Course no: Physics 11Course title: “Mechanics and Heat”Course description: (4, 3, 3)

This course covers the following topics.

All topics in mechanics and heat.Course outline:

Chapter Ⅰ: Introduction to Newtonian mechanicsChapter Ⅱ: Vector and ScalarChapter Ⅲ: Composition and Resolution of ForcesChapter Ⅳ: Equilibrium and ForcesChapter Ⅴ: KinematicsChapter Ⅵ: Newton’s law of motion Chapter Ⅶ: FrictionChapter Ⅷ: Work, Energy, and PowerChapter Ⅸ: Simple Machine

Course reference: * Modern Physics with Study Guide

By: Arthur Belset* University Physics

By: Freedman, Young, and Zemanzky* Textbook in Physics

By: Physics Committee

Course Grading System:Major Exam → 90%Assignment / Seatwork → 10%

100%Course Regulation:

1. Get 50% on its equivalent in order to pass.2. Incurred of 20% of the total no. of hours will automatically DROPPED.3. Late 15 minutes before time consideredABSENT.4. Wearing of uniform is strictly implemented.5. Wearing of sandos, sandals, slippers are not allowed.6. No earing for the boys.7. Others see in students hand book.

Chapter Ⅰ: “Introduction to Newtonian mechanics”

Chapter Objectives:At the end of this chapter, one should be able to:

1. Appreciate the relevance of physics.2. Explain why Physics is an experimental science.3. Show the proper presentation of magnitude by rounding off numbers.4. Use one system of units to another system interchangeably.5. Use prefixes multiples of SI units.

1.1 Physics is the science of matter and energy. Its field is restricted to those phenomena in which the nature of bodies is not changed. Physics is a precise science and its natural language is mathematics.

1.2 Parts of Physics. Physics is classified into groups according to the dominant form of energy involved. The divisions most frequently used are: Mechanics, Heat, Sound, Magnetism, Electricity, Light, and Nuclear.

1.3 Physics as an Experiment Science. By experimental science we should an arrangement of facts and law derived from those facts, and descriptive of certain foiled of knowledge. Our understanding of the physical world has as its foundation experimental measurements and observations, on these are based our theories our facts and deepen our understanding of nature.

1.4 Scientific Method . The coupling of observations, reason and experiment.1.5 Measurement .Measurement essentially a comparison. To measure is to compare a magnitude

with another of the same unit to as certain how many times the second is contained in the first.1.6 Fundamental Concepts .

1.6.1 Matter. The main manifestation of a matter are: it occupies space, it possesses inertia, is subjected to gravitation.

1.6.2 Mass . The quantity of matter that contains a body.1.6.3 Force .Is a push or pull exerted on a body.1.6.4 Motion . Is the change in position of a body with respect to another body in time, or of

some of its parts with respect to another.1.7 Fundamental Magnitudes. Fundamental magnitudes are those in terms of which all other

magnitudes can be expressed.1.8 Systems of Units . System of units set of standards for some or all the basic units. In a system of

units that evolved in English- speaking countries, the so – called English System, the units of length, mass and time are the foot, slug, and second respectively. The modernized version of the metric system, based on atomic standards is called the SI System, from the French SYSTEME INTERNATIONAL d ‘ UNITS (international system of units).

1.9 Unit conversions . Units are multiplied and divided just like ordinary symbols and are treated in a equation in exactly the same way as algebraic quantities.

2.0 Significant Figures. If a mass is accurately recorded as 56.0 kg. It means that the true mass lies between 55.95 and 56.05 kg. The accurate known digits, plus one uncertain digit, are called significant figure. The 5 and 6 which are accurate and the 0 which is uncertain. 2.1 Scientific Notation (Power – of – Ten Notation). When writing numbers especially those involving many zeroes before or after the decimal point. It is convenient to use the scientific notation using power. Example: 5,980,000,000,000,000,000,000,000kgs. → 5.98 x 1024 kgs. 2.2Rounding of Data. When the number desired to be dropped is from 0 to 4, drop the number and if from 6 to 9, add one to the preceding number. If it is exactly 5, the odd – even rule should be followed, that is if the number that proceeds is even, retain the number and add one if it is odd. 2.3 Fundamental Units. Fundamental units are the standards selected for each of the fundamental magnitudes.

