of the piston with increasing numbers of weights. You may click on either the stack of weights on the bench to move one up
to the piston, or the stack on the piston to return one to the bench.10.2.2 Charles’ law
Activity 10.4: Demonstrating Charles’s Law
by Expanding and Contracting a Balloon
Materials:
⚫ Erlenmeyer flask or retort flask
⚫ Water
⚫ Electric heater ( Hot plate)
⚫ Balloon
Procedure:
Carry out the steps from A to F shown below and answer the questions:
a) Add a small amount of water to an Erlenmeyer flask as shown in Fig.10.4.
Fig. 10.4: Erlenmeyer (retort flask)
b) Place the flask on a hot plate or burner as shown in Fig 10.5.
Fig. 10.5: Flask on a hot plate
c) Put the open end of a balloon over the opening of the flask as shown in Fig. 10.6.
Fig. 10.6: The balloon knotted on the flask’s opening
d) Observe the expansion of the balloon and record your observations.
e) Move the flask to an ice bath as in Fig 10.7.
Fig. 10.7: The flask in the ice bath
f) What happens to the balloon?
Questions:
1. What is causing the balloon to expand in step c?
2. What is causing the balloon to contract in step e?
3. Discuss and explain this effect in your own words?Charles’ law or law of volumes (French chemist Jacques Charles 1787,
relating volume and temperature at pressure constant):
The volume of a given amount of gas is directly proportional to absolute
temperature when pressure is kept constant.
The process where the temperature and volume change at constant
pressure is called Isobaric process.10.2.3 Gay-Lussac’s law
Activity 10.5: How to Crush a Can with Air Pressure
You can crush a soda can using nothing more than a heat source and
a bowl of water. This is a great visual demonstration of some simple scientific principles, including air pressure (Pressure law) and the concept of a vacuum. The experiment can be performed by teachers
as a demonstration, or by mature learners under supervision.Using the following steps a) to e), do the experiment and notice the
changes and observations.
a) Pour a little water into an empty soda can as in Fig. 10.8.
Fig. 10.8: Soda can
b) Prepare a bowl of ice water as in Fig.10.9.
Fig. 10.9: Bowl of ice water
c) Heat the can on the stove as in Fig.10.10.
Fig. 10.10: Heating the soda can
d) Use the tongs to turn the hot can upside down into the cold water.
Fig. 10.11: The hot can in the cold water
e) What happens when you heat the can of water?
Questions:
1. What happens inside the can when being heated?
2. Why when inverted in cold water does the can collapse?
3. Discuss and explain the phenomena of crushing can.
Pressure law or Third gas law (French chemist Joseph Guy-Lussac in
1809, relating temperature and pressure at constant volume):
“At constant volume, the pressure of a gas is directly proportional to the
absolute temperature” P∝ T
The process where the pressure and temperature change at constant
volume is called Isochoric process.
The pressure expansively of a gas at constant volume is defined as the
fraction of its pressure at 00C by which the pressure of a fixed mass of gas
increases per Kelvin rise in temperature.
10.3 Ideal gas
10.3.1 Definition
We can define an ideal gas as one which obeys Boyle’s law exactly and
whose internal energy is independent of its volume.
At normal ambient conditions such as standard temperature and pressure,
most real gases behave qualitatively like an ideal gas. Generally, deviation
from an ideal gas tends to decrease with higher temperature and lower
density, as the work performed by intermolecular forces becomes less
significant compared to the particles’ kinetic energy, and the size of the
molecules becomes less significant compared to the empty space between
them.
The ideal gas model tends to fail at lower temperatures or higher pressures,
when intermolecular forces and molecular size become important. At some
point of low temperature and high pressure, real gases undergo a phase
transition, such as to a liquid or a solid.10.3.2 Ideal gas law
Activity 10.6: Investigating ideal gas
Study carefully the Fig. 10.12 and answer to the following
questions:
Fig. 10.12: Ideal gas containers
Questions:
1. What is the difference between molecules of gas contained in
the volume 1 and volume 2?
2. Where is gas pressure greater between the two containers?
3. Comment on the relationship between the volume change
and pressure change in each case.
The combined gas law or general gas equation is formed by the
combination of the three laws, and shows the relationship between the
pressure, volume and temperature for a fixed mass of gas:
1. If temperature and pressure are kept constant, then the volume
of the gas is directly proportional to the number of molecules of
gas.
2. If the temperature and volume remain constant, then the pressure
of the gas changes is directly proportional to the number of
molecules of gas present.
3. If the number of gas molecules and the temperature remain
constant, then the pressure is inversely proportional to the volume.4. If the temperature changes and the number of gas molecules
are kept constant, then either pressure or volume (or both) will
change in direct proportion to the temperature.10.3.3 Dalton’s law of partial pressure
Activity 10.7: Partial pressure experiment
Use the information in table 10.1, to discuss and explain the effect of
partial pressure of two or more gases.
Partial Pressure
Table 10.1: Dalton’s Law of Partial Pressures
The total pressure in a gas mixture is the sum of the partial pressures of
each individual gas.
Table 10.2: Dalton’s law of partial pressure
Examples
1. 10g of nitrogen gas and 10g of helium gas are placed together
in a 10l container at 25oC. Calculate the partial pressure of each
gas and the total pressure of the gas mixture.
Calculate the molesof each gas present: n = mass ÷ molar mass
Table 10.3: Molar mass
Calculate the total moles of gas present = 0.4 + 2.5 = 2.9 mol
Calculate the total gas pressure assuming ideal gas behaviour:
PV = nRT
P = n × R × T ÷ V
n = 2.9 mol
R = 8.314
T = 25oC = 25 + 273 = 298K
V = 10l
P = 2.9 × 8.314 × 298 ÷ 10 = 718kPa (7 atm)Partial pressure of nitrogen = n(N2) ÷ n(total) x total pressure
Partial pressure of nitrogen = 0.4 ÷ 2.9 x 718kpa = 99kPa (0.9
atm)
Partial pressure of helium = n(He) ÷ n(total) x total pressure
Partial pressure of helium = 2.5 ÷ 2.9 x 718 = 619kPa (6.1atm)
2. At 15oC, 25mL of neon at 101.3 kPa (1 atm) pressure and 75
mL of helium at 70.9kPa (0.7 atm) pressure are both expanded
into a 1l sealed flask. Calculate the partial pressure of each gas
and the total pressure of the gas mixture.
Since the temperature and moles of each gas is constant, the
pressure exerted by each gas is inversely proportional to its volume
(Boyle’s Law).
PiVi = PfVf
Pf = PiVi ÷ Vf
Partial pressure Neon = 101.3kPa x 25 x 10-3L ÷ 1l = 2.5 kPa
Partial pressure Helium = 70.9kPa x 75 x 10-3L ÷ 1l = 5.3 kPa
Total pressure = 2.5 + 5.3 = 7.8kPa
10.3.4 Density of gases
Density has the units of mass per unit volume
Answer:
Note: The standard atmosphere (atm) of pressure is approximately
equal to air pressure at earth mean sea level and is defined
as: 1 atm = 101325 Pa = 1013.25 mbar = 760 Torr10.3.4 Avogadro’s Law
10.4 Unit 10 assessment
1. One atmosphere is equal to (Choose the correct answer):
a) 760 cm Hg
b) 760 mm Hg
c) 101325 mm Hg
d) 8.314 mm Hg
2. Which of the following quantities is not necessary to describe a
gas? (Choose the correct answer)
a) Volume
b) Temperature
c) Amount
d) Pressure density
3. Dalton’s Law of Partial Pressures states that (Choose the correct
answer):
a) The total pressure exerted by a mixture of gases can be
determined using the Ideal Gas Constant.
b) The pressure of a gas is inversely proportional to both the
temperature and number of moles of the gas.
c) The total pressure exerted by a mixture of gases is equal to the
sum of the pressures exerted by the individual components in the mixture.d) You must take into consideration the vapour pressure of the solvent that you are using.
4. Convert centigrade temperature to Kelvin (Choose the correct answer):
a) oC - 273.15
b) oC + 273.15
c) oF - 273.15
d) oF + 273.15
5. 125cm3 of gas are collected at 150C and 755 mm of mercury
pressure. Calculate the volume of the gas at s.t.p ( T= 297 K,
V= 22.4 l/mol, P= 1 atm).6. When tested in a cool garage at 120C a motor tyre is found
to have a pressure of 190kPa. Assuming the volume of the air
inside remains constant, what would you expect the pressure
to become after the tyre has been allowed to stand in the sun
so that the temperature rises to 320C? Atmospheric pressure =
100kPa.
7. Pure helium gas is contained in a leakproof cylinder containing a
movable piston. The initial volume, pressure, and temperature of
the gas are 15l, 2.0 atm, and 310K, respectively. If the volume
is decreased to 12l and the pressure is increased to 3.5atm, find
the final temperature of the gas.
8. Determine the volume of 1 mol of any gas at STP, assuming it
behaves like an ideal gas.9. A gas occupies a volume of 25.8l at 170C and under 690mm
Hg. What volume will it occupy at 345K and under 1.85atm?
10. Oxygen is collected in a bottle, turned upside down, containing
water at 270C. The barometer pressure measured in the bottle
is 757 torr. Calculate the partial pressure of O2 knowing that the
vapour pressure of water is 19.8 mm Hg at 270C.
11. We mix 200ml of N2 at 250C and under 250 torr with 350 ml
of 02 at 250C and under 300 torr, the final pressure is 300 torr
and the final volume is 300ml. What will be the final pressure at
250C?- Opened: Friday, 12 June 2020, 1:29 PMClosed: Friday, 12 June 2020, 1:29 PM
Key unit competence
The learner should be able to describe methods of magnetization and
demagnetization.
My goals
By the end of this unit, I will be able to:
describe a magnet.
describe the magnetic properties of iron and steel.
describe the methods of magnetising and demagnetising of materials.
explain the use of keepers in storing magnets.
explain magnetic shielding.
explain magnetisation using the domain theory.
make temporary and permanent magnets.
describe methods demagnetising magnets.
explain demagnetisation using the domain theory.
discuss methods of storing magnets.
list and explain the applications of magnets.
understand the social, economic, and technological implications of using magnets.
recognise the role of electromagnets in making electrical devices.Key concepts
1. What is needed to describe magnetism of a body?
2. How to magnetise or demagnetise a body?
3. When is a body magnetised or demagnetised?
Vocabulary
Magnetisation, demagnetisation, hammering, stroking, heating.Reading strategy
As you read this unit, put emphasis on all the key terms, their meaning
and their structures. Study the diagrams so that you master how things
work. Do all the experiments in this book about magnetisation.11.1 Structure of an atom
Matter is made of tiny particles called atoms. An atom is made of
three particles; electron, proton and neutron. These particles are called
fundamental particles of an atom or sub atomic particles.
Fig. 11.1: Atomic structure of a Carbon atom
Electron (e–) - Electron is denoted by ‘e’ and is a negatively charged
particle. The charge of an electron is equal to -1.6x10-19 Coulomb which
is negative charge.
The relative mass of electron is 1/1836 of the mass of a Proton. Electrons
revolve round the nucleus of atoms.
Proton (p+) - Proton is denoted by ‘p’ and is a positively charged particle.
The charge of proton is 1.6x10 -19 coulomb of positive charge and it is
considered as unit a positive charge. Thus absolute charge over a proton
is equal to +1.The mass of a proton is equal to 1.6x10 -27 kg and considered equal to
1 as it is equal to the mass of 1 hydrogen atom. Proton is present in the
nucleus of an atom.
Nucleus – The centre of an atom is called nucleus. Nucleus comprises of
neutron and proton. The nucleus of an atom contains the whole mass of
an atom.11.2 Magnetism
Activity 11.1: Investigation of magnetism
Materials:⚫ Magnets ( Different types)
⚫ Ferromagnetic materials (such as nails, blades, iron filings)
⚫ A pointing compassProcedure:
⚫ Carefully note the position of the pointer in the pointing compass when it is alone and far away from any metal.
⚫ Then bring a nail, blade and magnet one by one close to the pointing compass.Questions:
1. Describe the observation found in each situation.
2. Discuss and explain the phenomena observed above.
Magnetism is a class of physical phenomenon that includes forces exerted
by magnets. It has its origin in electric currents and the fundamental
magnetic moments of elementary particles. These give rise to a magnetic
field that acts on other currents and moments. Magnetism, phenomenon
associated with the motion of electric charges can take many forms. It
can be an electric current in a conductor or charged particles moving
through space, or it can be the motion of an electron in atomic orbit.11.2.1 Fundamentals of magnetism
A permanent magnet is a piece of steel or a similar metal having its own
magnetic field. Under ideal conditions, it will retain its magnetic strength
for many years. Frequent drops, impacts or high temperatures weaken the
magnetism. A piece of iron, called a keeper, fits over the magnet’s poles,
helping it retain its magnetism during long periods in storage.11.2.2 Ferromagnetism
Activity 11.2: Distinguishing magnetic and non-magnetic materialsWith reference to the activity 11.1, explain what would happen if the
materials listed below were put close to a magnet
⚫ A Piece of wooden bar
⚫ Pieces of copper, aluminium, steel brass, iron
⚫ A pen
⚫ A piece of paper
Questions:
1. Does a magnet attract all metals? Why?
2. Does a magnet attract papers? Why?
3. Differentiate the materials used in the above activity.All permanent magnets exhibit what scientists call ferromagnetism.
Under the right conditions, a ferromagnetic metal piece acquires its
own field, becoming magnetised. Other kinds of metals, such as copperand aluminum, are not ferromagnetic. A magnet keeper is a piece of
ferromagnetic material, and is made up of soft iron.11.2.3 Magnetic Domain
Activity 11.3: Investigating magnetic domain
Materials:
⚫ A permanent bar magnet
⚫ Iron filings
⚫ A stiff plain piece of paper
Procedure:
⚫ Place the magnet on a flat surface and put the stiff plain paper on it.
