Introduction:
The atom has a nucleus which contains two kinds of tiny particles namely protons and neutrons. Orbiting around the nucleus are even smaller particles called Electrons.
Protons, neutrons and electrons are very different from each other, they have their own properties. Protons have POSITIVE charge. Electrons have negative charge and neutrons have no charge, they are neutral. The charge of one proton is numerically equal to the charge of one electron.
When the number of protons in an atom equals the number of electrons, the atom has no overall charge, it is neutral. The protons and neutrons in the nucleus are held together very tightly, while the electrons are held loosely. They can move from one atom to another. An atom that looses electrons has more positive charge than negative charge. It is positively charged i.e. a DONOR atom. An atom that gains electrons has more negative charge than positive charge; it is negatively charged i.e. RECEPTOR ATOM.
Static electricity is the form of electricity produced by the electric charges at rest.
EFFECTS OF STATIC ELECTRICITY.
CONDUCTORS AND INSULATORS.
Some materials hold their electrons very tightly, electrons do not move through them very well, these are insulators e.g. plastic, cloth, glass, dry hair.
Other materials have some loosely packed electrons, which move through them very easily, these are conductors e.g. metals.
NB- When two different materials are rubbed, the electrons may be transferred from one to another. The more rubbing, the more electrons move and the larger, the static charge that builds up. Not the rubbing that causes electrons move but it is simply the contact between two different materials, rubbing just increases the contact area between them.
FUNDAMENTAL LAW OF ELECTROSTATICS.
Different charges attract or push towards each other, same charges repel or push away from each other, hence ”LIKE CHARGES REPEL AND UNLIKE CHARGES ATTRACT.”
METHODS OF CHARGING.
Charging process is either adding or subtracting electrons to/from the body.
For a charge of a given sign, it is necessary to start with an inducing charge of the opposite kind. E.g., a negatively charged rod is brought near the end A of the conductor AB.
Some of the electrons in the conductor are repelled towards B by the excess of electrons on the negative rod creating a negative induced charge at B and a positive induced charge at A. The conductor AB is then earthed by touching it with the finger, the free electrons will be repelled to earth. The finger is then removed from the conductor and after the conductor AB is positively charged.
Consider two brass spheres A and B placed together so that they touch each other and thus form in effect a single conductor. When a charged rod is brought near to A, a + charge is induced on A and – charge on B.
Keeping a charged rod in position, more sphere B a short distance from A.
A charged rod is then removed, when sphere A is tested for charge, it is found to posses positive charge and sphere B contains negative charge.
When two bodies are rubbed against each other, they acquire static charges.
Example:When plastic pen is rubbed with dry hair, the plastic pen acquires negative charge and dry hair acquires positive charge.
When glass rod is rubbed with silk cloth, the glass rod acquires positive charge and the silk cloth acquires negative charge.
THE GOLD LEAF ELECTROSCOPE.
It is a device used to detect and test small electric charges.
Mode of action:
By direct contact of charging by induction, the gold leaf. Within the electroscope indicates the relative amount an object is charged. The leaf will spread out more and more as it gains more and more charges either positive or negative.
When a charged body is brought near the brass cap of the electroscope RISE OR FALL of the gold leaf occurs. DIVERGENCE OR CONVERGENCE of the leaf indicates that the body on the brass cap contains static charges.
Suppose the gold leaf electroscope is charged negatively. When a charged body is brought near the brass cap of the electroscope, if the leaf DIVERGES (rises), from the brass plate, indicates that the body on the cap contains negative charge, or if the gold leaf CONVERGES to the brass plate, the charge on the cap of the electroscope is +ve charge.
If the gold leaf rises or falls very fast, the object on the brass cap contains some static charges in large numbers. Otherwise the body contains static charges in small numbers
DISTRIBUTION OF CHARGES ON THE CONDUCTORS.
Different shapes of conductors have different distribution of charges. The charge density of charged conductors of various shapes such as:
-Spherical.
-Cylindrical.
-Pear shaped.
-Cone.
These can be tested using a PROOF PLANE which consists of metal disk with and insulating handle.
HOW TO TEST THE DISTRIBUTION OF CHARGES.
To test the distribution of charge on the conductor, touch a part of the conductors with a proof plane, then touch the cap of the leaf which diverges and the amount of divergence is directly proportional to the charges collected by the proof plane which is also directly proportional to the charge density at the part of conductor.
Charge on spherical conductions is UNIFORM.
The charges of hollow conductor are distributed on the outside surface ONLY. There is no charge inside the hollow sphere.
More charges seem to be concentrated on sharp ends of conductors than on flat surfaces.
The charges concentrate more on corner points than on flat surfaces of the cylindrical conductors.
ELECTROPHORUS.
It is a device which consists of two metal sheets separated by insulator. The lower metal sheet is typically grounded for best performance and is attached to the insulating material, the upper metallic disc has an insulating handle attached to its centre such that the electrophorus can be dissected without grounding the top place.
MODE OF ACTION.
The top of the insulator is rubbed with cloth creating static charges on the surface of the insulator. The electrons are actually exchanged between the insulator and the cloth through a chemical reaction of the dissimilar materials, the rubbing increases the surface area over which the materials make contact. In order to separate the metal sheets. MECHANICAL WORK is rubbed against the force of attraction between the charges on the metal sheets, and this work becomes transferred to ELECTRIC ENERGY.
LIGHTNING CONDUCTOR.
The device which protects tall buildings from lightning strokes by providing an easier path for the current to flow to earth than through the building. It consists of thick copper wire strip of very low resistance connected to the ground below. A good connection to the tegmental plate deep in damp earth. In the event of a direct lightning strike, the current in the conductor may be so great as to melt or vaporize the metal. The damage to the building will nevertheless be limited.