“Basic Units of the SI System”Quantity Unit Name Unit Symbol

Length meter mMass kilogram KgTime second sElectric Current ampere ATemperature degree kelvin KAmount of substance mole molLuminous intensity candela cd

“SI Derived Units with Special Names”Quantity Unit Name Unit Symbol

Frequency hertz Hz (s−1)

Force newton N ( Kg .m /s2)Pressure pascal Pa ( N /m2)Energy, Work joule J (N.m)Power watt W (J/s)Electric Charge coulomb C (As)Electric Potential volt V (W/A)Capacitance farad F (C/V)Electric Resistance ohm Ω(V/A)Magnetic Flux Intensity (flux density) tesla T (Web./ A)Inductance henry H (Web./ A)

“Prefixes for Multiple of SI Units”Prefixes Symbol Value Prefixes Symbol ValueExa E 1018 Deci d 10−1

Peta P 1015 Centi c 10−2

Teta T 1012 Milli m 10−3

Giga G 109 Micro μ 10−6

Mega M 106 Nano n 10−9

Kilo K 103 Pico p 10−14

Hector h 102 Femto f 10−15

Deka da 101 Atto a 10−18

“Conversion Factors”1 mile = 1609.42 m. 1atm = 1.013 x 105 Pa1 meter = 3.28 ft. = 14.7 psi1 mile = 5280 ft. 1 calorie = 4.186 J1 inch = 2.54 cm. 1 horse power = 146 watts1 foot = 12 inches 1 gallon = 3,785 liters1 yard = 3 ft. 1 cubic meter = 1000 liters1 light year = 9.461 x 1015m. 1 radian = 57.30 degree1 hectare = 10,000 sq. m. 1 hour = 60 minutes1 acre = 43,560 sq. ft. 3,600 seconds1 kg = 2.2 lb. (at Sea level)

Example:1. Express 27.5 ft./s in kph (km./hr).2. Convert 15.5 mgd (million gallons per day) to liters.

Chapter Ⅱ: (“Vectors and Scalar”)Chapters Objectives

All the end of this chapter, one should be able to:1. Distinguish whether a quantity is vector or scalar.2. Resolve a vector into its components.3. Represent the vector graphically.4. Use trigonometric in solving vector problems.

2.1 Vectors and Scalars. Vectors are quantities that have both magnitude and direction. Vectors can be added and subtracted, not algebraically or arithmetically but geometrically, according to the rules to be given in each particular case.Examples of Vector and Scalar Quantities:

“Vector”Symbol Name Examples displacement 15 m, N30° Ev velocity 60 m/s, eastF force 45 newtons, 120°a acceleration 5.0 m/s2

“Scalar”Symbol Name Examples distance 15 m.v speed 60 m/st time 60 sE energy 200 J

2.2 Graphical Representation of a Vector. A vector quantity can be represented graphically by means of directed straight line or arrows. The length of the line representsthe magnitude of the vector. The geometrical line is also the line of action and the arrow head of the line shows the direction and the point of application.

2.3 Resultant and Equivalent. Resultant is that single vector which would have the same effect as several others combined. Equilibrant is a vector equal to the resultant but opposite in direction and therefore capable of neutralizing the common effect of several vectors.

2.4 Resolution of Vectors. Resolution is the process of finding the components of a vector which is given as a resultant.

2.5 Component of vectors. A vector may be considered as the resultant of its component vectors. Consider the displacement vector in Fig. 2.1 a coordinate system has been drawn that has an x – axis pointing east and y – axis pointing north. The component of F is a vector F x that points along the x – axis and whose length equals the projection of A on that axis as shown in Fig. 2.2.

The y – component of A is the projection of A on the y – axis and is labeled F y in Fig. 2.2.

To determine the magnitude of components, a brief review of trigonometry is helpful. Consider the right triangle shown in figure 2.3. Recall the definitions of the sine, cosine and tangent functions.

sin θ=opposite sidehypotenuse

= AC

cosθ=¿ adjacent sisehypotenuse

=BC

¿

tanθ= opposite sideadjacent side

= AB

Pythagorean Theorem: C2=B2+ A2

Chapter Ⅲ: (“Composition and Resolution of Forces”)Chapter Objectives:

At the end of this chapter, one should be able to:1. Distinguish between collinear and concurrent forces.2. Determine the resultant of concurrent and collinear forces.3. Use suitable scale in represent vectors.4. Calculate the resultant of concurrent forces by analytical and graphical methods.

Note: Operations with vectors will be facilitated if applied to a particular case such as forces and whatever we say about forces may be applied to other kinds ofvectors.3.1 Resultant of Collinear Forces. Two or more forces are collinear if they have the same line of action. The resultant will be equal to the algebraic sum of all the forces each taken with is respective signs. Normally forces to the right and north bound are given positive signs while their opposites the negative signs.