⚫ Sprinkle the iron fillings all over the stiff paper.
⚫ Gently, move the magnet in various directions.
Question:
Discuss and explain the changes you are observing.
Fig. 11.2: Magnetic domain
In all ferromagnetic materials, microscopic bits of metal, called domains;
have tiny magnetic fields. If their magnetic north and south poles line up,
they cooperate and form a large field around the whole object. Impacts and
heat scramble the orientation of the domains, weakening the field. Long
periods of time also weaken magnets. During storage, a keeper reinforces
the magnetic field, maintaining its strength for longer periods of time.11.2.4 Magnet Shapes
Fig. 11.3: Shapes of magnets
Permanent magnets come in a variety of shapes: bars, cylindrical,
horseshoe, (See Fig 11.3). Regardless of shape, every magnet has exactly
one North and one South Pole, located magnetically at opposite ends of
the field. Lines of magnetic force exit the magnet at the North Pole, curve
around and re-enter it at the South Pole, and pass through the magnet’s
material to the North Pole, forming a continuous loop. A horseshoe magnet
has its north and south poles near each other, one pole at each end of the
“U” shape. It makes an ideal candidate for a keeper, as it lies across both
poles, forming a magnetic bridge between them.11.2.5 Magnetic Circuit
Fig. 11.4: A Magnetic circuit
A magnetic field holds its strength best when the entire magnetic loop, or
circuit, passes through a ferromagnetic metal at all points. A horseshoe
magnet has an air gap between its two poles; the soft iron keeper closes
this gap (see Fig. 11.4). A bar magnet, left by itself, will lose its strength
over several months. Though a bar magnet has no “keeper,” if you lay two
bars side by side, with the north pole of one touching the south pole of
the other, they form a magnetic loop in iron and preserve the strength of both magnets.
11.3 Magnetisation and demagnetisation
Activity 11.4: Magnetisation by induction
⚫ Bring each of the metallic materials (copper, zinc, steel, soft
iron, alluminium, soft iron) you wish to magnetise in close proximity to a strong magnet.
⚫ Then use the magnetised item to pick up drawing pins, paperclips or iron filings.
⚫ If, after removing the magnet for a few minutes, there is a little adhering to the items; it will not be useful as a permanent magnet.Gentle tapping, for example with a pencil, should leave some magnetised items still in contact.
Re-magnetising a magnet is often necessary if the magnet has been mistreated. Occasionally magnets are required to be made from pins etc. in order to make compasses. Also there are often requests to make a tool (e.g. screwdriver) magnetic so that it complies with a desired function.
Sometimes a tool may have inadvertently become magnetic with unwanted consequences.There are a few methods of effecting the magnetisation of an object. However, it is important to make sure that the object is of the “right stuff”. Trying to make a permanent magnet from a
rod not capable of retaining the magnetism will just waste time. So, what materials hold magnetism? Obviously any “old magnets” will be useful, certain steels and some iron based rods such as nails and the
steel shafts of screwdrivers can also hold magnetism. Magnetically “soft” iron will magnetise but will lose the magnetism very quickly. This makes it ideal for electromagnets. Some stainless steels have very poor retention of the field so should not be used.11.3.1 Electromagnet and electric bell
Electromagnet
Activity 11.5: Making an electromagnetMaterials:
⚫ Metal rod like a long nail
⚫ A long conducting wire
⚫ A 12V battery or power supply
⚫ A Switch
⚫ Bulb
⚫ Paperclip
Procedure:
⚫ Coil the wire on the metal rod.
⚫ Connect one end to the positive pole and another to the negative pole.
⚫ Make sure that you have connected the bulb in series with the coil.
⚫ Switch on, approach the paperclip and observe.
Questions:
1. What changes have you observed by approaching the rod to the paperclip?
2. Discuss and explain how you would use the same techniques in real life.A simple electromagnet consists of a coil of insulated wire wrapped around
an iron core. A core of ferromagnetic material like iron serves to increase
the magnetic field created.
The strength of the magnetic field generated is proportional to the amount
of current through the winding magnetic field produced by a solenoid (coil
of wire). This drawing shows a cross section through the center of the coil.
The crosses are wires in which current is moving into the page; the dots
are wires in which current is moving up out of the page.
Fig. 11.5: The parts and formation of an electromagnet
An electromagnet is a type of magnet in which the magnetic field is
produced by an electric current. The magnetic field disappears when the
current is turned off. Electromagnets usually consist of a large number
of closely spaced turns of wire that create the magnetic field. The wire
turns are often wound around a magnetic core made from soft iron; the
magnetic core concentrates the magnetic flux and makes a more powerful magnet.
The main advantage of an electromagnet over a permanent magnet is
that the magnetic field can be quickly changed by controlling the amount
of electric current in the winding. However, unlike a permanent magnet
that needs no power, an electromagnet requires a continuous supply of
current to maintain the magnetic field.
Electromagnets are widely used as components of other electrical devices,
such as motors, generators, relays, loudspeakers, hard disks, MRI
machines, scientific instruments, and magnetic separation equipment.
Electromagnets are also employed in industries for picking up and moving
heavy iron objects such as scrap iron and steel.Electric bell
Many objects around you contain electromagnets. They are found in
electric motors and loudspeakers. Very large and powerful electromagnets
are used as lifting magnets in scrap yards to pick up, and then drop, old
cars and other scrap iron and steel.Electric bells like the ones used in most schools also contain an electromagnet.
Fig. 11.6 The Electric bell
The cycle repeats as long as the switch is closed. When the switch is
pushed close, the circuit is completed and current flows through the
electromagnetic coil.11.3.2 Project 13.10 Make your own electric bell
Description
When you build your own model for Science Activity Workshop¸ remember
that the magnetic motor is the most important part of your door bell. The
idea is to create a circuit which will cause the electromagnet to generate
a magnetic field. As a result, when an iron object is placed close to this
magnet it is attracted towards the magnetic field. On pressing the bell,
we are actually completing this circuit and thus eventually generating the sound.Kit Contents:
⚫ Nut / Bolts
⚫ Cardboard base
⚫ Iron strip
⚫ Thumb tack
⚫ Safety pin
⚫ Copper wire
⚫ 9-v battery with cap
⚫ Sand paper
You may also need scissors, a screw driver and cello-tape and an adult to
help you.11.3.3 Demagnetisation by hammering
Fig. 11.7: Hammering a magnet
11.3.4 Magnetisation by stroking
This is, historically, the oldest form of consistently creating magnet. This
produces magnets that are not as strong as the electrical methods. There
are two methods which have traditionally been given the names “single
stroke” and “double stroke".
Fig. 11.8: Single stroke and double stroke North will be at the
LHS (left hand side) of the barIn the “single stroke” method, a magnet is drawn over the rod so as to
go completely over the length of the rod. The magnetic domains are then
pulled into alignment as the magnet passes. A useful method of realising
the polarity of the induced magnetism is to consider the bar as a compass
and which way it would point to the magnet as it finally leaves the bar.
Best results are obtained after about twenty passes when the magnet is
taken in a big loop away from the bar in between passes.
In the “double stroke” method two magnets are used at the
same time in what may be thought of as a mirroring action. This method produces a stronger magnet than the single stroke method.
Beware of polarity: If this method was to be done using two similar poles
facing the bar it is possible to create a bar magnet with two like poles at
either end! These are termed “consequent poles”.11.3.5 Demagnetisation by Heating
This method can create a magnetised bar without any apparent magnet
being present (i.e. just using the earth’s field). A stronger field may, of
course, be used by placing the cooling bar between two magnets or in
an electrically created field. Care should be taken that the heated bar
is thermally isolated from the field magnets, so as not to destroy their
properties.
Fig. 11.9: Heating a magnet
It may be of interest to try heating an old, weak magnet (all the paint will
be burned off!) to red hot using a pair of tongs in a Bunsen flame and then
placing it on a piece of heat mat with a rare-earth magnet underneath.
Demagnetisation can be achieved by allowing the bar to cool in an East-
West orientation shielded from magnetic influences.11.3.6 Electrical method
Activity 11.5: Investigating the electric method of magnetisation
⚫ Two cell battery (1.5V each).
⚫ A 40cm insulated wire.
⚫ A nail.
⚫ A drawing pin or paper clip.
Procedure:
⚫ Coil the wire around the nail.
⚫ Connect one end to the positive pole and another to the negative pole of the cell battery as in Fig. 11.8.
⚫ Slowly move the nail to the drawing or paper clip.
Question:
Discuss and explain the observation of the changes made.Modern methods of magnetisation and demagnetisation tend to use
electrical methods as it is easily manufactured and controlled. A current
passing through a coil will produce a magnetic field. The strength of the
field is proportional to the current schematic showing conventions and many coils are required.
Fig. 11.10: Magnetisation by electric method
The polarity of the field is easily seen by examining the path of the
conventional current in the coil. If looking at the end of the coil the current
is going clockwise, it will produce the “south seeking” pole. A capital
“S” has the ends following the clockwise rotation. Similarly the other end
will be anticlockwise. This produces the “north seeking” pole. A (albeit
rounded) capital “N” has the ends following an anticlockwise rotation.
These coils can be bought or made or even by modifying some “coil gun” circuits.11.3.7 Demagnetising electrically
Fig. 11.11: Demagnetisation by electric method
This involves taking the bar through the coil and an alternating current is
used to create a field that swamps the existing field in a magnet. This is
gradually reduced. This eventually becomes negligible. Alternatively the
rod can be drawn out and away from a constant amplitude alternating the field.The whole of the coil (240 turns), should be used. Nails placed in the coil
will be felt to vibrate as the voltage is brought up to 6V. The nails can be
moved so that all of their length passes through the centre of the coil at
this voltage. The voltage is then slowly reduced to zero. The nails should
now only have minimal magnetisation.11.4 Magnetic keepers
If two magnets are placed side by side there will be mutual repulsion or
attraction. This weakens the strength of the magnet. To prevent this, bar
magnets are placed side by side with opposite poles near. A soft iron piece
called a keeper is placed across the poles as shown in the figure. This soft
iron piece provides a path for the magnetic field lines to form a continuous
loop. Thus it helps in preserving the magnetic field.
Fig. 11.12: Magnetic keepers
11.5 Magnetic shielding
Equipment sometimes requires isolation from external magnetic fields. In
these cases shields made of high magnetic permeability metal alloys can
be used, such as sheets of Permalloy and Mu-Metal, or with nanocrystalline
grain structure ferromagnetic metal coatings. These materials don’t block
the magnetic field, as with electric shielding, but rather draw the field
into themselves, providing a path for the magnetic field lines around the
shielded volume. The best shape for magnetic shields is thus a closed
container surrounding the shielded volume. The effectiveness of this type
of shielding depends on the material’s permeability, which generally drops
off at both very low magnetic field strengths and at high field strengths
where the material becomes saturated. So to achieve low residual fields,
magnetic shields often consist of several enclosures one inside the other,
each of which successively reduces the field inside it.
Fig. 11.13: Magnetic shielding
Because of the above limitations of passive shielding, an alternative used
with static or low-frequency fields is active shielding; using a field created
by electromagnets to cancel the ambient field within a volume. Solenoids
and Helmholtz coils are types of coils that can be used for this purpose.11.6 Unit 11 assessment
1. Complete the following statement:
____are the charged parts of an atom.
a) Only electrons
b) Only protons
c) Neutrons only
d) Electrons and neutrons
e) Electrons and protons
f) Protons and neutrons2. TRUE or FALSE
a) An object which is positively charged contains all protons and
no electrons.
b) An object which is negatively charged could contain only
electrons with no accompanying protons.
c) An object which is electrically neutral contains only neutrons.
3. Identify the following particles as being charged or uncharged.
If charged, indicate whether they are charged positively or
negatively. (n = neutron, p = proton, e = electron)
4. The amount of charge carried by a lightening bolt is estimated
at 10 Coulombs. What quantity of excess electrons is carried by the lightening bolt?
5. Respond to the following learner statement: “A positively charged
object is an object which has an excess of positive electrons”.
6. A magnet attracts a piece of iron. The iron can then attract
another piece of iron. On the basis of domain alignment, explain
what happens in each piece of iron.
7. Why does hitting a magnet with a hammer cause the magnetism to be reduced?8. Examine the parts of an electric bell shown in the figure below.
Discuss how it works when it is switched on. Make an electric bell and exhibit it.
9. What causes the magnetism in a magnet? A bar magnet is heated. State the effect on magnetic properties.
10. What is the difference between temporary magnets and permanent magnets?Key unit competence
The learner should be able to explain applications static charges.
My goals
By the end of this unit, I will be able to: explain and describe distribution of electric charges and metallic conductors.
explain electric force, electric field and electric potentials.
discuss applications of electrostatic.
define electric field strength.
relate electric field patterns and charge distribution on conductors of different shapes.
evaluate applications of electrostatics in other fields (agriculture, environment, industry).
list the applications of electrostatics.
identify possible hazards related to electrostatic and how to avoid them.Key concepts
1. What is needed to describe electrostatics?
2. When and where can you observe the effects of electrostatics?
3. How can you avoid an electric circuit from your house?
4. How do you know the amount of electric potential from a charged body?
5. How do you calculate/tell/gauge the potential difference?
6. What do charge distributions refer to?
7. Describe different applications of electrostatics.Vocabulary
Electric field, electric potential, lightening arrestors, paint spray, laser
printer, electrostatic precipitator, photocopy machine.
Reading strategy
Read all the details of this unit. Perform calculations and do experimental
work about an electric field. Hence, describe different application of
electrostatics. Participate in all activities, be keen on procedures and observations.Activity 12.1: Investigating the electric charges on a rubbed balloon
Materials:
⚫ 2 inflated balloons with string attached
⚫ Your hair
⚫ Aluminium can
⚫ Woolen fabricProcedure:
⚫ Rub the 2 balloons one by one against the woolen fabric, and then try moving the balloons together.