CAPACITORS AND CAPACITANCE.
A capacitor is a device used to store electric charges. Its symbol is _____^{+}| |^{–}______
The dielectric materials or insulators are placed between the two parallel plates.
CAPACITANCE OF A CAPACITOR.
The charge of Q on either plate of a capacitor is directly proportional to the potential difference between the plates.
Q = CV
Where C is the constant of proportionality called the capacitance of a capacitor.
I.e. C = Q/V
Capacitance C is expressed in Farads, F.
A capacitance of a capacitor is the ratio of the charge on either plate for the potential difference between the plates.
If Q is expressed in coulombs and V in volts then c will be expressed in coulombs per volts or (CV^{-1}) or Farads, F.
Note- 1 CV^{-1 }= 1F
Not only that, but also there are other commonly used SI units such as:
A is the given number. If the number is 4, then 4 x 10^{-6 } F
B is the given number. If the given number is 6 then 6 x 10^{12 } F.
CAPACITANCE OF A PARALLEL PLATE CAPACITOR.
Let Q be the magnitude of charge on each plate.
If the surface area of either plate is A, the distance between the two plates is d.
ARRANGEMENT OF CAPACITORS IN CIRCUITS.
Capacitors can be arranged in either series or parallel.
The capacitors are said to be in series of the polarities of the capacitors alternate. For capacitors in series the charge on each plate is SAME. Applying the relation C = Q/V.
For C1 = Q/V1 -> V1 = Q/C1
C2 = Q/V2 -> V2 = Q/C2
C3 = Q/V3 -> V3 = Q/C3
C4 = Q/V4
C5 = Q/V5
C6 = Q/V6
The total potential difference:
VT = V1 + V2 + V3 + V4
= Q/C1 + Q/C2 + Q/C3 + Q/C4
= Q [1/C1 + 1/C2 + 1/C3 + 1/C4 ]
VT/Q = 1/C1 + 1/C2 + 1/C3 + 1/C4
VT/Q = 1/CT
i.e.: 1/CT = 1/C1 + 1/C2 + 1/C3 + 1/C4
Note: The reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of individual capacitors.
For series, arrangement of capacitors, the equivalent capacitance is always LESS than the smallest capacitance.
Since the p.d is the same across each capacitor, the current divides, then the charge in each capacitor is said to be different.
The total charge QT = Q1 + Q2 + Q3
Using the relation C = Q/V
Q= CV
Q1= C1V
Q2= C2V
Q3= C3V
Thus: QT = C1V + C2V + C3V
QT = V (C1 + C2 + C3)
= QT/V = C1 + C2 + C3
Formulae: = CT = C1 + C2 + C3
Example:
Calculate the equivalent capacitance in the circuit with capacitances 2,4,1.2.
CT= C1 + C2 + C3
= 1.2 + 2 + 4
= 7.2 MF
CHARGING AND DISCHARGING A CAPACITOR.
Charging:
The positive terminal of the battery is connected to the +ve terminal of the capacitor and vice versa (Negative terminal to –ve capacitor).
The voltage builds up between X and Y and opposes the battery voltage charging stops when these voltages are equal, the electron flow i.e. charging current is then zero.
NB- As the time passes, the charges on the capacitor grow.
Discharging:
When the conductor is connected across the charged capacitor, there is a brief flow of electrons from negative plate to +ve plate i.e. from Y to X. The charge stored by the capacitor falls to zero as does the voltage across it.
As the time passes, the charges decrease:
FACTORS AFFECTING CAPACITANCE (of a capacitor).
USES OF CAPACITORS.
CURRENT ELECTRICITY.
Current is the quantity of electricity per second passing past a point.
I = Q/T
The SI unit of current is Ampere, A.
1 ampere = 1 coulomb per second.
Ampere is denoted by ——-A——–
In metals the electricity is carried by electrons. An electron has the magnitude of -1.6 x 10^{-19 }
Calculate the current produced by the charge of magnitude 2.7 x 10^{-18 } passed in 5 seconds.
I = Q/T
= 0.54 X 10^{-18}/5
I = 0.54 X 10^{-18 }Ampere.
POTENTIAL DIFFERENCE.(P.d)
A potential difference, P.d is required to make a current flow. The p.d is measured in volts, V.
The P.d between two points is numerically equal to the energy change in joules when one coulomb is transferred from one point to the other. The instrument used to measure the P.d is voltmeter denoted by V
The instruments used to measure P.d are connected in parallel with the resistor concerned.
RESISTANCE
The opposition to the flow of electric charges in the given conductor, denoted by R. The SI unit of resistance is ohm, Ω
OHM’S LAW.
“The current flowing in a conductor is directly proportional to the P.d provided that the physical state of the conductor remains unchanged.
V = IR
OHMIC CONDUCTORS.
The graph is a straight line passing through the origin slope = change in V/change in I
= ∆ V/ ∆ 1
R = ∆V/∆1
∆ V -> 0 and ∆ I -> 0
R = V/I
V = IR
Example:
Calculate the resistance created when the current of 0.4A is produced by p.d 6v
V = 1R
R = V/I
R = 6v/0.4
= 15 Ω
CONNECTIONS OF SIMPLE ELECTRIC CIRCUITS.
Electric circuit:
Is a continuous path along which electric charges flow from one point to another. The resistors can be connected in either series or parallel electric circuits.
Consider a simple electric circuit below with resistors R1, R2 and R3.
For series arrangement:
-The current through each resistor is same.
-The P.D’s are different.