3.2Resultant of Concurrent Forces. Concurrent forces are forces whose lines of action all pass through a common point.

3.2.1 Graphical Solution.3.2.1.1Parallelogram Method. The forces are supposed to act simultaneously. Construct

the parallelogram by drawing the other two sides parallel to the forces. The diagonal will then represent the resultant both in magnitude and direction.

3.2.1.2 Polygon Method. The method consists in beginning at any convenient point and drawing to scale each vector arrow in succession. The tail end of each arrow is attached to the tip end of the preceding one. The resultant is the arrow with its tale end at the origin and its tip end at the last vector added.

If the diagram is a closed polygon, the forces are in equilibrium.3.2.2 Analytical Method.

3.2.2.1 Law of Cosines. If two sides F1∧F2 of a triangle are given and the included

angleθ, the third side F3 may be found by the law:

However, F3 is not the resultant of F1 and F2 if they are forces. To find the vector sum, graph the forces in succession. The angleθ’ then is the supplement of angleθ. Now the cosine ofθ’ is equal but opposite in sign to that of angleθ. Since cos θ = - (cosθ), then expression for the side oppositeθ’ which is the resultant, becomes:

3.2.2.2 By Components. Each vector is resolved into its x and y components, with negatively directed components taken as negative. The algebraic sum of all the x or horizontal components is the x - component of the resultantFR. In similar fashion is the y – component found. Knowing the components, the magnitude of the resultant is given by:

FR2= ∑ F x

2 + ∑ F y2

The direction of FR is determined by the relation:

tan¿|∑ F y

∑ Fx| ; θ=tan−1|∑ F y

∑ F x|

Example:1. Calculate the resultant of the three forces acting on a point as shown in the figure.

Chapter Ⅳ: (“Equilibrium and Torque”)Chapters Objectives:

At the end of this chapter; one should be able to:1. Determine whether an object is in stable, unstable or neutral equilibrium.2. Draw the free – body diagram that shows the forces acting on an object. 3. Discus the steps in solving problem in statics.4. Determine the tension on the cords where a body is suspended.5. Determine the center of gravity of an object.

4.1 Equilibrium. Equilibrium is a state or condition wherein there is no change in the motion of a body.4.1.1 Conditions for Equilibrium.

4.1.1.1 First Condition. No accelerated translation. The vector sum of all forces acting on the body must be zero; therefore the vector diagram must form a closed polygon.

∑ F x=∑ F y=04.1.1.2 Second Condition. No accelerated rotation. The algebraic sum of the moments about a

point taken either outside or inside the body must be zero.4.2 Types of Equilibrium. The equilibrium of a body may be stable, unstable or neutral.

When a body returns to its original position after being slightly disturbed, the equilibrium is stable. A body is in unstable equilibrium if it tends to get as far as possible from its original position when disturbed. If a body remains in any position in which it finds itself with no tendency to return to its original position or to go still farther away from it is in neutral equilibrium.

4.3 Torque. Torque which is also called as the “moment of the force” is the ability of a force to produce rotation. It is equal to the product of the force by its moment arm – the perpendicular distance from its line of action to the fixed point.

Torque = (force) (moment arm)τ=F l

In the SI system, the unit of torque is the newton meter (N.m).4.4 Composition of Moments. Moments can be added algebraically, by qualifying them with

opposite signs, as they tend to rotate clockwise or counter clockwise. The resultant of several moments is their algebraic sum.

4.5 Weight of an Object. Weight is the force with which gravity pulls downward upon an object and it is not the same everywhere. Weight becomes lesser as the object moves farther from the center of the attracting body. Therefore, an object weighs lesser when it is on a mountain top than when at sea level.

Weight = (mass) (acceleration of gravity)w = mg

4.6 Center of Gravity. The center of gravity is the point at which the whole mass of a body may be assumed to be concentrated without changing the action of gravity on it.

The center of gravity may be outside the body.4.7 Steps in Solving Problems in Statics.

4.7.1 Make an illustration of the whole structure being considered.Include all of the known information bin the illustration and identify in symbols unknown quantities you wish to determine.

4.7.2 Draw a free – body diagram for one part or object of the structure. The part of the structure chosen is usually one on which many of the known forces and at least one of the unknown forces act. The conditions of equilibrium must be satisfied by this object because it is at rest.

4.7.3 Superimpose an x – axis and y – axis on the free – body diagram. These axes may be oriented in any direction. Once you decide on the placement of the axes and the origin of the coordinates, all future calculations must be consistent with that choice.