Questions:
1. Do the balloons want to or are they unattracted to each other?
⚪ Rub 1 of the balloons back and forth on your hair then slowly pull it away.
2. Ask someone nearby what they can see or if there's nobody else
around try looking in a mirror and discuss your observations.
⚪ Put the aluminium can on its side on a table, after rubbing the balloon on your hair again hold the balloon close to the can and watch as it rolls towards it, slowly move the balloon away from the can and it will follow.
3. Discuss and explain the observations made.Electrostatics is a branch of physics that deals with the phenomena and
properties of stationary electric charges.
12.2 Static Electricity
A nylon garment often crackles when it is taken off. We say it has become
charged with static electricity. The crackles are caused by tiny electric
sparks which can be seen in the dark. Pens and combs made of certain
plastics become charged when rubbed on the sleeve and can attract scraps
of paper. Those materials are electrified, possess an electric charge or are electrically charged.
There are two kinds of charges in nature; negative and positive charges.Like charges (+ and + or – and -) repel while unlike charges (+ and -) attract
The net amount of electric charge produced in any process is zero. This is
known as the Law of Conservation of Electric Charge.
If one object or one region of space acquires a positive charge then an
equal amount of negative charge will be found in neighbouring areas or objects.12.3 Electric field
The Concept of Electric FieldActivity 12.2: Investigation of electric field
Materials:
⚫ One battery cell (1.5V)
⚫ A conducting wire
⚫ 5 magnetic needles
⚫ A slotted cardboard
Procedure:
⚫ Arrange the materials as shown in Fig. 12.1.
⚫ Remove the battery and note the changes on needles.
⚫ Reconnect the battery and note the changes on needles.
Question:
What is the main cause of the directions change when the battery is
connected?
Fig. 12.1: Electric field effect on the magnetic needlesWhen a small charged particle is located in the area surrounding a
charged object, the charged particle experiences a force in accordance
with Coulomb’s Law. The space around the charged object where force is
exerted on the charged particle is called an electric field or electrostatic field.
Theoretically, an electric field due to charge extends to infinity but its
effect practically dies away very quickly as the distance from the charge increases.12.3.1 Electric Intensity or Field Strength
The intensity of an electric field at any point is determined by the force
acting on a unit positive charge (+1C) placed at that point.
If a positive point charge Q0 (also called test charge) is placed at any point
in an electric field and it experiences an electric force F', the electric field
vector E' at the point is the electric force divided by the magnitude of the
test charge Q0
Electric field is an electric force per unit charge. The following points should note:
a) The electric field is a force per unit charge, it is therefore a
vector, it has both magnitude and direction.
b) The electric field can be described (drawn) in terms of lines of
force. Where the lines of force are close together, the intensity
(strength) is high and where the lines are widely separated the intensity (strength) will below.12.2.2 Electric Intensity at a point in Electric Field
The magnitude of the electric field at any point due to a point charge can
be calculated using Coulomb’s Law.
a) Lets consider point charge Q, where Q is positive.
The electric field E' at point P (distance d away) due to an isolated
+Q in a medium of permittivity can be calculated by imaging a very small charge +Q0 to be placed at P. By Coulomb’s Law.
b) Now consider point charge Q where Q is negative (-Q)
The electric field E' at point P (distance d) due to an isolated –Q in a
medium of permittivity can be calculated by imaging a very small charge +Q0 at P.
Conclusion: The electric field due to a point charge always points
away from the positive charge but towards the negative charge.12.4 Electric field lines
Electric field lines (or line of force in an electric field) are an imaginary line
drawn through an area or place. The line is drawn in such a way that the
direction of its tangent is the same as the direction of the electric field at that point.
a) Electric field lines produced by a single positive point charge.
The electric field lines always point away from positive chargeb) Electric field lines produced by a single negative point charge.
The electric field lines always point towards a negative charge.c) Electric field lines produced by two equal and opposite point charges.
Note: The number of lines leaving the positive charge equals the
number entering the negative charge.d) Electric field lines for two equal and positive charges.
e) Electric field lines for two equal and negative charges.
12.5 Electric field strength due to the distribution (superposition) of electric field
Consider many point charges, Q1, Q2, Q3, …Qn and the electric field caused
by the individual point charges E1, E2, E3, ... En respectively.
The resultant electric field at a point P is the vector sum of the field at P
due to each point charge distribution.
12.6 Electric potential
Every charge has an electric force which extends theoretically up to infinity.
Let us consider an isolated charge +Q fixed in space:
If a test charge Qo is placed at infinity, the force on it due to charge +Q is zero.
If the test charge Qo at infinity is moved towards +Q a force of repulsion
acts on it and hence work is required to be done to bring it to a point like
A. The work done by the electric force does not depend on the path taken
by charge Qo, it is only dependent on the initial and final position, i.e. the
electric force is conserved and the work done can be expressed in terms
of potential energy, U.
The work done in bringing Qo from infinity to A is given by:
The electric potential V at a point distance d from the charge Q is the
electric potential energy U per unit charge associated with a test charge
Qo placed at that point:
Hence, electric potential at a point in an electric field is the amount of
work done in bringing a unit of positive charge from infinity to that point, i.e.
Where W is the work done to bring a charge of Coulomb from infinity to
the point of consideration.
Unit: The SI unit of electric potential is volt (V) and may be defined as:
“The potential at a point in an electric field is 1 volt if 1 joule of work
is done in bringing a unit of positive charge from infinity to that point
against the electric field.”12.6.1 Electric Potential Difference
In practice we are more concerned with potential difference (p.d.) between
two points rather than their individual absolute potential.
The potential difference, p.d., between two points may be defined as:
“The potential difference between two points is the amount of work
done in moving a unit of positive charge (+1C) from the point of lower
potential to the point of higher potential.”
Consider two points A and B in the electric field of a charge +Q as shown here:
The potential, V1, at B means that V1 joules of work have been done in
bringing a unit of positive charge from infinity to B.
Let the extra work to bring the unit of positive charge (+1C) from B to A
be in joules, therefore:
potential at A = V2 = V1 + WThe p.d. between A and B is equal to V2 – V1 = V1 + W – V1 = W
The SI unit of p.d. is Volt and may be defined as:
“The p.d. between two points is 1V if 1 joule of work is done in bringing a
unit of positive charge (+1C) from the point of lower potential to the point
of higher potential.”
Let V be the potential difference between point A and B:V = V2 – V1By definition:
So the work done is W = Q(V2 – V1)
Which is the work done for bringing a positive charge Q from B to A.12.6.2 Potential of a Charged Sphere
Fig. 12.2: Field lines of a charged sphereConsider an isolated sphere of radius
r metres placed in air and charged uniformly with Q Coulombs. The field
has spherical symmetry i.e., lines of force spread out normally from the
surface and meet at the centre of the sphere if projected backward. Outside
the sphere, the field is exactly the same as though the charge Q on the
sphere were concentrated at its centre.(i) Potential at the Sphere Surface
Due to the spherical symmetry of the field, we can imagine the
charge Q on the sphere concentrated at its centre, O. The problemthen reduces to finding the potential at a point r metres from a charge Q.
(ii) Potential Outside the Sphere
Consider a point P outside the sphere. Let this point P be at a
distance of D metres from the surface.
Then potential at P:
(iii) Potential Inside the Sphere
Since there is no electric flux inside the sphere, the electric field
inside the sphere is zero. Hence all points inside the sphere are at
the same potential as the points on the surface.
12.7 Relationship between E and V
The effect of any charge distribution can be described either in terms of the
electric field or in terms of the electric potential. Electric potential is often
easier to use since it is a scalar, whereas electric field is a vector.
Fig. 12.3: Electric field between two charged plates
12.8 Charge distribution
12.8.1 The Electric Field due to Continuous Charge Distributions
Fig. 12.4: Partial charge distribution12.8.2 Electric flux
Fig. 12.5: Electric flux through a flat areaWhere E is the electric field (having units of V/m), E is its magnitude, S is
the area of the surface.12.8.3 Gauss’s Law
The total of the electric flux out of a closed surface is equal to the
charge enclosed divided by the permittivity ( εv).
(Where εv= 8.85 × 10–12 P/m)
Fig. 12.6: Electric flux through a sphere (Gauss law)The electric flux through an area is defined as the electric field multiplied
by the area of the surface projected in a plane perpendicular to the field.
Gauss’s Law is a general law applying to any closed surface. It is an
important tool since it permits the assessment of the amount of enclosed
charge by mapping the field on a surface outside the charge distribution.
For geometries of sufficient symmetry, it simplifies the calculation of the electric field.12.8.4 Common Solid conductor shapes and charge distributiona) The Electric Field of a Conducting Sphere
Fig. 12.7: The Electric field of a Conducting Sphere
b) Distribution of Charge over the Surface of a Closed Conductor (Pear)The charge around the conductor is measured using a proof plane.
The proof plane is a small brass disc on an insulated handle.
Fig. 12.8: Distribution of charge on a pear-shaped closed conductorIt is first earthed to ensure there is no charge on the disc. The
proof plane is then touched onto the surface of the conductor. The
proportion of charge acquired by the disc when it is touched ontothe surface of the conduction is proportional to the surface charge
density of the conductor. The charge can be measured by touching
the brass disc of the proof plane on the inside of a metal can standing
on the top of the cap of a gold leaf electroscope. The deflection is an
indication of the charge at the surface at that point.
Fig. 12.9: Transfer of charges to a proof planeIf this process is repeated at various positions on the conductor,
it is found that the least charge is found where the radius of the
curvature of the conductor is greatest and that the charge is greatest
where the surface of the conductor has the smallest radius of the
curvature, i.e. where it is most pointed. If the radius of the curvature
is very small, the density of charge can be large enough for charge
leakage to occur creating a wind of ions.c) Distribution of Charge over the Surface of a cylindrical Conductor
Fig. 12.10: Cylindrical conductord) Distribution of Charge over the Surface of a sharp point
Conductors are materials in which charges can move freely. If
conductors are exposed to charge or an electric field, their internal
charges will rearrange rapidly. For example, if a neutral conductor
comes into contact with a rod containing a negative charge, some
of that negative charge will transfer to the conductor at the point
of contact. But the charge will not stay local to the contact point
-- it will distribute itself evenly over the surface of the conductor.
Once the charges are redistributed, the conductor is in a state of
electrostatic equilibrium. It should be noted that the distribution
of charges depends on the shape of the conductor and that static
equilibrium may not necessarily involve an even distribution of
charges, which tend to aggregate in higher concentrations around
sharp points. This is explained in Fig. 12.11;
Fig. 12.11: Electrical Charge at a Sharp Point of a ConductorForces between like charges at either end of the conductor are
identical, but the components of the forces parallel to the surfaces
are different. The component parallel to the surface is greatest on
the flattest surface and therefore moves charges away from one
another more freely. This explains the difference in concentration of
charge on flat vs. pointed areas of a conductor.
Similarly, if a conductor is placed in an electric field, the charges
within the conductor will move until the field is perpendicular to
the surface of the conductor. Negative charges in the conductor will
align themselves towards the positive end of the electric field, leaving
positive charges at the negative end of the field. The conductor thus
becomes polarised, with the electric field becoming stronger near
the conductor but disintegrating inside it. This occurrence is similar
to that observed in a Faraday cage, which is an enclosure made of a
conducting material that shields the inside from an external electric
charge or field or shields the outside from an internal electric charge or field.12.9 Application of electrostatics
12.9.1 Point discharge (Lightening)
Activity 12.3: Lightening description
Take the case of the phenomena of lightening and thunderstorm,
and then answer to the following questions.
Questions:
1. What do you think are lightening and thunderstorm?
2. Suggest the causes of thunderstorm and lighting.
3. Discuss and explain how one can create a protection from lightening and thunderstorm.
Fig. 12.12: LighteningLightening is a sudden electrostatic discharge during an electrical
storm between electrically charged regions of a cloud (called intra-cloud
lightening or IC), between that cloud and another cloud (CC lightening), or
between a cloud and the ground (CG lightening).
The charged regions in the atmosphere temporarily equalise themselves
through this discharge referred to as a strike if it hits an object on the
ground, and a flash if it occurs within a cloud. Lightening causes light
in the form of plasma, and sound in the form of thunder. Lightening
may be seen and not heard when it occurs at a distance too great for the
sound to carry as far as the light from the strike or flash. For the case of
RWANDA, it is taken as the world’s place with high potential of lightening
and thunderstorms.Activity 12.4: Lightening situation in the world
Read the article below (Fig. 12.13) and answer to the following questions.
Questions:
1. Which part of Rwanda is mostly affected by lightening?
2. Discuss the consequences of lightening in Rwanda.
3. Discuss and comment on how scientists discovered that in
Rwanda there is more lightening than other places in the world.
4. How would Rwandans protect themselves from lightening.
Fig. 12.13: The comparison of lightening in Rwanda and other countries in the world12.8.2 Lightening Arrestors
Activity 12.5: Visiting lightening arrestors
Visit any building which has lightening arrestors and answer the
following questions.
Questions:
1. Why do people use lightening arrestors on their houses buildings?
2. How and when do you think lightening arrestors work?
Fig. 12.14: Lightening ArrestorsA lightening arrester is a device used on electrical power systems and
telecommunication systems to protect the insulation and conductors of
the system from the damaging effects of lightening. The typical lightening
arrester has a high-voltage terminal and a ground terminal. When a
lightening surge (or switching surge, which is very similar) travels along
the power line to the arrester, the current from the surge is diverted through
the arrestor, in most cases to earth.