The total P.d’s VT = V1 + V2 + V3
Ohm’s law: V = 1R1, V100 = 1R100, V619 = 1R619 etc.
And VT = 1RT
Hence 1RT = 1R1 + 1R2 + 1R3 + 1R4
1RT = 1 (R1 + R2 + R3 + R4)
RT = R1 + R2 + R3 + R4
For equivalent, effective or total resistance for series. The effective resistance is always greater than the greatest value of resistance.
Example 1:
Find the equivalent resistance in the circuit with resistances 20,5 and 12.
RT = R1 + R2 + R3
= 20 + 5 + 12
= 37 Ω
NB- The effective resistance in series is greater than the greatest value of the resistance in a whole circuit. Economically, it is so expensive to run series connection.
Consider the circuit below with the resistors R1, R2 and R3 in parallel. The current through each resistor in parallel is different.
The p.d across each resistor is SAME.
The total current IT = I1 + I2 + I3
From Ohm’s law: V = 1R or I = V/R, I1 = V/R1, I4 = V/R4, IT = V/RT
V/RT = V/R1 + V/R2 + V/R3
V/RT = V [1/R1 + 1/R2 + 1/R3]
1/RT = 1/R1 + 1/R2 + 1/R3.
The equivalent resistance in parallel connection.
The equivalent resistance in parallel connection is always less than the smallest value of resistance in the whole circuit.
Special case.
Consider two resistors R1 and R2 in parallel.
1/RT = 1/R1 + 1/R2
1/RT = R1 + R2/R1 R2
1/RT = 1/R1 + 1/R2 = R2 +R1/R1 R2
RT = R1 R2/R1 + R2
FACTORS AFFECTING THE RESISTANCE OF A CONDUCTOR.
The resistance is directly proportional to the length of the conductor. The longer the conductor, the higher the resistance and vice versa.
Resistance is directly proportional to the temperature i.e. R = Temperature. Due to the fact that when the temp increases the charge carriers will acquire more kinetic energy which result in the increase in number of collisions which hinder the free movement of electric current from one end to the other.
The resistance is inversely proportional to the cross sectional area of the conductor. R = 1/A.
The materials have different conducting abilities which result into different resistance. Some conductors conduct electric currents easily and the other not.
Magnetism simply means Magnesia or mineral. In the 4^{th} century B.C the Greek philosopher Thales who observed that a piece of iron was attracted to a natural mineral iron are called Lodestone (Fe_{3 }0_{4}). This behaviour is therefore referred to as magnetism. The lodestone had a tendency of attracting certain metals more strongly at specific points on its surface than at others. These materials are called magnets.
MAGNETS.
A magnet is a substance which can identify North-south directions. The substance has north-south pole.
Types of Magnets.
Compass needle.
The end pointing North is North seeking pole.
The end pointing South is South seeking pole.
PROPERTIES OF MAGNETS.
THE FUNDAMENTAL LAW OF MAGNETISM.
Like magnetic poles repel and unlike magnetic poles attract.
MAGNETIC AND NON-MAGNETIC MATERIALS.
The materials/substances which can be attracted towards magnets are called MAGNETIC SUBSTANCES. The materials are themselves not magnets because they don’t attract each other e.g. iron, steel, cobalt, and nickel.
The materials which can’t be attracted towards a magnet are NON-MAGNETIC MATERIALS. E.g. wood, plastic, glass, zinc, carbon and rubber.
MAGNETIZATION.
Is the process of changing a magnetic material like steel, iron, cobalt to a magnet.
There are 3 main methods:
Stroke the piece of steel with the north pole. Start at end A of the magnet and lift the magnet clear at end B of each stroke. Do this about 10 times in the same direction. Finally end A is magnetized. For double touch, use two opposite poles to stroke at the same time
A magnetic material is magnetized by bringing it into contact with a magnet.
Disadvantages:
Wind about 1 metre of insulated copper wire around a large nail AB, then connect the ends of the wire to the battery.
These magnets are called electromagnets.
ELECTROMAGNETS.
These are used in making devices such as electric bells, relay doors etc.
DEMAGNETIZATION.
Is the process of removing magnetism from a material.
Place the magnet in East-West direction then heat it up and hammer it for some time.
Coil a wire round the magnet and then connect the two ends of the wire to the two terminals of the battery. Then the connections to the battery terminal are quickly interchanged continuously for several minutes.
THE DOMAIN THEORY OF MAGNETISM.
The small portions of a magnetic material which behave like a magnet. These small portions are called DOMAINS. For a magnetic material, domains have North and south magnetic poles aligning in different directions.
For this material to be magnetized, its domains have to be aligned in such a way that their North poles face one common direction and south poles face another common direction.
Magnetic field: The region around a magnet where it has a magnetic effect.
In any magnet there are several invisible lines that extend from north to south pole through the outside of the magnet and then move back to the north pole through the inside of the magnet forming closed loops. These imaginary lines are called magnetic lines of force which form magnetic field of a magnet.
Note – When two similar poles are placed close to each other, there is a point somewhere between them at which the magnetic field lines do not pass. This point is called neutral point. OR When a pair of S-poles are kept close to each other.
Note – A neutral point occurs because the magnetic lines of forces from similar poles do not join each other but repel and avoid each other.
A neutral point can’t form between two unlike poles because the magnetic field lines from one pole join the lines from the other pole.
THE EARTH’S MAGNETISM.
The earth has a magnetic field that extends from deep below earth’s surface to the sky. The magnetic north pole of the earth corresponds to real magnetic south pole and the magnetic south pole corresponds to real north pole.
-The vertical plane containing the axis of a freely suspended magnet at rest under action of earth’s field at any place is magnetic meridian.