In telegraphy and telephony, a lightening arrestor is placed where wires
enter a structure, preventing damage to electronic instruments within and
ensuring the safety of individuals near them. Smaller versions of lightening
arresters, also called surge protectors, are devices that are connected
between each electrical conductor in power and communication systems
and the Earth. These prevent the flow of the normal power or signal
currents to ground, but provide a path over which high-voltage lightening
current flows, bypassing the connected equipment. Their purpose is to
limit the rise in voltage when a communication or power line is struck by
lightening or is near to a lightening strike.If protection fails or is absent, lightening that strikes the electrical system
introduces thousands of kilovolts that may damage the transmission lines,
and can also cause severe damage to transformers and other electrical or
electronic devices. Lightening-produced extreme voltage spikes in incoming
power lines can damage electrical home appliances or even cause death.Components
Fig. 12.15: The Simple spark gap device diverts lightening
strike to the ground (earth)A potential target for a lightening strike, such as a television antenna,
is attached to the terminal labelled A in the photograph. Terminal E is
attached to a long rod buried in the ground. Ordinarily no current will
flow between the antenna and the ground because there is extremely high
resistance between B and C, and also between C and D. The voltage of
a lightening strike, however, is many times higher than that needed to
move electrons through the two air gaps. The result is that the electrons
go through the lightening arrester rather than travelling on to the television
set and destroying it.A lightening arrester may be a spark gap or may have a block of a
semiconducting material such as silicon carbide or zinc oxide. “Thyrite”
was once a trade name for the silicon carbide used in arresters. Some
spark gaps are open to the air, but most modern varieties are filled with
a precision gas mixture, and have a small amount of radioactive material
to encourage the gas to ionize when the voltage across the gap reaches a
specified level. Other designs of lightening arresters use a glow-discharge
tube (essentially like a neon glow lamp) connected between the protected
conductor and ground, or voltage-activated solid-state switches called varistors or MOVs.Lightening arresters built for power substation use are immense devices,
consisting of a porcelain tube several feet long and several inches in
diameter, typically filled with discs of zinc oxide. A safety port on the side
of the device vents the occasional internal explosion without shattering the porcelain cylinder.
Lightening arrestors are rated by the peak current they can withstand, the amount of energy they can absorb.12.9.3 Paint spraying, Photocopy Machines/
Xerography and Laser Printers
12.8.3.1 Paint Spraying
Fig. 12.16: A Paint sprayerThe paint is propelled through the gun, rubbing against the side, and gaining a static electric charge.
Fig. 12.17: The function paint sprayingSince the paint particles all have the same charge, they repel each other.
This helps to distribute the paint particles evenly and get uniform coverage.
Usually the object being painted is metal and grounded but almost any
product can be finished electrostatically. A metal object may need to be
placed behind the object to create a ground or it can be sprayed with a
conductive primer. The paint particles have a charge so they are attracted
to the opposite charge of the object being painted. This makes the particles
less likely to stay in the air.
Electrostatic spray painting has distinct advantages. It creates a strong
bond. It also covers a three dimensional object more evenly with a good
edge and wrap around coverage. It saves paint by using the least amount
of paint since it has a high transfer efficiency. Also the finish will look
better because it has a uniform paint thickness.There are some disadvantages as well. Material to be sprayed must be
conductive or made conductive for bonding. Care has to be taken with
this equipment as guns can be delicate and bulky. It requires the use of
grounding as improper usage can be a safety or fire hazard. Lastly it is
more expensive to apply than regular spray painting.12.8.3.2 Photocopy Machines /XerographyActivity 12.6: Printers and photocopier⚫ Visit a printing stationary shop, where they do printing and photocopy and then discuss about the following questions.
⚫ Tell the shop keeper to show you different parts of a printer and photocopier.
Questions:
1. List and describe the different machines you have seen.
2. Discuss and explain how a printer and photocopier function.
3. Explain and discuss the functions of the following parts in
application of electrostatics:
a) Scanner b) Cartridge
c) Fuse d) Toner
4. Differentiate a printer from a photocopier.
Fig. 12.18: PhotocopierA photocopier (also known as a copier or copy machine) is a machine that
makes paper copies of documents and other visual images quickly and
cheaply. Most current photocopiers use a technology called xerography, a dry
process that uses electrostatic charges on a light sensitive photoreceptor to
first attract and then transfer toner particles (a powder) onto paper in the
form of an image. Heat, pressure or a combination of both is then used to
fuse the toner onto the paper. (Copiers can also use other technologies such as ink jet, but xerography is standard for office copying.)How it works (using xerography)
Fig. 12.19: The function of a PhotocopierPhotocopying is widely used in the business, education, and government
sectors. While there have been predictions that photocopiers will
eventually become obsolete as information workers increase their
use of digital document creation, storage and distribution, and rely less
on distributing actual pieces of paper, as of 2015, photocopiers continue
to be widely used. In the 2010s, there is a convergence in some highend
machines between the roles of a photocopier, a fax machine, a scanner,
and a computer network-connected printer. As of 2015, some high-end
machines can copy and print in colour.Schematic overview of the xerographic photocopying process:
1. Charging: The cylindrical drum is electrostatically charged by a
high voltage wire called a corona wire or a charge roller. The drum
has a coating of a photoconductive material. A photoconductor
is a semi-conductor that becomes conductive when exposed to
light.
2. Exposure: A bright lamp illuminates the original document, and
the white areas of the original document reflect the light onto the
surface of the photoconductive drum. The areas of the drum that
are exposed to light become conductive and therefore discharge
to the ground. The area of the drum not exposed to light (those
areas that correspond to black portions of the original document)
remains negatively charged.
3. Developing: The toner is positively charged. When it is applied
to the drum to develop the image, it is attracted and sticks to
the areas that are negatively charged (black areas), just as paper
sticks to a balloon with a static charge.4. Transfer: The resulting toner image on the surface of the drum
is transferred from the drum onto a piece of paper with a higher
negative charge than the drum.
5. Fusing: The toner is melted and bonded to the paper by heat and pressure rollers.
A negative photocopy inverts the colours of the document when
creating a photocopy, resulting in letters that appear white on a
black background instead of black on a white background. Negative
photocopies of old or faded documents sometimes produce
documents which have better focus and are easier to read and study.12.8.3.3 Application of Electrostatics in Laser Printer
Activity 12.7: The printing process with laser printer
Observe the process shown in Fig.12.20 and then try to interrupt the
laser printer when it has started printing and ask someone to help
in opening the printer and remove the cartridge; and check on the
cylinder of the cartridge.
Questions:
Comment and discuss your observations.
Fig. 12.20: The xerography process
The xerography process:
a) The photoconductive surface of the drum is positively charged.
b) Through the use of light source and lenses, an image is formed on
the surface in the form of positive charges.
c) The surface containing the image is covered with a negatively charged
powder, which adheres only to the image area.
d) A piece of paper placed over the charged powder migrate to the
paper. The paper is the heat-treated to fix the powder.
e) A laser printer operates similarly except that the image is produced
by turning a laser beam on and off as it sweeps across the seleniumcoated drum.Laser printing is an electrostatic digital printing process. It produces
high-quality texts and graphics (and moderate-quality photographs) by
repeatedly passing a laser beam back and forth over a negatively charged
cylinder called a “drum” to define a differentially-charged image. The drum
then selectively collects electrically charged powdered ink (toner), and transfers the image to paper, which is then heated in order to permanently fuse the text and/or imagery. As with digital photocopiers and multifunction/ all-in-one inkjet printers, laser printers employ a xerographic printing process. However, laser printing differs from analog photocopiers in that the image is produced by the direct scanning of the medium across the printer’s photoreceptor. This enables laser printing to copy images more quickly than most photocopiers.12.10 Van de Graff generator, electrostatic precipitator
12.10 The Van de Graff Generator
When a charged conductor is placed in contact with the inside of a
hollow conductor, all of the charge conductor is transferred to the hollow
conductor. In principle, the charge on the hollow conductor and its electric
potential can be increased without limit by repetition of the process.This type of generator is used extensively in nuclear physics research.
A schematic representation of the generator is given in Figure 12.21.
Charge is delivered continuously to a high-potential electrode by means of
a moving belt of insulating material. The high-voltage electrode is a hollow
conductor mounted on an insulating column. The belt is charged at a point
by means of a corona discharge between comb-like metallic needles and a
grounded grid. The needles are maintained at a positive electric potential
of typically 104 V. The positive charge on the moving belt is transferred to
the hollow conductor by a second comb of needles at point B. Because the
electric field inside the hollow conductor is negligible, the positive charge
on the belt is easily transferred to the conductor regardless of its potential.
In practice, it is possible to increase the electric potential of the hollow
conductor until the electrical discharge occurs through the air. Because
the “breakdown” electric field in air is about 3 x 106 V/m, a sphere 1m in
radius can be raised to a maximum potential of 3 x 106 V. The potential
can be increased further by increasing the radius of the hollow conductor
and by placing the entire system in a container filled with high-pressure gas.
Fig. 12.21: The Van de Graff generatorVan de Graaff generators can produce potential differences as large as 20
million volts. Protons accelerated through such large potential differences
receive enough energy to initiate nuclear reactions between themselves
and various target nuclei. Smaller generators are often seen in science
classrooms and museums. If a person insulated from the ground touches
the sphere of a Van de Graaff generator, his or her body can be brought
to a high electric potential. The hair acquires a net positive charge, and
each strand is repelled by all the others. The result is a scene such as that
depicted in the Fig. 12.22.
In addition to being insulated from the ground, the person holding the sphere is safe in this demonstration because the total charge on the sphere is very small (on the order of 1μC). If this amount of charge accidentally passed from the sphere through the person to the ground, the corresponding
current would do no harm.Activity 12.8: Threadlike flows of electric wind
When placed in a strong e-field, human hair, eyelashes, and other
sharp objects create tiny coronas which emit "electric wind". These
invisible flows of air are extremely narrow and rapid, and their effects
can be made visible by using dry-ice fog. (Fig. 12.22)
Materials:
⚫ A VDG ( Van de Graaf ) machine
⚫ Wire and tape (or clip-leads)
⚫ Tray of warm water sitting on an insulator
⚫ Chips of dry ice
⚫ Dark paper (submerged in the water for contrast)Procedure:
⚫ Drop several CO2 chips in the water so that a thin layer of fog forms.
⚫ Use tape and a wire to connect the tray to the sphere of your VDG. Charge the tray with respect to the ground.
⚫ Move your hand slowly over the fog, keeping your hand a few centimeters above it. You'll see small mysterious furrows being carved in the fog by the invisible, narrow threads of "electric wind."
⚫ If your hands are extremely clean (no sharp microscopic defects), try wetting your fingers and brush them across fuzzy clothing to pick up some microscopic lint. Or instead try waving
a torn bit of paper over the mist. The sharp paper fibers seem to generate these "threads" of charged air fairly well. If humidity is very low, then perhaps the paper should be made moist.
⚫ Wave your hand fast, and the spots in the mist will follow your
hand's motions. Pull your hand back, and the spots still appear.
⚫ Form a "thread", then wave a charged object near it. The spot in
the mist moves, indicating that the "thread" is being deflected.
⚫ Use a soda straw to blow hard across a "thread". The corresponding spot in the mist will move only a small amount.⚫ Drop some short (1cm) pieces of hair onto the charged water surface. They will stand on one end, emit "threads" upwards, and narrow flows of entrained mist will be seen to project
upwards from the fog layer.
Fig.12.22: The Van de Graff Experiment12.10.2 The Electrostatic Precipitator
Fig. 12.23: The Electrostatic precipitatorOne important application of electrical discharge in gases is the electrostatic precipitator. This device removes particulate matter from combustion gases, thereby reducing air pollution. Precipitators are especially useful in coal-burning power plants and in industrial operations that generate large
quantities of smoke. Current systems are able to eliminate more than 99% of the ash from smoke.Figure 12.23 shows a schematic diagram of an electrostatic precipitator. A high potential difference (typically 40 kV to 100 kV) is maintained between a wire running down the center of a duct and the walls of the duct, which are grounded. The wire is maintained at a negative electric potential with respect to the walls, so the electric field is directed toward the wire. The values of the field near the wire become high enough to cause a corona discharge around the wire; the discharge ionizes some air molecules to
form positive ions, electrons, and such negative ions as O2. The air to be cleaned enters the duct and moves near the wire. As the electrons and negative ions created by the discharge are accelerated toward the outer wall by the electric field, the dirt particles in the air become charged by collisions and ion capture. Because most of the charged dirt particles are negative, they too are drawn to the duct walls by the electric field. When the duct is periodically shaken, the particles break loose and are collected
at the bottom. In addition to reducing the level of particulate matter in the atmosphere (compare Figs. 12.23b and c), the electrostatic precipitator recovers valuable materials in the form of metal oxides.Activity 12.9: Investigating electrostatic precipitator
Materials:
⚫ ½ Teaspoon ground black pepper (or small bits of paper)
⚫ 1 balloon
⚫ 1 sheet of white paper
OR
⚫ Ground black pepper
⚫ 1 piece of plastic (PVC) tubing, 90-150cm long, 6.5cm in diameter
⚫ 1 plastic shopping bag.
⚫ 1 sheet of white paper.Procedure:
1. Give each learner a balloon and some black pepper on a sheet of paper.
2. Ask the students to blow up their balloons and rub them on their hair or a piece of cloth.
3. Hold the balloon over the pepper on the paper. What happens to the pepper? See Fig. 12.24
Fig.12. 24: Materials required for the balloon model electrostatic precipitator. Close-up of pepper grains on the balloon surface.4. Record your observation.
5. Pull the plastic grocery bag through the PVC tube so that the
inside edge of it becomes charged with static electricity.
6. Gently pour some pepper through the tube while holding it
over a piece of white paper. Some pepper should stick to the
inside of the tube (in the same way that it was attracted to
the balloon).
7. Note your observations.
Questions:
1. Does the electrostatic precipitator remove all of the particulates?
2. How does this compare to the efficiency of a wet scrubber? Which one is better?
3. If the precipitators are more efficient, why would you ever want to use a wet scrubber?12.11 Unit 12 assessment
1. Two small spheres spaced 35.0cm apart have equal
charge. How many excess electrons must be present on each
sphere if the magnitude of the force of repulsion between them is
2.20 × 10–21N?2. A point charge q1 = –7 nC is at the point x1 = 0.6m, y1 = 0.8
m, and a second point charge q2 = 4 nC is at the point,. Find
the magnitude and direction of the net electric field at the origin.
x2 = 0.6 m and y2 = 0m
3. What must the charge (sign and magnitude) of a particle of mass 5g be for it to remain stationary when placed in a downwarddirected electric field of magnitude 800nC?
4. What is the magnitude of an electric field in which the electric
force on a proton is equal in magnitude to its weight?
5. A particle has a charge of –8.00nC. Find the magnitude and
direction of the electric field due to this particle at a point 0.5m directly above it.
6. Two particles having charges of 0.70nC and 12nC are separated by a distance of 2m. At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?7. Each square centimeter of the surface of an infinite plane sheet
of paper has 4 × 106 excess electrons. Find the magnitude and direction of the electric field at a point 6.00cm from the surface of the sheet, if the sheet is large enough to be treated as an
infinite plane.
8. What is the strength and direction of the electric field 3.74cm on
the left hand side of a 9.1 nC negative charge?
9. At what distance from a negative charge of 5.536nC would the
electric field strength be 1.90 x 105nC?
10. If it takes 88.3J of work to move 0.721C of charge from a
positive plate to a negative plate, what is the potential difference
(voltage) between the plates?
11. Two parallel oppositely charged plates are 5.1cm apart. The potential difference, in volts, between the plates is 44.6V. Find the electric field strength between them.12. Explain Van De Graff generator?
13. What do you mean by a lightening conductor?
14. What is lightening and thunderstorms?
15. What is Electrostatic Shielding?
16. What are Conductors?
17. What will be the electric field intensity due to a group of charges?
18. What are Electric Lines of Force?- Key unit competenceThe learner should be able to describe arrangement of resistors in a simple electric circuit.
My goalsBy the end of this unit, I will be able to: arrange resistors in a simple electric circuit.
explain the magnetic effect of an electric current.
explain how grounding, fuses, and circuit breakers protect people against electrical shocks and short circuits.
state and explain the effect of electric current.
analyse arrangement of resistors in a simple electric circuit.
construct a simple electric circuit with resistors in series and parallel, ammeter and voltmeter.
Illustrate the effect of electric current.
apply knowledge of safety to prevent circuits from overheating devices (fuses and circuit breakers).
predict what would happen in a house without a fuse or circuit breakers with overloaded electric circuit.
measure electric current and potential difference using an ammeter and voltammeter.Key concepts
1. What do you understand by an electric resistor?
2. What is an electric circuit?
3. How to make your own electric circuit.
4. What are materials needed to form a complete simple circuit?
5. Describe different arrangement of resistors in electric circuits.
6. How do you recognise the domestic electric energy use?
Vocabulary
Simple electric circuit, electric potential, potential difference, electric bell,
electromagnet, electrolysis, heat effect, chemical effect.Reading strategy
After you have read each section, pay attention to every detail especially
the diagrams, illustrations and instructions for every practical work.
Perform calculations on electricity and set up your own circuit diagrams.13.1 Simple circuit elements
Activity 13.1: Making a simple electric circuit
Materials:
⚫ 6-volt battery
⚫ 6-volt incandescent lamp
⚫ Jumper wires
⚫ Breadboard
⚫ Terminal strip
With the following given instructions, use provided materials and present your observations.Schematic diagram and illustration:
Fig. 13.1: Diagram and illustration of a simple electric circuitInstructions:
a) Connect the lamp to the battery as shown in the illustration
above (Fig. 13.1)
b) Connect the lamp as shown in Fig. 13.2 from step (A) to step (D)
Fig. 13.2: Illustration of break in electric circuitc) Using your multimeter set to the appropriate “DC volt” range as
shown in Fig.13.3 below, measure voltage across the battery,
across the lamp, and across each jumper wire. Familiarise
yourself with the normal voltages in a functioning circuit.
d) Now, “break” the circuit at one point and re-measure voltage
between the same sets of points, additionally measuring voltage
across the break like this:
Fig. 13.3: Voltage drop displayQuestions:
1. Which voltages measure the same as before?
2. Which voltages are different since introducing the break?
3. How much voltage is manifest, or dropped across the break?
4. What is the polarity of the voltage drop across the break, as indicated by the meter?
Re-connect the jumper wire to the lamp, and break the circuit in another
place. Measure all voltage “drops” again, familiarising yourself with the voltages of an “open” circuit.Activity 13.2: Using a breadboard in an electric circuit
Materials: Use the same materials as in activity 13.1
⚫ Essential configuration needed to make a circuit.
⚫ Normal voltage drops in an operating circuit.
⚫ Importance of continuity to a circuit.
⚫ Working definitions of “open” and “short” circuits.
⚫ Breadboard usage.
⚫ Terminal strip usage.Procedure:
⚫ Construct the same circuit on a breadboard (see Fig.13.4),
taking care to place the lamp and wires into the breadboard in
such a way that continuity will be maintained.
Fig. 13.4: An Electric circuit using a breadboard⚫ Experiment with different configurations on the breadboard,
plugging the lamp into different holes. If you encounter a
situation where the lamp refuses to light up and the connecting
wires are getting warm, you probably have a situation known
as a short circuit, where a lower-resistance path than the lamp
bypasses current around the lamp, preventing enough voltage
from being dropped across the lamp to light it up. Here is an
example of a short circuit made on a breadboard:
Fig. 13.5: Shorting wire on a breadboard
Fig. 13.6: A Short circuit⚫ Here there is no “shorting” wire present on the breadboard, yet
there is a short circuit, and the lamp refuses to light. Based on
your understanding of breadboard hole connections, can you
determine where the “short” is in this circuit?
Short circuits are generally to be avoided, as they result in very high
rates of electron flow, causing wires to heat up and battery power
sources to deplete. If the power source is substantial enough, a
short circuit may cause heat of explosive proportions to manifest,
causing equipment damage and hazardous to nearby personnel.
This is what happens when a tree limb “shorts” across wires on a
power line: the limb being composed of wet wood acts as a lowresistance
path to electric current, resulting in heat and sparks.
Fig. 13.7: Terminal Strip13.2 Arrangement of resistors13.2.1 Series circuits
Activity 13.3: Investigating series connection
Materials:
⚫ Battery cells
⚫ Three torch light bulbs
⚫ Conducting wires
Procedure:
⚫ Arrange the battery cells as in fig. 13.8 below.
⚫ Connect all the three bulbs in series and switch on.
⚫ Remove one bulb and note your observation.
⚫ Arrange the circuit to have two bulbs, and then one bulb and note your observations.
Fig. 13.8: A series circuitQuestions:
1. What happens in the circuit with three bulbs when one bulb is removed?
2. What happens when the circuit has two bulbs?
3. What happens when the circuit has one bulb only?A series circuit is a circuit in which resistors are arranged in a chain, so the
current has only one path to take. The current is the same through each
resistor. The total resistance of the circuit is found by simply adding up the
resistance values of the individual resistors:Equivalent resistance of resistors in series: R = R1 + R2 + R3 + ...
A series circuit is shown in the diagram above. The current flows through
each resistor in turn.Example:
Find the equivalent resistance in the circuit below:
R = R1 + R2 + R3
R = 3kΩ + 10kΩ + 5kΩ
R = 18kΩ13.2.2 Parallel circuits
Activity 13.4: Investigating parallel connectionMaterials:
⚫ Battery cells
⚫ Three torch light bulbs
⚫ Conducting wires
Procedure:
⚫ Arrange the battery cells as in fig. 13.9 below.
⚫ Connect all the three bulbs in parallel and switch on.
⚫ Remove one bulb and note your observations.
⚫ Remove the second bulb and note your observations.PICTORIAL DRAWING OF A PARALLEL CIRCUIT
Fig. 13.9: A parallel circuitQuestions:
1. What happens in the circuit with three bulbs when one bulb is removed?
2. What happens when the circuit has two bulbs?
3. What happens when the circuit has one bulb only?A parallel circuit is a circuit in which the resistors are arranged with their
heads connected together, and their tails connected together. The current
in a parallel circuit breaks up, with some flowing along each parallel branch
and re-combining when the branches meet again. The voltage across each
resistor in parallel is the same.
The total resistance of a set of resistors in parallel is found by adding up
the reciprocals of the resistance values, and then taking the reciprocal of
the total:
A parallel circuit is shown in the diagram above. In this case the current
supplied by the battery splits up, and the amount going through each
resistor depends on the resistance. If the values of the three resistors are:R1 = 8Ω .R2 = 8Ω and R3 = 4ΩThe total resistance R is found by
Note: that the currents add together to 5A, the total current.13.2.3 Circuits with series and parallel components (In mixture)
If we combined a series circuit with a parallel circuit we produce a Series-
Parallel circuit which makes the arrangement of resistors in mixture.
.R1 and R2 are in parallel and R3 is in series with R1 II R2.
The double lines between R1and R2 are a symbol for parallel.
We need to calculate R1 II R2 first before adding R3.
Many circuits have a combination of series and parallel resistors.
Generally, the total resistance in a circuit like this, is found by reducing
the different series and parallel combinations step-by-step to end up with
a single equivalent resistance for the circuit. This allows the current to be
determined easily. The current flowing through each resistor can then be
found by undoing the reduction process.General rules for doing the reduction process include:
1. Two (or more) resistors with their heads directly connected together and their tails directly connected together are in parallel, and they can be reduced to one resistor using the equivalent resistance equation for resistors in parallel.
2. Two resistors connected together so that the tail of one is connected to the head of the next, with no other path for the current to take along the line connecting them, are in series and can be reduced to one equivalent resistor.
Finally, remember that for resistors in series, the current is the same for
each resistor, and for resistors in parallel, the voltage is the same for each one.13.3 Electric potential and electric potential differenceActivity 13.5: Investigating electric potential
Materials:
⚫ Voltmeter
⚫ Ammeter
⚫ Wire
⚫ Two or three battery cells
⚫ A resistor
Procedure:
⚫ Arrange the simple electric circuit, comprising the above materials.
⚫ The Voltmeter should be parallel to the resistor.
⚫ The Ammeter should be in series with the resistor.
⚫ Use one cell, note and read the value given by the voltmeter.
⚫ Use two cells in series, note and read the value given by the voltmeter.
⚫ Use three cells in series, note and read the value given by the voltmeter.Questions:
1. What do you think the voltmeter is measuring?
2. What is that quantity referred to?
3. Discuss and explain in groups why the number of cells
increase as the voltmeter also changes the value.
4. Repeat the procedure given above but arrange cells in parallel
then discuss and explain the results obtained.Voltage, electric potential difference, electric pressure or electric tension
(formally denoted ΔV or ΔU, but more often simply as V or U, is the
difference in electric potential energy between two points per unit electric
charge. The voltage between two points is equal to the work done per unit
of charge against a static electric field to move the test charge between
two points and is measured in units of volts (V) (a joule per coulomb).
Voltage can be caused by static electric fields, by electric current through
a magnetic field, by time-varying magnetic fields, or some combination
of these three.
Fig. 13.10: A VoltmeterA voltmeter is used to measure the voltage (or potential difference)
between two points in a system; often a common reference potential such
as the ground of the system is used as one of the points. A voltage may
represent either a source of energy (electromotive force), or lost, used, or
stored energy (potential drop).Electrical Energy and Electrical Potential
⚫ In order to bring two like charges near each other, work must be done.
⚫ In order to separate two opposite charges, work must be done.
Remember that whenever work gets done, energy changes form. Electrical
potential energy could be measured in Joules just like any other form of energy.
Since the electrical potential energy can change depending on the amount
of charge you are moving, it is helpful to describe the electrical potential
energy per unit of charge. This is known as electrical potential.Note: this sounds very similar to electrical potential energy, but it is not!)
The energy per unit of charge is often called voltage so it is symbolised by
the capital letter V. Work or energy can be measured in Joules and charge
is measured in Coulombs so the electrical potential can be measured in
Joules per Coulomb which has been defined as a volt.
Example:In this example the amount of work done by the person is 30J, this is also
the amount of electrical potential energy that is possessed by all three
charges together. The electrical potential (not energy) is the amount of energy per unit of charge.
At the original position of the charges they have no energy, so they also
have no electrical potential or 0 volts. Once they are pulled apart, they
have an electrical potential of 10 volts. We could say that the electrical
potential difference from one point to the other is 10 volts.Keep in mind that the electrical potential describes the amount of energy
per unit of charge. This means that when one of the charges is released,
the electric field will do 10 Joules of work on the charge so it will have a
kinetic energy of 10 Joules the instant before it strikes the negative charge.
Remember that 1 Coulomb of charge is a large amount. Also keep in mind
that a Joule is a fairly large unit for work. These units don’t work well if we
are dealing with a small amount of charge. That is why we sometimes talk
about the elementary charge (charge on 1 electron or proton) as another
unit of charge. When we use the elementary charge (e) we need a smaller
unit to measure energy or work in. The unit of the electron volt (eV) was
developed. The electron volt is not a smaller unit for volts!!! It is a smaller
unit for energy. An electron volt is the amount of energy an electron gains
after being accelerated by one volt.Notice that if we look at the equation again for potential difference but use
units of elementary charges (e) and electron volts (eV), we still get units of
volts (V) when we are done.