-The vertical plane containing the earth’s axis of rotation at any place is called Geographic meridian.
-The angle between direction of alignment of the magnet and the geographic axis is magnetic declination.
-The horizontal direction in the magnetic meridian is magnetic dip-angle of dip.
APPLICATION OF EARTH’S MAGNETIC FIELD.
USES OF MAGNETS.
-The forces which produce a turning effect when applied on the objects are called moments. They allow the objects to rotate about the pivot.
-Moment of force is the size of the force applied on the object to make it rotate.
Moment of force = Force x perpendicular distance from the pivot.
The unit of moment of force is Newton metre – Nm.
-The direction of the rotational motion is either clockwise or anticlockwise.
FACTORS AFFECTING THE TURNING EFFECT.
Example:
When opening a door, a large force can be produced by a comparatively small force provided the perpendicular distance from the pivot is long.
.——————————–.
ß————d————-à f
The longer the perpendicular distance, the smaller the force (F).
.——–.
ß-d-à
The shorter the perpendicular distance, the larger the force (F).
When using a wheel spanner, the force is applied at the far end of the spanner leaving a large allowance between the hand and the turning point, pivot or fulcrum.
Moment of force about a point is the product of the force and the perpendicular distance of its line of action from the point.
MF =F X D
COMBINING PARALLEL FORCES.
Any turning effect causes the rotational motion either positive sense i.e. anticlockwise moment or negative sense i.e. clockwise moment.
If rod AB is free to rotate/turn about the point 0. F1 has +ve moment and F2 has –ve moment about 0. If rod AB has no tendency to turn either clockwise or anticlockwise then +ve and –ve moments are equal in magnitude.
The Principle of Moments.
If a body is in equilibrium under the action of forces which lie in one plane, the sum of clockwise moments is equal to the sum of anticlockwise moments about any moment in that plane provided that no external force acts on the system.
Total clockwise moment = Total anticlockwise moment.
The sum of anticlockwise moments = sum of clockwise moments.
F1X1 + F2X2 = F3X3 + F4X4
Example:
A 100W object is suspended 45cm from the pivot F of a light rod. If a weight W suspended 20.5cm from the pivot balances the 100W, determine the weight W.
Sum of anticlockwise moments = sum of clockwise moments.
100 x 45 = W X 20.5
4500 = 20.5W
W = 45000/20.5
= 220N
CENTRE OF GRAVITY OF A BODY.
The centre of gravity of a body is the point where the force of gravity seems to act.
The centre of gravity of a body is defined as the point through which the resultant of the weights of all the particles of the body acts.
Note- To locate the centre of gravity of the object, when the body is freely suspended in a vertical plane, the centre of gravity and the suspension point lie on the same vertical line.
TYPES OF EQUILIBRIUM.
The three types of equilibrium are:
The equilibrium is the state of balance of the body. The following are the conditions for equilibrium of an object.
The stable equilibrium.
An object is in stable equilibrium when it is slightly displaced, the weight of the object tends to increase the displacement of the object in unstable equilibrium lowers the centre of gravity of the object.
The neutral equilibrium.
A neutral equilibrium of an object is when it is slightly displaced, it remains in its new position without changing the centre of gravity.
APPLICATIONS OF STABILITY.
The position of the centre of gravity of an object has a significant influence on its stability. The lower the centre of gravity, the more the likely is stable equilibrium.
It is any device used to simplify work.
These devices have few or no moving parts. They help to move objects closer, apart or to raise them to different levels e.g. crowbar, pulley, screw jack etc.
In a simple machine, a force EFFORT, E is applied at one point to overcome another force LOAD, L acting at another point.
Simple machines make works easier by allowing one to move a load over an increased distance with less effort.
Mechanical advantage = M.A
M.A of a simple machine is the ratio of the output force to the input force.
M.A = LOAD/EFFORT.
It is a unitless ratio.
Generally the simple machines are designed in such a way that the applied force, EFFORT is less than the load to be overcome.
A man uses a force of 100N to lift a box weighing 500kg. Calculate MA.
W = 500kg x 10
MA = Load/effort
= 5000N/1000N
MA = 5
Note – The lever of a simple machine magnified by man’s weight five times to lift the box.
Example:
A certain machine is designed in such a way that a force of 150N is used to lift a load of 600N, what is the M.A?
MA = L/E
= 600N/150N
M.A = 4
M.A of a simple machine indicates the factor to which the machine will increase or reduce the applied force.
Example:
E = 10N, MA = 4.
MA = L/E
4 = L/10
L = 10 x 4
= 40N
M.A depends on the friction.
Velocity ratio (V.R)
V.R of the simple machine is the ratio of the distance moved by effort to the distance moved by load.
VR = Effort distance/Load distance
It is a unitless ratio.
Suppose the effort arm moves down a distance of 100cm while the load is raised through 25cm at the same time.
VR = ED/LD
= 100cm/25cm
= 4
Example:
In a simple machine, a force of 10N moves down a distance of 5cm in order to raise a load of 100N through a height of 0.5cm. Find VR.
VR = ?
ED = 5cm
LD = 0.5cm
VR = 5cm/0.5cm = 10
VR = 10
EFFICIENCY OF THE SIMPLE MACHINE.
Efficiency of a simple machine is the ratio of the work output to the work input, expressed as % e.g. E = Work output/Work input x 100%
But work output = load x load distance x 100%
And work input = effort x effort distance
E = L/E x L.D/E.D x 100%
E = MA x 1/VR x 100%
E = M.A/V.R x 100%
In a perfect machine, work input is equal to work output, however no machine is ideal, all practical machines have losses mainly due to friction, part of the input work is used to overcome the friction thus the work output is always less than the work input i.e. efficiency of a machine is less than 100%
Example:
A machine having a V.R of 5 requires 600J of work to raise a load of 400N. If the load moved through the distance of 0.5m. Calculate M.A and efficiency of the machine.
Data:
Work input = 600J
Work output = load distance x load
= 400N x 0.5m
= 200J
Efficiency = ?
M.A = ?
V.R = 5
E = M.A/VR x 100%
= Work output/work input x 100%
E = 200/600 x 100
= 33.3%
33.3 = M.A/5 x 100
M.A = 33.3 x 5/100
= 33.3/20
= 1.665
LEVERS.
A lever is a rigid body which when in use turns about a fixed point called a fulcrum (pivot) used to shift heavy loads.
Levers exist in first, second and third class.
The classification depends on the position of the pivot with respect to load and effort.
First class levers.
It has the pivot between load and effort.
Examples:
See-saw
Claw hammer
Crowbar
Scissors.
Second class levers.
It has the load between the effort and fulcrum.
Examples:
Wheelbarrow
Opener
Nut cracker
Third class levers.
It has the effort between the load and fulcrum.
Example:
Tongs
Spade
Fishing rods.
Mechanical Advantage of a lever.
By principal of moments.
Sum of anticlockwise moments = Sum of clockwise moments
Consider 1^{st} class lever.
Load x load arm = Effort x effort arm
Load = effort x effort arm/load arm
Load/effort = effort arm/load arm
But L/E = M.A
M.A = Effort arm/load arm
For 2^{nd} class levers, effort arm is longer than load arm, thus M.A > 1
For 3^{rd} class load arm is longer than effort arm, thus M.A < 1
Note – Third class levers waste part of the force applied since the effort needed to lift the load needs to be greater than the load.
V.R of levers.
Consider 1^{st} class levers.
By principle of moments
Sum of anticlockwise moments = Sum of clockwise moments
Load x load arm = effort x effort arm
Load x load arm/effort = effort arm
Load/effort = effort arm/load arm
Load/effort = V.R
M.A = V.R
-For perfect levers, the efficiency is 100%
-But for imperfect levers efficiency is less than 100% due to FRICTION.
PULLEYS.
It is a grooved wheel which is free to turn about an axle fixed in a frame. The rope passes over the pulley and is held in position by the groove. The load is attached to one end while effort is applied to the other end. Pulleys are useful for lifting loads vertically. They are also used to change the direction of applied force.
SINGLE FIXED PULLEY.
An effort E is applied to one end of the rope so as to raise the load, L. It is used to lift small objects like a flag up a pole.
When the effort is applied, tension is developed in the rope so that the load is raised up.
If frictional forces are negligible, the tension in the rope is equal to the applied force (effort) which is equal to the load raised.
M.A of pulley = L/E
= E/E = 1.
M.A of fixed pulley system = 1.
Note- A single fixed pulley can’t be used to lift load which is greater than the weight of the person lifting the load.
V.R of single fixed pulley
V.R = Effort distance/load distance.
But for a single fixed pulley effort distance equals to load distance.
V.R = Effort distance/Effort distance = 1
SINGLE MOVEABLE PULLEY.
The tension, T in the rope is same throughout and it is equal to the effort, E. Since the load is supported by 2 sections of the rope, then:
L = 2T (Neglect friction weight of pulley and rope.)
M.A = Load/effort
= 2/T
M.A = 2
V.R = Effort distance/load
= x/1/2 x
V.R = 2
Fixed pulley:
V.R = 1
M.A = 1
Movable pulley:
V.R = 2
M.A = 2
The block and tackle pulley system.
When two or more pulleys are fixed in a frame, a block is formed, fixed independently on separate axles. Two blocks can be connected by a single rope passing round on all the pulleys to form a block and tackle system. The lower block is free to move, these are used in cranes and elevators for moving heavy loads.
Note – The tension in each section of the rope is equal to the effort. The total tension in the rope sections supporting the movable block is equal to the load.
Load = 4 x E
The number of sections the rope supporting the movable block is equal to the number of pulleys in the system.
Load = E x 4
L = 4E
M.A = L/E
E x No. of pulleys/E
M.A = No. of pulleys
V.R = No. of pulleys
V.R = Effort distance/load distance
= x/1/4x
V.R = 4
For perfect block and tackle pulley system, M.A = V.R = No. of pulleys
For imperfect block and tackle pulley system
M.A < V.R
E = M.A/V.R x 100%
Example:
A block and tackle pulley system has 4 pulleys, if a load of 200N is raised by using a force of 75N, find M.A of efficiency of the system.
VR = 4
E = 75N
L = 200N
MA = L/E
= 200/75
= 8/3
= 2.67
E = MA/VR x 100
= 2.67/4 x 100
= 267/4
= 67%
Block and tackle pulley system:
Inclined plane.
A sloping plane surface usually a wooden plank used to raise heavy loads by pulling or pushing them along the surface of the plane.
V.R of inclined plane = Length of plane/vertical height
MA = L/E
Note – Load x load distance = Effort x effort distance
Load/effort = effort distance/load distance
M.A = V.R
MA = VR only when the plane is 100% efficient. But there is always frictional force resisting the movement of the load up the plane. Thus MA < VR because part of the effort applied on the load is used to overcome the frictional force.
Example:
A force of 600N is used to move 3000N up on inclined plane. Slant height is 18m and vertical height is 3m. Find:
a.) V.R and M.A
VR = 18m/3m
= 6
MA = L/E
= 3000/600
= 5
b.) Efficiency:
= MA/VR x 100%
= 5/6 x 100
= 83.3%
Screw and screw jack.