Remember that:1e = 1.6 x 10-19C this leads us to: 1eV = 1.6 x 10-19JThis portion of the unit contains many easy places to develop
misunderstandings. They have all been addressed earlier on this page but
here is a quick list of them. Please go out of your way to keep these from
becoming a misunderstanding for you:⚫ Electric Potential Energy is not the same as Electrical Potential.
⚫ Electrical Potential can also be described by the terms, potential
difference, voltage, potential drop, potential rise, electromotiveforce, and EMF. These terms may differ slightly in meaning depending on the situation.
⚫ The variable we use for potential difference is V and the unit for
potential difference is also V (volts). Don’t let that confuse you
when you see V = 1.5V
⚫ The electron volt is not a smaller unit of the volt, it’s a smaller unit of the Joule.13.4 Ohm’s law
Activity 13.6: Investigating Ohm’s law
Materials:
⚫ Voltmeter
⚫ Ammeter
⚫ Wire
⚫ Five battery cells
⚫ A resistor
Procedure:
⚫ Arrange the simple electric circuit, comprising the above materials.
⚫ The Voltmeter should be parallel to the resistor.
⚫ The Ammeter should be in series with resistor.⚫ Use one cell, note and read the value given by the ammeter and voltmeter.
⚫ Use two cells in series, note and read the value given by the voltmeter and the ammeter.
⚫ The Three cells in series then four and then five but each time read and record the different values of current and voltage.
⚫ Record your readings in a table like the one below.Ohm’s law states that the current through a conductor between two points
is directly proportional to the voltage across the two points. Introducing
the constant of proportionality, the resistance, one arrives at the usual
mathematical equation that describes this relationship:where I is the current through the conductor in units of amperes, V is
the voltage measured across the conductor in units of volts, and R is the
resistance of the conductor in units of ohms. More specifically, Ohm’s law
states that the R in this relation is constant, independent of the current.
The law was named after the German physicist Georg Ohm, who, in a
treatise published in 1827, described measurements of applied voltage
and current through simple electrical circuits containing various lengths of
wire. He presented a slightly more complex equation than the one above
(see History section below) to explain his experimental results. The above
equation is the modern form of Ohm’s law.13.5 Energy and power
Electrical appliances at home transfer energy from the main supply to
heat and light our homes as well as to operate our appliances such as TV,
Microwave and Computers etc.
The energy used is constant so a TV will use double the amount of energy
in two hours as it will in one hour. The power of an electrical appliance
tells us how much electrical energy it transfers in a second.
Power, P is measured in watts (W) where:1 W = 1 J/s (joule/second).Appliances used for heating have a much higher rating than those used to
produce light or sound.
The amount of energy transferred from the main appliance depends on
the power rating of the appliance and the time for which it is switched on.
Energy transferred from electricity is worked out by:Energy = power × time
E = P × tEnergy, E is measured in:⚫ Joules (J) when the power is in watts and the time, t, is in seconds.
⚫ Kilowatt hours (kWh) when the power is in kilowatts and the time, t, is in hours.
Example: A 800W toaster is switched on for one minute. The energy used is:E = 800 W × 60 s
E = 48000 JExample: A 1.8 kW kettle used for 5 minutes:13.5.1 Paying for Electricity
The units on an electricity bill, measured by an electricity meter, are
kilowatt hours. The cost of each unit of electricity varies. The electricity
bill is calculated by working out the number of units used and multiplying
by the cost of a unit.Cost of electrical energy used
= power in kW × time in hours × cost of one unitorCost = number of kWh used × cost of one unitExample: The 1kW microwave is used for half an hour and the cost of a
unit is 234 Rwf:
Cost = 1kW × 0.5 hours × 234 Rwf/kW h
Cost = 117 RwfPower, Current and Voltage
The main voltage in the UK is 230V. Electrical power depends on the
current and the voltage:
Power = current × voltage
P = I × V
Power is measured in watts (W), current, I, in amps (A) and voltage, V, in
volts (V).
A torch with a 3.0V battery has a current of 0.4A.
Its power is: P = 3.0 × 0.4 = 1.2W13.6 Magnetic effects of electric current
In the previous Chapter on ‘Electricity’ we learnt about the heating effects
of electric current. What could be the other effects of electric current? We
know that an electric current-carrying wire behaves like a magnet. Let us
perform the following Activity to reinforce it.Activity 13.7: Investigation of magnetic effect of electricity⚫ Take a straight thick copper wire and place it between the points X and Y in an electric circuit, as shown in Fig. 13.11
⚫ Place a small compass near to this copper wire. See the position of its needle.
⚫ Pass the current through the circuit by inserting the key into the plug.
⚫ Observe and discuss the change in the position of the compass needle.Fig. 13.11: The Compass needle is deflected on passing an
electric current through a metallic conductorWe see that the needle is deflected. What does it mean? It means that
the electric current through the copper wire has produced a magnetic
effect. Thus we can say that electricity and magnetism are linked to
each other. Then, what about the reverse possibility of an electric effect
of moving magnets? In this unit we will study magnetic fields and such
electromagnetic effects. We shall also study about electromagnets and
electric motors which involve the magnetic effect of electric current, and
electric generators which involve the electric effect of moving magnets.13.6.1 Magnetic field due to a current-carrying conductor
In Activity 13.6, we have seen that an electric current through a metallic
conductor produces a magnetic field around it. In order to find the direction
of the field produced let us repeat the activity in the following way.Fig. 13.12: Compass needle is deflected on passing an electric
current through a metallic conductorActivity 13.8: The magnetic field on a
conductor
Materials:
⚫ A long straight copper wire,
⚫ Two or three cells of 1.5V each,
⚫ A plug key.Procedure:
⚫ Connect all of them in series as shown in Fig. 13.12 (a).
⚫ Place the straight wire parallel to and over a compass needle.
⚫ Plug the key in the circuit.
⚫ Observe the direction of deflection of the north pole of the needle. If the current flows from north to south, as shown in
Fig. 13.12 (a), the north pole of the compass needle would move towards the east.
⚫ Replace the cell connections in the circuit as shown in Fig.
13.12(b). This would result in the change of the direction of
current through the copper wire, that is, from south to north.
⚫ Observe the change in the direction of deflection of the needle.
You will see that now the needle moves in opposite direction,
that is, towards the west [Fig. 13.12(b)]. It means that the
direction of the magnetic field produced by the electric current is also reversed.13.6.2 Magnetic Field due to a Current through a Straight Conductor
What determines the pattern of the magnetic field generated by a current
through a conductor? Does the pattern depend on the shape of the
conductor? We shall investigate this with an activity.
We shall first consider the pattern of the magnetic field around a straight
conductor carrying current.Activity 13.9: Magnetic field along a straight conductor
Materials:
⚫ Battery (12 V),
⚫ a variable resistance (or a rheostat),
⚫ an ammeter (0–5 A),
⚫ a plug key,
⚫ a long straight thick copper wire.Procedure:
⚫ Insert the thick wire through the centre, normal to the plane of a
rectangular cardboard. Take care that the cardboard is fixed and
does not slide up or down. Connect the copper wire vertically
between the points X and Y, as shown in Fig. 13.13(a), in series
with the battery, a plug and key.
⚫ Sprinkle some iron filings uniformly on the cardboard. (You may
use a salt sprinkler for this purpose.)
⚫ Keep the variable of the rheostat at a fixed position and note the
current through the ammeter.
⚫ Close the key so that a current flows through the wire. Ensure
that the copper wire placed between the points X and Y remains vertically straight.
⚫ Gently tap the cardboard a few times. Observe the pattern of the iron filings. You will find that the iron filings align themselves showing a pattern of concentric circles around the copper wire
(Fig. 13.13 (b)).Questions:
1. What do these concentric circles represent?
2. How can the direction of the magnetic field be found?
3. Does the direction of the magnetic field lines get reversed if
the direction of current through the straight copper wire is reversed? Check it.
Fig. 13.13: The electric field around a long conducting wire(a) Fig 13.13 (a) shows a pattern of concentric circles indicating the
magnetic field around a straight conducting wire. The arrow in the circles show the direction of the field lines.
(b) What happens to the deflection of the compass needle placed at a given point if the current in the copper wire is changed? To observe this, vary the current in the wire. We find that the deflection in the
needle also changes. In fact, if the current is increased, the deflection also increases. It indicates that the magnitude of the magnetic field produced at a given point increases as the current through the wire increases.
(c) What happens to the deflection of the needle if the compass is moved from the copper wire but the current through the wire remains the same? To see this, now place the compass at a further
point from the conducting wire. What change do you observe? We see that the deflection in the needle decreases. Thus the magnetic field produced by a given current in the conductor decreases as the
distance from it increases.13.6.3 The Right-Hand Thumb Rule
A convenient way of finding the direction of the magnetic field associated
with a current-carrying conductor is:
Imagine that you are holding a current-carrying straight conductor in your
right hand such that the thumb points towards the direction of current.
Then your fingers will wrap around the conductor in the direction of the
field lines of the magnetic field, as shown in Fig. 13.14. This is known as the right-hand thumb rule.
Fig. 13.14: The Right-hand thumb ruleSolution: The current is in the east-west direction. Applying the right-hand
thumb rule, we get that the direction of magnetic field at a point below
the wire is from north to south. The direction of magnetic field at a point
directly above the wire is from south to north.Trial activities
⚫ Draw magnetic field lines around a bar magnet.
⚫ List the properties of magnetic lines of force.
⚫ Why don’t two magnetic lines of force intersect each other?13.6.4 Magnetic Field due to a Current through a
Circular Loop
We have so far observed the pattern of the magnetic field lines produced
around a current-carrying straight wire. Suppose this straight wire is bent
in the form of a circular loop and a current is passed through it. How
would the magnetic field lines look like? We know that the magnetic
field produced by a current-carrying straight wire depends inversely on
the distance from it. Similarly at every point of a current-carrying circular
loop, the concentric circles representing the magnetic field around it would
become larger and larger as we move away from the wire (Fig. 13.15).
By the time we reach at the centre of the circular loop, the arcs of these
big circles would appear as straight lines. Every point on the wire carrying
current would give rise to the magnetic field appearing as straight lines at
the center of the loop. By applying the right hand rule, it is easy to check
that every section of the wire contributes to the magnetic field lines in the
same direction within the loop.
Fig. 13.15: Magnetic field lines of the field produced by a
current-carrying circular loopThis rule is also called Maxwell’s corkscrew rule. If we consider ourselves
driving a corkscrew in the direction of the current, then the direction of the
corkscrew is the direction of the magnetic field.We know that the magnetic field produced by a current-carrying wire at a
given point depends directly on the current passing through it. Therefore,
if there is a circular coil having 'n' turns, the field produced is n times as
large as that produced by a single turn. This is because the current in each
circular turn has the same direction, and the field due to each turn then
just adds up.Activity 13.10: Magnetic field around a conducting loop
Materials:
Take a rectangular cardboard having two holes. Insert a circular coil
having large number of turns through them, normal to the plane of the
cardboard.
Procedure:
⚫ Connect the ends of the coil in series with a battery, a key and a rheostat, as shown in Fig. 3.16.
⚫ Sprinkle iron filings uniformly on the cardboard.
⚫ Plug the key.
⚫ Tap the cardboard gently a few times. Note the pattern of the iron filings that emerges on the cardboard.
Fig. 13.16: A Magnetic field produced by a current carrying
circular coil13.6.5 A Magnetic Field around a current-carrying SolenoidA coil of many circular turns of insulated copper wire wrapped closely in
the shape of a cylinder is called a solenoid. The pattern of the magnetic
field lines around a current-carrying solenoid is shown in Fig. 13.17.
Compare the pattern of the field with the magnetic field around a bar
magnet (Fig. 13.17). Do they look similar? Yes, they are similar. In fact,
one end of the solenoid behaves as a magnetic North Pole, while the other
behaves as the South Pole. The field lines inside the solenoid are in the
form of parallel straight lines. This indicates that the magnetic field is the
same at all points inside the solenoid. That is, the field is uniform inside
the solenoid.
A strong magnetic field produced inside a solenoid can be used to
magnetise a piece of magnetic material, like soft iron, when placed inside
the coil (Fig. 13.18). The magnet so formed is called an electromagnet.
Fig. 13.17: Field lines of the magnetic field through and around
a current-carrying solenoid
Fig. 13.18: A current-carrying solenoid coil is used to
magnetise the steel rod inside it – an electromagnet13.7 The Heating effect of electricity
Activity 13.11: Investigating the heat effect of an electric current
Materials:
⚫ An Electric kettle or electric iron or electric heater.
⚫ Water
Procedure:
⚫ Take water in a kettle and connect it on a wall socket. Or,
⚫ Connect the iron on the wall socket, or
⚫ Connect the electric heater and lay it in the bucket of water.
Questions:
1. What changes are you observing?
2. Discuss and explain where the heat is coming from.In our daily life we use many devices where the electrical energy is
converted into heat energy, light energy, chemical energy or mechanical
energy. When an electric current is passed through a metallic wire like the
filament of an electric heater, oven or geyser, the filament gets heated up
and here electrical energy is converted into heat energy. This is known as
the ‘heating effect of current’.
Potential difference is a measure of work done in moving a unit charge
across a circuit. Current in a circuit is equal to the amount of charge
flowing in one second.
Therefore, the work done in moving ‘Q’ charges through a potential
difference ‘V’ in a time ‘t’ is given by:Work done = potential difference x current x time
W = V × I × tThe same can be expressed differently using ohm’s law.
According to ohm’s law V = IR
Therefore work can be expressed as:W = VIt
or W = (IR)It = I2 Rt
Thus, heat produced is directly proportional to the resistance, to the time
and to the square of the current.13.7.1 Application of the Heating Effect of Current
The heating effect of current is utilised in electrical heating appliances
such as the electric iron, room heaters, water heaters, etc. All these
heating appliances contain coils of high resistance wire made of nichrome
alloy. When these appliances are connected to power supply by insulated
copper wires then a large amount of heat is produced in the heating coils
because they have high resistance, but a negligible heat is produced in
the connecting wires because the wires are made to have low resistance.Fig. 13.19: An Electric fuse and Iron as heat application of current effectThe heating effect of electric current is utilised in electric bulbs for producing
light. When electric current passes through a thin high resistance tungsten
filament of an electric bulb, the filament becomes white hot and emits light.