The screw consists of a cylinder with a spiral ridge which runs around it. The spiral ridge is known as thread, T. The distance between two adjacent threads is Pitch, P.
Screw jack is a weight lifting machine acting by screw.
Effort is applied at the end of the handle and the load L is made to act at the top of the screw. When the handle makes one complete turn, the effort E raises the load through a distance equal to the pitch of the screw. If R is the length of the handle then VR = Circumference of circle of radius/pitch of screw.
VR = 2∏R/P
MA = L/E
For practical situation, MA < VR
Efficiency = MA/VR x 100%
I.e. W < 100%
Example:
A screw jack with a pitch of 0.2cm and the handle of length 50cm is used to lift a car of weight 1.2 x 10^{4. }If the efficiency is 30%, Find:
a.) VR and MA
VR = 2∏R/P
= 2 x 22/7 x 50/0.2
= 2200/1.4
= 1571.4
MA = W = MA/VR x 100
= 30 = MA/1571.4 x 100
0.3 = MA/1571.4
MA = 471.4
b.) Effort
E = 1.2 x 10^{4}/471.4
= 12000/471.4
= 25.5N
WHEEL AND AXLE.
Consists of wheel and code mounted with the same axis of rotation. Radius of wheel is greater than radius of axle. The effort is applied to a string wound round the wheel while the load is attached to a string wound round the axle in opposite direction to that of the string on the wheel.
Wheel:When the wheel completes one turn, the axle rotates once, thus the effort and load move distances 2∏R and 2∏r respectively.
V.R = Effort distance/load distance
= 2∏R/2∏r
VR = R/r
Wheel/axle
MA = L/E
But M.A < V.R due to friction in bearings.
Examples:
Wheel spanner the brace and gear wheels.
Example:
A wheel and axle of efficiency 80% is used to raise a load of 2000N. If the radius of wheel is 50cm and of axle is 2cm, find:
a.) MA
E = MA/VR x 100%
80 = MA/25 x 100
0.8 = MA/25
= 0.8 x 25
= 20
b.) VR
= R/r
= 50/2
= 25
c.) Effort required to overcome load.
MA = L/E
20 = 2000/E
E = 100N
Hydraulic press:
When effort is applied downwards on the effort piston of radius, r, the load piston of radius, R, lifts the load L.
Note – The pressure on the effort piston equals the pressure on the load pistol.
E/∏r^{2 }= L/∏r^{2}
L/E = ∏r^{2}/∏r^{2}
MA = R^{2}/ R^{2}
MA = (R/r)^{2}
If the friction is neglected:
Work done by effort = work done on load
E x X = lxy
x/y = l/e
VR = L/E = R^{2}/ r^{2}
^{ }
MA < VR due to friction
E < 100%
E is less than 100% due to friction between the cylinder and walls.
Distance and displacement.
Distance is the length covered by a moving object from one point to another without considering the direction.
The distance is a scalar quantity, i.e. quantity only.
The SI unit of distance is METRE M. Other units are km, cm and mm.
Displacement is the length covered by a moving object from a fixed point to another in a specific direction.
The displacement is a vector quantity i.e. both magnitude and direction. The SI unit is Metre. Other units are km, cm and mm.
Speed and velocity.
Speed is the distance moved by an object per unit time.
Speed = Distance moved/time used
Velocity = S/T
SI unit of speed is metre per second (m/s) or km/h.
Speed is a scaler quantity.
Speed is the rate of change of distance moved with time.
Velocity is the displacement moved per unit time.
V = Displacement moved/time taken
V = S/T
The SI unit of velocity is m/s or km/h. Velocity is a vector quantity.
Example:
A body covers 480m in 6 seconds. Find speed.
S = D/T
S = 480/6
S = 80m/s
Note – A body is said to move with uniform speed if it moves through equal distances at equal intervals of time.
ACCELERATION.
Due to traffic conditions, a driver can apply brakes in order to decrease the speed of the car, or to stop it, he may also use an accelerator in order to increase its speed. Car is said to accelerate if velocity increases with time and is said to decelerate if speed decreases with time.
Acceleration is rate of change of velocity with time.
Acceleration = Change in velocity/time taken
A = ∆V/T
Where ∆V = Final velocity – initial velocity
∆V = V – U
A = V – U/T
SI unit of acceleration is m/s ^{2 } or km/hr^{2}
^{ }
Retardation is the negative acceleration (Deceleration).
If rate of change of velocity is constant, then acceleration is uniform.
Example:
A car with a velocity of 90km/hr is uniformly retarded and brought to rest after 10s. Find acceleration.
Initial velocity, U = 90km/hr
90 x 10/3600
= 25 m/s.
Final velocity, V = 0 m/s
Time taken = 10s
Acceleration = v – u/t
= 0 – 25/10
= -25/10
A = -2.5m/s^{2}
Retardation = 2.5m/s^{2}
^{ }
Note- the –ve sign means that the car is retarding.
Equations of uniformly accelerated motion.
1^{st} equation of motion:
In practical reality, the bodies move with changing velocities, i.e. they are accelerated with time. If a body is accelerated from rest i.e. U = 0m/s
V = final velocity
U = initial velocity
A = acceleration
T = time
A = v – u/t
At = v – u
V = u + at … i (equation 1)
2^{nd} equation of motion:
Area under velocity – Time graph represents the distance moved by an object.
S = Area of ∆ABC + Area of rectangle ACTO
= (½ bh) + (lw)
= ½t (v-u) + t x u
= ½t [u + at – u] + ut
= ½t (at) + ut
= ½ at^{2} + ut^{2}
S = Ut + ½at^{2}
S = Distance
3^{rd} equation of motion:
V = u + at
V^{2 }= (u + at)^{2}
V^{2 }= (u + at) (u + at)
= u^{2} + uat + uat + a^{2 }t^{2}
= u^{2 }+ 2uat + a^{2 }t^{2}
= u^{2 }+ 2a (ut + ½at^{2})
But ut + ½at^{2 }= S
V^{2 }= U^{2 }+ 2as
Example:
A body moving with velocity of 30m/s is accelerated uniformly to a velocity of 50m/s in 5 seconds. Find acceleration.
U = 30
V = 50
T = 5
V = u + at
50 = 30 + a(5)
5a/5 = 50-30/5
= 20/5
= 4m/s^{2}
S = 30 x 5 + ½ x 4 (5)^{2}
= 150 + 2 x 25
= 150 + 50
= 200m
THE VELOCITY TIME GRAPHS.
The area under the velocity time graph represents the total distance travelled by the moving object.
Area:Use the mathematical formula to find the area of the geometrical figures formed e.g.:
Triangle = ½ x b x h
Trapezium = ½ (a + b) h
Example:
A car accelerates from a speed of 80m/s to 120m/s in 1 minute. It then moves with the same speed for 20secs and then finally stops after 2 minutes. Draw the velocity time graph. Calculate:
a.) Deceleration of the car.
= 0 – 120/120 = -120/120
= -1m/s^{2}
Deceleration = 1m/s^{2}
^{ }
b.) Total distance moved by the car.
Area of A = ½ (a + b) h
= ½ (80 + 120) 60
= 200 x 30
= 6000m
Area of B = l x w
= 120 x 20
=2400m
Area of C = ½bh
= ½ x 120 x 120
= 7200m
Total = 7200 + 2400 + 6000 = 15600m
= 15600m
Acceleration Due to gravity.
Consider a feather and an iron ball. When these two bodies are allowed to fall down in presence of vacuum, they fall down at the same time, but in presence of air resistance a feather falls down more slowly than an iron ball because the feather is affected by air resistance while the iron ball is considered to be falling freely.
NB- The objects are pulled down towards earth’s centre by acceleration due to gravity. G = 9.8m/s^{2} or 10m/s^{2}
When an object is thrown vertically upwards with an initial velocity, U, at the maximum height reached, the velocity of object becomes zero.
Equations of motion of a free fall.
When an object falls freely from rest, a = g = 9.8/10m/s^{2 }and U = 0m/s
From V = U + at
V = 0 + gt
V = gt
S = Ut + ½ at^{2}
S = ½gt^{2}
V = U^{2} + 2as
= 0 + 2gs
V^{2 }= 2gs or V =√2gs
NB – When an object is thrown vertically upwards, an acceleration a = -g
From V = U + at
V = U – gt
S = Ut + ½at^{2}
S = Ut – ½gt^{2}
^{ }V^{2 }= U^{2 }+ 2as
V^{2 }= U^{2 }– 2gs
Example:
A body is released from rest at a certain height above the ground, if the boy strikes the ground with velocity of 60m/s, find height from which the body was released and time.
U = 0m/s
V = 60m/s
A = 10/ms^{2}
^{ }
V^{2 }= 2gs
S = V^{2}/2g = 60^{2}/2 x 10
= 3600/20
=180m
V = gt
T = v/g
= 60/10
= 6 secs.
Inertia.
Is the tendency of a body to remain in a state of rest unless acted upon by an external force. This is the resistance of an object that has to change its state of motion.
NEWTON’S FIRST LAW OF MOTION.
“Everybody will continue to be in a state of rest provided that no external force acts on the body causing it to change that state.”
NB- Newton’s first law of motion is sometimes known as Law of Inertia.
Examples:
Demonstration:
MOMENTUM.
Momentum of the body is the product of mass and its velocity.
Momentum = mass x velocity
P = MV
If mass is in kg, velocity in m/s, SI unit will be kgm/s.
The linear momentum can be changed if an external force is applied.
Linear momentum.
Linear momentum of the body is momentum of the body whose direction is constant and direction is fixed.
Example:
A car has mass 1500kg moves with velocity 20m/s. Find momentum.
P = MV
= 1500 x 20
= 30,000 kgm/s
NEWTON’S SECOND LAW OF MOTION.
“The rate of change of momentum of the body is directly proportional to the applied force and takes place in the direction of the force.”
Initial momentum = MU
Final momentum = MV
Change of momentum = MV – MU
Rate of change of momentum = MV – MU/ T
I.e. F = MV – MU/T
F = M (V-U)/T
V-U/T = Acceleration.
F = Ma
F = KMa
When F =1N, M 1kg and a = 1m/s, then K = 1
Thus: F = ma
Example:
Find net force of a car whose mass is 2000kg and acceleration is 4m/s
F = ma
F = 2000 x 4
= 8000N
AN IMPULSE OF A FORCE.
Consider F = ma
Ba = v – u/T
F = m (v – u/t)
FT = m (v – u)
FT = MV – MV
The quantity FT is known as an impulse of a force which is numerically equal to the change in momentum, the SI unit of impulse is Ns (Newton second.)
Example:
If a can has a net force of 600N, find the impulse of force after 20 seconds.
Ft = I
I = 600 x 20
=12000Ns
NEWTON’S THIRD LAW OF MOTION.
“To every action there is an equal and opposite reaction.”
The book presses the table with the force (Mg), the table supports the book with the upward force.
Note:
Weight -> Action forces.
Upward force -> Reaction force.
Example:
A person jumping from the boat, the jumping action produces a backward force on the boat, thus widens the gap between the boat and the land. The person may fall into the water.