An ‘electric fuse’ is an important application of the heating effect of current.
When the current drawn in a domestic electric circuit increases beyond a
certain value, the fuse wire gets over heated, melts and breaks the circuit.
This prevents fire and damage to various electrical appliances.13.8 The Chemical effect of the electric currentActivity 13.12: Investigating chemical effect of electric currentMaterials:
⚫ Three battery cells Bulb
⚫ Switch
⚫ Long conducting wire
⚫ Water
⚫ Table salt
⚫ Beaker
⚫ Two metal electrodesProcedure:
⚫ Pour water in beaker and mix it with table salt.
⚫ Arrange the circuit as shown in Fig. 13.20
⚫ Switch on and make sure the bulb is lighting (to prove that the current is passing).
⚫ Take like two or three minutes and observe the change on the liquid.
Questions:
1. What changes did you observe?
2. Discuss and explain the changes in the liquids.
Fig. 13.20: The Chemical effect of current in electrolysisThe passage of an electric current through a conducting solution causes
chemical reactions. This is known as the chemical effect of electric
current. Some of the chemical effects of electric current are the following:
⚫ Formation of bubbles of a gas on the electrodes.
⚫ Deposition of metal on electrodes.
⚫ Change in colour of solutions.13.8.1 ElectrolysisThe process of decomposition of a chemical compound in a solution when
an electric current passes through it is called electrolysis. The solution that
conducts electricity due to the presence of ions is called an electrolyte.
Two electrodes are inserted in the solution and are connected to the
terminals of a battery with a switch in between. This arrangement is called
an electrolytic cell. The electrode that is connected to the positive terminal
of the battery is called the anode, and the other connected to the negative
terminal is called the cathode. The electrolyte contains ions, which are
charged. The positively charged ions are called cations and the negativelycharged ions are called anions. Cations, being positively charged, get
attracted to the negatively charged cathode and move towards it. On the
other hand, anions, being negatively charged, get attracted towards the
positively charged anode and move towards it. This is how ions move
in an electrolyte and thus conduct electric current. A chemical reaction
takes place at the electrodes. The reaction depends on the metals of
which the electrodes are made and the electrolyte used. As a result of this
reaction, we may observe bubbles at the electrodes due to the production
of gases, deposition of metal on the electrodes or change in the colour of the electrolyte.During electrolysis, the concentration of the electrolyte remains unchanged; the number of electrons extracted at the cathode is equal to the number of electrons supplied at the cathode; since metal atoms are deposited on the cathode, the mass of the cathode increases and the mass of the anode decreases by an equal amount.Electrolysis is used in refining and extraction of metals from impure
samples. This process is called electro-refining. It is also useful in coating
one metal with another. This process is called electroplating.13.9 Unit 13 assessment
1. A combination circuit is shown in the diagram at the right.
Use the diagram above to answer the following questions about
electric current.
a) The current at location A is _____ (greater than, equal to, less
than) the current at location B.
b) The current at location B is _____ (greater than, equal to, less
than) the current at location E.
c) The current at location G is _____ (greater than, equal to, less
than) the current at location F.
d) The current at location E is _____ (greater than, equal to, less
than) the current at location G.
e) The current at location B is _____ (greater than, equal to, less
than) the current at location F.
f) The current at location A is _____ (greater than, equal to, less
than) the current at location L.
g) The current at location H is _____ (greater than, equal to, less
than) the current at location I.
h) Use the diagram above to answer the following questions about
the electric potential difference. (Assume that the voltage drops
in the wires themselves in negligibly small.)
i) The electric potential difference (voltage drop) between points
B and C is __ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points J and K.
j) The electric potential difference (voltage drop) between points
B and K is _____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points D and I.
k) The electric potential difference (voltage drop) between points
E and F is ____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points G and H.
l) The electric potential difference (voltage drop) between points
E and F is ____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points D and I.
m) The electric potential difference (voltage drop) between points
J and K is ____ (greater than, equal to, less than) the electric
potential difference (voltage drop) between points D and I.n) The electric potential difference between points L and A is
_____ (greater than, equal to, less than) the electric potential
difference (voltage drop) between points B and K.2. A battery whose emf is 20V and internal resistance of 1Ω is
connected to three resistors according to the diagram in the
Figure below. Determine:
a) the potential difference to the battery terminals;
b) the current through each resistor and voltages across each
resistor;
c) the power supplied by the emf;d) the power dissipated in each resistor.
3. Use the concept of equivalent resistance to determine the
unknown resistance of the identified resistor that would make
the circuit’s equivalent.Diagram A
4. A parallel pair of resistance of value of 3Ω and 6Ω are together
connected in series with another resistor of value 4Ω and a
battery of e.m.f. 18 V as shown on the fig. (a) below. Calculate
the current through each resistor.
5. (a) Find the equivalent resistance between points a and b in
the figure below.
(b) A potential difference of 34.0V is applied between points a
and b. Calculate the current in each resistor.
6. Use your understanding of equivalent resistance to complete the
following statements:
a) Two 3Ω resistors placed in series would provide a resistance
which is equivalent to one ___ Ω resistor.
b) Three 3Ω resistors placed in series would provide a resistance
which is equivalent to one ___ Ω resistor.
c) Three 5Ω resistors placed in series would provide a resistance
which is equivalent to one ___ Ω resistor.
d) Three resistors with resistance values of 2Ω, 4Ω and 6Ω are
placed in series. These would provide a resistance which is
equivalent to one ___ Ω resistor.
e) Three resistors with resistance values of 5Ω, 6Ω and 7Ω are
placed in series. These would provide a resistance which is
equivalent to one _____ Ω resistor.
f) Three resistors with resistance values of 12Ω, 3Ω and 21Ω
are placed in series. These would provide a resistance which
is equivalent to one _____ Ω resistor.7. As the number of resistors in a series circuit increases, the overall
resistance __________ (increases, decreases, remains the same)
and the current in the circuit __________ (increases, decreases,
remains the same).
8. Consider the following two diagrams of series circuits. For each
diagram, use arrows to indicate the direction of the conventional
current. Then, make comparisons of the voltage and the current
at the designated points for each diagram.
9. As more and more resistors are added in parallel to a circuit,
the equivalent resistance of the circuit ____________ (increases,
decreases) and the total current of the circuit ____________
(increases, decreases).
10. Which adjustments could be made to the circuit below that
would decrease the current in the cell? List all that apply.
a) Increase the resistance of bulb X.
b) Decrease the resistance of bulb X.
c) Increase the resistance of bulb Z.
d) Decrease the resistance of bulb Z.
e) Increase the voltage of the cell (somehow).
f) Decrease the voltage of the cell (somehow).
g) Remove bulb Y. Key unit competence
The learner should be able to explain the working principle of basic
electronic devices.
My goals
By the end of this unit, I will be able to:
define an electronic device.
identify symbols of electronic components.
name different electronic components.
outline the working principle of basic electronic devices.
mention the importance of electronic devices in everyday life.
appreciate the important role of electronic devices in life.
demonstrate knowledge in analysing and modeling physical processes.Key concepts
1. What do you understand by the term electronic?
2. How can you differentiate electronics from electricity?
3. Illustrate different electronic components.
4. What is a motherboard?
5. Describe different examples of electronic devices.Vocabulary
Electronics, motherboard, forwarding bias, reverse bias, transistor, diode,
capacitor, inductor, electronic devices.Reading strategy
After you read each section, pay attention to the paragraphs that contain
definitions of key terms. Use all the information to explain the key terms
in your own words and make drawings of different electronic components
and their functions and suggest different applications on motherboards.15.1 Definition of electronics
Activity 15.1: Investigation about what electronics is.
Take a dictionary and discuss on the following key concepts.
⚫ Electronics
⚫ Electricity
⚫ Conductors
⚫ Semi-conductors
Questions:
1. Give the difference between electricity and electronics.
2. Give the difference between conductors and semi-conductors.
3. When listening to the radio, where do you think voices are
coming from?
4. When watching a television where do you think images are
coming from and how are they coming on your television set?Electronics is the branch of science that deals with the study of flow and
control of electrons (electricity) and the study of their behaviour and effects
in vacuums, gases, and semi-conductors, and with devices using such
electrons. This control of electrons is accomplished by devices that resist,
carry, select, steer, switch, store, manipulate, and exploit the electron.In fact, Electronics is the branch of physics that deals with the
emission and effects of electrons and with the use of electronic
devices. Electronics refers to the flow of charge (moving electrons)
through non-metal conductors; mainly semi-conductors, whereas
electrical refers to the flow of charge through metal conductors.
The Difference between Electronics and Electrical is that: Electronics
deals with flow of charge (electron) through non-metal conductors (semiconductors)
while Electrical deals with the flow of charge through metal conductors.Semi-conductor devices: This is a conductor made of semi-conducting
material. Semi-conductors are made up of a substance with electrical
properties intermediate between a good conductor and a good insulator.
A semi-conductor device conducts electricity poorly at room temperature,
but has increasing conductivity at higher temperatures. Metalloids are
usually good semi-conductors.
Example: Flow of charge through silicon which is not a metal would come
under electronics whereas flow of charge through copper which is a metal
would come under electrical.15.2 Illustration of standard symbols of some electronic components
Activity 15.2: Identifying the electronic devices
Use the provided devices in Figures 15.1 and 15.2 to answer the following questions.
Fig. 15.1: Some electronic devices
Fig. 15.2: Electronic components
Questions:
1. Discuss and write the names of the electronic devices in Fig. 15.1.
2. Discuss the function of each component in Fig. 15.2.
3. Why are devices in Fig. 15.1 recognised to be electronic devices?
Electronic components are basic electronic elements or electronic parts
usually packaged in a discrete form with two or more connecting leads
or metallic pads. Electronic components are intended to be connected
together, usually by soldering to a printed circuit board (PCB), to create an
electronic circuit with a particular function (for example an amplifier, radio
receiver, computer, oscillator) (see Fig 15.1). Some of the main Electronic
components are: resistors, capacitors, semi-conductors (transistor, diode,
operational amplifier, etc) transformers and others (See Fig. 15.2).
Electronic Components are of 2 types: Passive and ActiveActivity 15.3: Identification of electronic components
Search the internet or use a dictionary plus the provided electronic
components in table 15.1 to define and depict the function of each
component and then answer the following questions.Materials:
Motherboard of an electronic device (such as a radio receiver)
Questions:
1. Identify and list down the components that are on the motherboard.
2. Draw their symbols and comment on them.
3. Write down in two columns, which ones are passive and
which ones are active electronic devices.
4. Take any motherboard to find and identify its electronic components.Passive electronic components are those that do not have gain
or directionality. They are also called Electrical elements or electrical
components. E.g. resistors, capacitors, diodes, inductors.
Active electronic components are those that have gain or directionality.
E.g. transistors
Here are some of the Electronic Components and their functions in
electronics and electrical (Table 15.1 below):Table 15.1: Electronic component
15.2.1 Electronic component name acronyms
Electronic Component Name Acronyms: Here is a list of Electronic
Component name abbreviations widely used in the electronics industry:
Activity 15.4: Identifying electronic components on a motherboard
Take the mother board of a radio receiver and try to identify the different
electronic components in Fig. 15.3.
Fig. 15.3: Motherboard
Alternatively referred to as the mb, mainboard, mobo, mobd, backplane
board, base board, main circuit board, planar board, system board, or
a logic board on Apple computers, the motherboard is a printed circuit
board that is the foundation of any electronic device. For a computer, it is
located at the bottom of the computer case. It allocates power to the CPU,
RAM, and all other computer hardware components. Most importantly,the motherboard allows hardware components to communicate with one another.
15.3.1 Ordinary diode
Activity 15.5: Different types of diodes
Using a dictionary and the internet, plus the provided diodes in
Fig 15.4, define and depict the function of each type of diode and
answer the following questions.Materials:
Motherboard of an electronic device (such as a computer).Questions:
1. Write down the application of each type of diode.
2. Take any motherboard to find and identify different types of diodes on it
Fig. 15.4: Types of diodes
In electronics, a diode is a two-terminal electronic component that
conducts primarily in one direction (asymmetric conductance); it has low
(ideally zero) resistance to the flow of current in one direction, and high
(ideally infinite) resistance in the other. A semi-conductor diode, the most
common type today, is a crystalline piece of semi-conductor material
with a p–n junction connected to two electrical terminals. A vacuum tube
diode has two electrodes, a plate (anode) and a heated cathode. Semiconductor
diodes were the first semi-conductor electronic devices.15.4 Current–voltage characteristic
Fig. 15.5: I–V (current vs. voltage) characteristics of a p–n junction diode
A semi-conductor diode’s behaviour in a circuit is given by its current–
voltage characteristic, or I–V graph (Fig. 15.5). The shape of the curve
is determined by the transport of charge carriers through the so-called
depletion layer or depletion region that exists at the p–n junction
between differing semi-conductors. When a p–n junction is first created,
conduction-band (mobile) electrons from the N-doped region diffuse into
the P-doped region where there is a large population of holes (vacant
places for electrons) with which the electrons “recombine”.15.4.1 Forwarding and reverse biasing
are actively being created in the junction by, for instance, light (LEDs);
If an external voltage is placed across the diode with the same polarity as
the built-in potential, the depletion zone continues to act as an insulator,
preventing any significant electric current flow (unless electron–hole pairs
this is called the reverse bias phenomenon. However, if the polarity of the
external voltage opposes the built-in potential, recombination can once
again proceed, resulting in a substantial electric current through the p–n
junction (i.e. substantial numbers of electrons and holes recombine at
the junction). Thus, if an external voltage greater than and opposite to
the built-in voltage is applied, a current will flow and the diode is said to
be “turned on” as it has been given an external forward bias. The diode
is commonly said to have a forward “threshold” voltage, above which it
conducts and below which conduction stops. However, this is only an
approximation as the forward characteristic is according to the Shockley
equation absolutely smooth (see Fig. 15.7)15.4.2 Rectifications
Activity 15.6: Defining rectifier
Use a dictionary and search the internet to identify the meaning
of rectifier and its function.