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM.
States that “When two or more bodies act upon one another, their total momentum remains constant provided that no external force acts on the system.”
Consider two bodies of masses M1 and M2.
Impulse of force due to: M1 is F1T and that due to M2 is M2T
F1T = M1V1 – M1U1
And
F2T = M2V2 – M2U2
So F1T = -F2T
M1V1 – M1U1 = – (M2V2 – M2U2)
M1V1 – M1U1 = -M2V2 + M2U2
M1U1 +M2U2 = M1V1 + M2U2
Total momentum before collision = Total momentum after collision.
Example:
A shot of mass 100kg leaves a cannon of mass 5 tons with velocity of 100m/s. Find the velocity of the recoil of the cannon.
M1U1 = M2U2
100 x 100 = 5000 x U2
10,000 = 5000U2
2 = U2
U2 = 2m/s
Note: For inelastic collision U1 > U2
M1U1 + U2M2 = (M1 + M2) V
Where V = common or joint velocity.
Example:
Two bodies of masses 50kg and 80kg moving in same direction with velocity 20m/s and 10m/s respectively. Calculate joint velocity after collision.
M1U1 + M2U2 = (M1 + M2) V
50 x 20 + 80 x 10 = (50 + 80) V
1000 + 800 = 130V
1800 = 130V
V = 1800/130
= 13.8m/s
Note:
Elastic collision is the collision whereby bodies don’t remain together after collision.
Inelastic collision is the collision whereby bodies join together after collision and move with common velocity.
Work is physical or mental activity.
In physics work is done when the point of application of a force moves.
I.e. Work = force x distance moved in the direction of the force.
Work done = F x d
Force is N thus work = joules.
Distance is M
Example:
Cameron pushes a vertical wall, the wall does not move thus he does not do any work on the wall.
Example:
A bag of rice which weighs 900N is lifted to a height of 2m. Calculate the work.
W = Fd
= 900 x 2
= 1800J
Energy is the ability to do work.
Forms of energy
Chemical energy.
Mechanical energy.
Heat energy.
Sound energy.
Electrical energy.
Nuclear energy.
Light energy.
Geothermal energy.
Wind energy.
Solar energy.
The SI unit of energy is JOULE.
MECHANICAL ENERGY:
Two types:
POTENTIAL ENERGY.
Is possessed by a body due to its position.
FACTORS AFFECTING POTENTIAL ENERGY.
Potential energy depends on mass of the body.
P.E = M
P.E = h
P.E = m = h
P.E = gmh
P.E = mgh
Example:
A ball of mass 0.5kg is kicked vertically upwards and rises to a height of 5m. Find the PE acquired by the ball.
Mass = 0.5kg
Height = 5m
Acceleration = 10m/s
P.E = mgh
= 0.5 x 10 x 5
Potential energy = 25J
KINETIC ENERGY.
Kinetic energy is the energy possessed by the body due to its motion.
K.E = ½mv^{2}
M = Mass of the body.
V = Velocity of the body.
Example:
A bus of mass 500kg is moving with velocity 20m/s. Calculate its K.E
KE = ½mv^{2}
= ½ x 500 x 20^{2}
= 250 x 400
= 100,000J
PRINCIPLE OF CONSERVATION OF ENERGY.
Energy can’t be created nor destroyed, it can only be changed from one form to another.
E.g.
PE to KE
Mechanical to electrical
Electrical to light
POWER.
Power is the rate for doing work or the rate at which energy is consumed.
I.e. Power = Energy consumed/time taken
or
Power = Work done/time taken
P = W.D/T
The SI unit of power is J/S or Watt.
Other units are kilowatt, megawatt and Gigawatt.
i.e. 1kw = 1000w
1mw = 1,000,000W
For engineering purposes horsepower (hp) is a practical unit of power.
i.e. 1 hp = 746w
1 hp = ¾kw
Example:
A man raised a bag of rice of mass 90kg from the ground to a height of 2m in 5 secs. Find the power.
90 x 10 = 900
P = W.D/T
900 x 2 /5
= 3600W
Sustainable sources of energy.
Energy sources are areas of origin of a particular kind of energy. These are:
Renewable sources:
These are sources, once used can be replaced.
Non renewable:
These are sources once used cannot be replaced.
Sustainable sources are used in production of electricity without destroying the environment.
These include:
NB – If people adhere to the policy of afforestation wood falls in this category and not deforestation alone. People should be keen on the increasing, conserving and restoring forest which are major sources of wood energy.
Is used more than any other renewable source for producing electricity.
In Tz, HEP stations are found in Mtera, Kidatu, Kihamsi etc.
HEP is clean and useful source of energy. It produces no waste products.
Importance of HEP.
Applications of HEP.
Used to drive machines in manufacturing goods.
In homes, schools, hospitals and offices.
Electric cookers, heaters use electricity.
Used to maintain some medical processes e.g. to run incubators.
Disadvantages:
Is the energy from the Sun. It can be converted to electricity by using solar cells, (photovoltaic cells, photo electric cells). The main surface of the solar panel is dull black to enhance max absorption of radiant energy from sun.
Applications.
Electricity can be trapped from blowing wind by building a tall tower with a large propeller on top (wind mill) at open places. Strong winds break trees and destroy houses.
Applications.
Disadvantages.
Sea waters are continuously moving. They are powerful sources of energy to drive generations. Tides moves huge amount of water hence source of energy.
Energy generated by flow of heat from the earth’s surface, associated with the areas of frequent earthquakes and high volcanic activity e.g. rift valley in Kenya is well structured to produce efficient geothermal systems for heating buildings, water and driving generators.