Materials
A telephone charger
An AC to DC converter
Procedure:Open the charger and identify the component of the motherboard inside it.
Question:
1. When charging a phone why does it require a specific charger plug?
2. Describe and explain different components found on that motherboard.
3. Discuss and comment on the importance of rectifiers.
Fig. 15.6: Half wave rectifier and full wave rectifier
The non-symmetric behaviour is due to the detailed properties of the p-n
junction. The diode acts like a one-way valve for current and this is a very
useful characteristic. One application is to convert alternating current (AC),
which changes polarity periodically, into direct current (DC), which always
has the same polarity. Normal household power is AC while batteries
provide DC; and converting from AC to DC is called rectification. Diodes
are used so commonly for this purpose that they are sometimes called
rectifiers, although there are other types of rectifying devices. Fig. 15.6
(a) shows the input and output current for a simple half-wave rectifier. The
circuit gets its name from the fact that the output is just the positive half
of the input waveform.Fig. 15.6 (b) shows a full-wave rectifier circuit which uses four diodes
arranged so that both polarities of the input waveform can be used at the
output. The full-wave circuit is more efficient than the half-wave one.15.4.3 Zener Diode
Fig. 15.7: Zener diode and its symbolA Zener diode (Fig. 15.7) allows current to flow from its anode to its
down with a reverse voltage but the voltage and sharpness of the knee
cathode like a normal semi-conductor diode, but it also permits current
to flow in the reverse direction when its “Zener voltage” is reached. Zener
diodes have a highly doped, p-n junction. Normal diodes will also break
are not as well defined as for a Zener diode. Also normal diodes are not
designed to operate in the breakdown region, but Zener diodes can reliably operate in this region.
Operation (voltage regulator)
Consider the current-voltage characteristic of a Zener diode with a
breakdown voltage of 17 volts. Notice the change of voltage scale between
the forward biased (positive) direction and the reverse biased (negative)
direction, as shown in the Fig. 15.8 below.
Fig. 15.8: I-V curve for Zener diodeA conventional solid-state diode allows significant current if it is reversebiased
15.4.6 Transistors
above its reverse breakdown voltage. When the reverse bias
breakdown voltage is exceeded, a conventional diode is subject to high
current due to avalanche breakdown. Unless this current is limited by
circuitry, the diode may be permanently damaged due to overheating.
A Zener diode exhibits almost the same properties, except the device is
specially designed so as to have a reduced breakdown voltage, the socalled
Zener voltage. By contrast with the conventional device, a reversebiased
Zener diode exhibits a controlled breakdown and allows the current
to keep the voltage across the Zener diode close to the Zener breakdown
voltage. For example, a diode with a Zener breakdown voltage of 3.2V
exhibits a voltage drop of nearly 3.2V across a wide range of reverse
currents. The Zener diode is therefore ideal for applications such as the
generation of a reference voltage (e.g. for an amplifier stage), or as a
voltage stabilizer for low-current applications.
Activity 15.7: Identifying a transistor
Materials
Motherboard of a radio receiver
Procedure:
Identify transistors from the motherboard among many electronic
components on it.Question:
1. According to its position, try to discern its function there.
2. Where else in other components is a transistor useful?
3. Comment and discuss the further use of a transistor.A transistor is a semi-conductor device used to amplify or switch electronic
signals and electrical power. It is composed of semi-conductor material
with at least three terminals for connection to an external circuit. A voltage
or current applied to one pair of the transistor’s terminals changes the
current through another pair of terminals. Because the controlled (output)
power can be higher than the controlling (input) power, a transistor can
amplify a signal. Today, some transistors are packaged individually, but
many more are found embedded in integrated circuits.Simplified operation
The essential usefulness of a transistor comes from its ability to use a
small signal applied between one pair of its terminals to control a much
larger signal at another pair of terminals. This property is called gain.
It can produce a stronger output signal, a voltage or current, which is
proportional to a weaker input signal; that is, it can act as an amplifier.
Alternatively, the transistor can be used to turn current on or off in a
circuit as an electrically controlled switch, where the amount of current is
determined by other circuit elements.
There are two types of transistors, which have slight differences in how
they are used in a circuit. A bipolar transistor has terminals labeled base,
collector, and emitter.A small current at the base terminal (that is, flowing between the base
and the emitter) can control or switch a much larger current between the
collector and emitter terminals. For a field-effect transistor, the terminals
are labeled gate, source, and drain, and a voltage at the gate can control
a current between source and drain.Transistor as an amplifier
Fig. 15.9: Amplifier circuit, common-emitter configuration with a
voltage-divider bias circuitModern transistor audio amplifiers of up to a few hundred watts are common and relatively inexpensive.
15.5 An example of electronic devices
Activity 15.8: Visiting an Electronic repair workshop
Use any available computer or phone visit a computer or phone
repair workshops to identify different components of a telephone and computer.Question:
Discuss and explain the components such as diodes, resistors,
capacitors, transistors and the function of:
1. A telephone.
2. A computer.
3. A radio receiver.15.6.1 Mobile phone
Fig. 15.10: Mobile phone motherboard
All mobile phones have a number of features in common, but manufacturers
also try to differentiate their own products by implementing additional
functions to make them more attractive to consumers. This has led to15.5.2 Computers
The motherboard is the main component of a computer. It is a large
rectangular board with integrated circuitry that connects the other parts
of the computer including the CPU, the RAM, the disk drives (CD, DVD,
hard disk, or any others) as well as any peripherals connected via the
ports or the expansion slots.
Components directly attached to or part of the motherboard include:
⚫ The CPU (Central Processing Unit), which performs most of the
calculations which enable a computer to function, and is sometimes
referred to as the brain of the computer. It is usually cooled by
a heatsink and fan, or water-cooling system. Most newer CPUs
include an on-die Graphics Processing Unit (GPU). The clock speed
of CPUs governs how fast it executes instructions, and is measured
in GHz; typical values lie between 1 GHz and 5 GHz. Many modern
computers have the option to overclock the CPU which enhancesreat innovation in mobile phone development over the past 20 years.performance at the expense of greater thermal output and thus a
need for improved cooling.
⚫ The chipset, which includes the north bridge, mediates
communication between the CPU and the other components of the
system, including main memory.
⚫ Random-Access Memory (RAM), which stores the code and data
that are being actively accessed by the CPU. RAM usually comes
on DIMMs in the sizes 2GB, 4GB, and 8GB, but can be much
larger.
⚫ Read-Only Memory (ROM), which stores the BIOS that runs
when the computer is powered on or otherwise begins execution,
a process known as Bootstrapping, or “booting” or “booting up.”
The BIOS (Basic Input Output System) includes boot firmware and
power management firmware. Newer motherboards use Unified
Extensible Firmware Interface (UEFI) instead of BIOS.
⚫ Buses connect the CPU to various internal components and to
expand cards for graphics and sound.
⚫ The CMOS battery, which powers the memory for date and time in
the BIOS chip. This battery is generally a watch battery.
⚫ The video card (also known as the graphics card), which processes
computer graphics. More powerful graphics cards are better suited
to handle strenuous tasks, such as playing intensive video games.15.5.3 Watches
Activity 15.9: Identifying the component of a digital watch
Take a digital wrist watch as in Fig. 15.11 and open it to see components inside on the motherboard. Discuss and explain functions of different components you have observed on the motherboard.
Fig. 15.11: Electronic digital watch
Common examples of signals generated by oscillators include signals
broadcast by radio and television transmitters, clock signals that regulate
computers and quartz watches, and the sounds produced by electronic
beepers. This crystal oscillator creates a signal with very precise
frequency, so that quartz clocks are at least an order of magnitude more
accurate than mechanical clocks.15.6 Working principle of basic electronic devices
Principles of Electronics presents a broad spectrum of topics, such as
A system can be described in terms of its expected, desired or intended
atomic structure, Kirchhoff’s laws, energy, power, introductory circuit
analysis techniques, Thevenin’s theorem, the maximum power transfer
theorem, electric circuit analysis, magnetism, resonance, control relays,
relay logic, semi-conductor diodes, electron current flow, and much
more. Smoothly integrate the flow of material in a non-mathematical
format without sacrificing depth of coverage or accuracy to help readers
grasp more complex concepts and gain a more thorough understanding
of the principles of electronics. This includes many practical applications,
problems and examples emphasising troubleshooting, design, and safety
to provide a solid foundation in the field of electronics.
In general, troubleshooting is the identification of the diagnosis of “trouble”
in the management flow of a corporation or a system caused by a failure of
some kind. The problem is initially described as symptoms of malfunction;
and troubleshooting is the process of determining and remedying the
causes of these symptoms.
behaviour (usually, for artificial systems, its purpose). Events or inputs
to the system are expected to generate specific results or outputs. (For
example selecting the “print” option from various computer applications
is intended to result in a hardcopy emerging from some specific device).
Any unexpected or undesirable behaviour is a symptom. Troubleshooting
is the process of isolating the specific cause or causes of the symptom.
Frequently the symptom is a failure of the product or process to produce
any results (nothing was printed, for example). Corrective action can then
be taken to prevent further failures of a similar kind.15.7 Unit 15 assessment
1. A semi-conductor is formed by ……… bonds.
2. A semi-conductor has …………. temperature coefficient of
A. Covalent
B. Electrovalent
C. Co-ordinate
D. None of the above
resistance.
A. Positive
B. Zero
C. Negative
D. None of the above3. The most commonly used semi-conductor is ………..
A. Germanium
B. Silicon
C. Carbon
D. Sulphur4. A semi-conductor has generally ……………… valence electrons.
A. 2
B. 3
C. 6
D. 45. When a pure semi-conductor is heated, its resistance …………..
A. Goes up
B. Goes down
C. Remains the same
D. Can’t say6. Addition of pentavalent impurity to a semi-conductor creates
many ……..
A. Free electrons
B. Holes
C. Valence electrons
D. Bound electrons7. A pentavalent impurity has ………. Valence electrons.
A. 3
B. 5
C. 4
D. 68. Addition of trivalent impurity to a semi-conductor creates many ……...........................
A. Holes
B. Free electrons
C. Valence electrons
D. Bound electrons
9. A hole in a semi-conductor is defined as …………….
A. A free electron
B. The incomplete part of an electron pair bond
C. A free proton
D. A free neutron10. In a semi-conductor, current conduction is due to ……..
A. Only holes
B. Only free electrons
C. Holes and free electrons
D. None of the above11. What is the basis for classifying a material as a conductor, semiconductor?
12. Differentiate semi-conductors, conductors and insulators on the basis of band gap.
13. Is a hole a fundamental particle in an atom?
14. Define a hole in a semi-conductor.
15. Why is it that silicon and germanium are the two widely used semi-conductor materials?16. Which of the two semi-conductor materials Si or Ge has larger conductivity at room temperature? Why?
17. Why does a pure semi-conductor behave like an insulator at absolute zero temperature?
18. What is the main factor for controlling the thermal generation and recombination?
19. In which bands do the movement of electrons and holes take place?
20. Discuss the mechanism by which conduction takes place inside the semi-conductor?Hypertext book
1. htt://www.mkpublishers.com
2. http://en.wikibooks.org/w/index.php?title=Physics_Exercises/
Kinematics
3. http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
4. http://fr.wikipedia.org/wiki/kinemarics
5. http://fr.wikipedia.org/wiki/phys#Recherche-d
6. http://pstc.aapt.org
7. http://www.answers.com/Topic/Physics
8. http://www.answers.com/topic/temperature-conversion
9. http://www.answers.com/Topic/Thermometer
10. http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/newtlaws
11. Http://www.howstuffworks.com/pulley.htm
12. http://www.newworldencyclopedia.org/entry/Freezing_point
13. http://www.school-for-champions.com/science/matterstates
14. www.iop.org
15. www.physicscetral.com
16. www.sciam.com
17. www.worldofteaching.com
18. http://www.dynamicscience.com.au/tester/solutions1/chemistry/
atomic%20structure/
19. https://www.boundless.com/physics/textbooks/boundlessphysics-
textbook
20. http://www.electronicsandyou.com/index.html#sthash.
ZoDTLzgP.dpufBibliography
1. ABBOT A.F, JOHN COCKCROFT, 1989, Physics, Fifth Edition,
Heinemann Educational Publishers.
2. Cunningham, William P, 2001, Environmental science, Mc
Graw-Hill, 6th Ed.
3. HEWITT P.G, SUCH0CKI J., HEWITT L.A., 1999, Conceptual
Physical Science, Second edition, Addison Wesley Longman
4. Holt, Rinehart and Winstone, 1999, Physics
5. MINEDUC, 2006, Ordinary level science curriculum (Biology,
Chemistry, Physics), NCDC, Kigali,
6. PATRICK T., 2004, Mathematics standard level, 3rd Edition,
Ibid Press, Victoria,
7. SILVER Burdett, GINN Inc, 1990, Physical Science, United
States of America,
8. WABWIRE S.H, KASIRYE S., 2007, MK secondary Physics,
student’s book 1, MK Publishers LTD, Kampala-Uganda.- MODEL QUESTIONS File Uploaded 26/04/20, 13:05
- OTHER QUESTIONS File Uploaded 27/04/20, 12:17