4.7.4 Satisfy the conditions of equilibrium. To do this, you must first resolve the known force into their components. You will also have to include in your equations symbols for the unknown components.

4.7.5 Substitute known information into the equations from step 4.7.4 and perform the necessary algebraic manipulation to solve for the unknowns.

Chapter Ⅴ: (“Kinematics”)Chapter Objectives:

At the end of this chapter; one should be able to:1. Distinguish between distance and displacement and between speed and velocity.2. Discuss the steps in solving problems in kinematics.3. Write the five basics motion equations for uniformly accelerated motion.4. Derive the equations for problems involving free – falli8ng bodies and projectiles.5. Apply the formula in problem – solving.5.1 Kinematics. Kinematics is defined as the quantitative description of an object’s motion.5.2 Kind of Motion. Motion may be simple if it is due to a single force; or compound, if it is

combined effect of several forces.5.3 Uniform Motion. This motion is due to an instantaneous force. It is characterized by a

constant velocity, that is, that is, the moving body will pass over equal distances per unit time.

S = V tWhere:

S = total space or distanced travelledV = the constant velocity t = time

5.3.1 Speed. Speed is the scalar value of velocity.5.3.1.1 Average Speed. Average speed V is defined as the distance travelled

divided by the time required to travel that distance.5.3.2 Average Velocity. The average velocity V of an object during some time

period is the object’s displacement (a vector) divided by the time.5.3.3 Instantaneous Velocity. Instantaneous velocity indicates how fast an object

moves at each instant of the time and the direction of that motion.5.4 Uniformly Accelerated Motion. This motion is due to a constant force, that is, a force

acting continuously. The result of this continuously action is acceleration.5.4.1 Acceleration. Acceleration may be defined as the increase of velocity (either in

speed or in action) per unit time. If it were a decrease, the result would be deceleration or negative acceleration.

a = change∈velocitytie taken

=V f −V i

tWhere:V f = final velocityV i = initial velocityt=¿ Time

Table:

1. S = V t

2. V=V f +V i

2

3. a=V f −V i

t

4. s=V i t+12

a t2

5. 2as=V f2−V i

2

5.5 Projectile. A projectile is an object that moves through space under the influence of the earth’s gravitational force.

Projectile motion then is described in terms of motion in the horizontal or x direction and independent by motion in the vertical or y direction. If air resistance can be ignored, the projectile’s horizontal motion continues at constant velocity because its horizontal acceleration is zero. The vertical motion, however, will have acceleration due to gravity.

5.6 Steps in solving problems in Kinematics. 5.6.1 Illustrate the situation described in the problem. Include in the diagram a

coordinate a coordinate axis or axes; the values of known quantities represented in terms of appropriate symbols; a symbol for the unknown quantity you wish to determine.

5.6.2 Divided the problems into parts. Each part can often be solved with relative ease even though the original problem might have seemed impossibly complicated.

5.6.3 Find one of the equations listed earlier in which the only unknown quantity in the equation is the one whose value you wish to calculate.

5.6.4 Rearrange the equations so that the unknown appears alone on the left and the known quantities on the right.

5.6.5 Substitute the values of the known quantities and calculate the value of the unknown including its unit.

5.6.6 Check your work.5.7 Gravitational Acceleration. Gravitational acceleration caused the vertical motion

of objects thrown into the air or of the objects falling through the air. If not for the resistive force of the air on these objects, this acceleration would be constant. This acceleration is caused by the gravitational force on the earth pulling down on the objects.

The magnitude of this acceleration has a value at the earth’s surface of

9.8m

s2.

g = 9.8m

s2

5.8 Factors that Affect air Resistance. 5.8.1 The speed of the object. The faster the objects move, at more air

resitanc4e affects its acceleration.5.8.2 The shape and surface area of an object.5.8.3 Air resistance is much more important for a light object, such as paper,

than for a heavier object of the sane size such as meatal sheet.Equation for Bodies in free – fall:

g t = V f −V i

h=V it +12

g t2

2g h=V f2−V i

2

Chapter Ⅵ: (“Newton’s Law of Motion”)Chapter Objectives:

At the end of this chapter, one should able to:1. Discuss the Newton’s law of motion.2. Use the second law of motion to solve problems involving force, mass and acceleration.3. Difference mass and weight.4. Discuss the applications of Newton’s law of motion.5. Determine the weight of an object with given mass at certain altitudes or distances above

earth’s surface or above sea – level. 6.1 Newton’s First Law: an object at rest will remain at rest; an object in motion will continue with constant velocity unless forces acted upon it. This also called as the LAW OF INERTIA.6.2 Newton’s Second Law: according to Newton, an object accelerates if acted on by an unbalance force. This acceleration depends on two factors:

1. The resultant force on the object.2. The mass or amount of matter of which the object is made.

The acceleration is directly proportional to the resultant force acting on the object’s

mass. If a group of forces act on an object of mass (m). the vector sum of these forces (∑ F ¿¿ causes the object to have an acceleration a, given by:

a = ∑ F

m or ∑ F=ma

The direction of a is in the direction of the resultant force.Newton’s second law is often written as F=ma. The force F on the left side of the equation,

however, represents the net or resultant force acting on the object.6.2.1 Units for Newton’s Second Law of motion. The unit of acceleration in the

SI system is the meter

second 2 (

m

s2). For the mass is the kilogram, and for the force is the Newton (N).

1 N = (1 kg) (1m

s2)

6.3 Newton’s Third Law. For every force exerted on one body, this is an equal oppositely directed force acting on some other body. This is the often called the “LAW OF ACTION AND REACTION”.

Newton indicated in this law that forces comes in pairs. The actions force is the exerted by one object on another. The reaction force is the force exerted by the other object on the first.

6.4 Tension on a String. Tension in a string is a particular case of Newton’s third law of motion. This is the back pulling or reaction exerted by a mass on the string cause of its resistance to motion.

6.5 Law of Universal Gravitation. The gravitational force exerted by amass m1 on another

mass m2 is proportional to the product of their masses and inversely proportional to the square of their separation r.

F = Gm1m2

r2

Where: G = is a constant called the universal gravitational constant.G = 6.67 x 10−11Nm2/kg2

Planetary Constants:

Planet Equation Planet Mean Mean DistanceRadius mass Density from the Sum

(km) (x 1024 kg) (kg

m3) (x10−11m)

Mercury 2,439 0.329 5440 .5834Venus 6,052 4.874 5240 1.0771Earth 6,378 5.980 5497 1.4960Mars 3,397 0.642 3900 2.2739Jupiter 71,398 1901.000 1300 7.7792Saturn 60,000 569.100 700 14.2718Uranus 27,900 87.069 1000 28.7232Neptune 24,300 87.069 1700 44.8800Pluto - - - 58.9424

CHAPTER Ⅶ; FRICTIONCHAPTER OBJECTIVES

At the end of this chapter one should be able to;1. Discuss the concept of friction.2. Discuss the methods in reducing friction.3. Explain the situation wherein friction is an advantage.4. Determine the minimum force necessary to move an object resting on a surface wherein

friction is 5. Compute problems on moving objects, subject to friction.7.1 FRICTION. A friction force opposes the motion of an object across a surface on which the

object rest or slides and is directed parallel to the surface of contact.Friction can never be eliminated completely, although it can be reduced to exceedingly small values in certain circumstances.Friction forces on objects always act in a direction opposite to the direction of motion.

7.1.1 CONCEPTS ON FRICTION. 7.1.1.1 Friction is independent of the velocity (below certain limits) between two sliding

bodies.7.1.1.2 If the normal force (force perpendicular to the surface) remains the same,friction is

independent on the area of the rubbing surfaces.7.1.1.3 The forces of friction is proportional to the total force pressing the surface against

another.7.1.1.4 The force of friction is slightly greater at the start that after the motion has begun.7.2 COEFFICIENT OF FRICTION. 7.2.1 Coefficient of static friction. The coefficient of static friction is defined as the ratio of

the force necessary to start motion to the force pressing the bodies together, that is, the normal force.

7.2.2 Coefficient of kinetic friction. The coefficient of kinetic friction is the ratio of the force necessary to move the body uniformly, to the normal force.

μs=FN

μK= F

N

TABLE 9

Surfaces μs μk

Rubber on concrete, dry 0.7-1.0 0.70Rubber on concrete, damp 0.7 0.5Steel on steel, dry 0.8 0.4Steel on steel, lubricated 0.15 0.08Steel on ice 0.10 0.06Steel on Teflon 0.04 0.04Giass on glass 0.94 0.4Wood on wood 0.50 0.25Bone on bone, dry - 0.30Bone on bone, lubricated - 0.003

7.3 ANGLE OF REPOSE. The angle of repose may be defined as the angle at which the body just begin to slide.7.4 ROLLING FRICTION. The friction of a solid rolling on a surface is less than the friction of a solid sliding over a surface. 7.5 METHODS TO REDUCE FRICTION.

7.5.1 Substitute rolling to sliding friction.7.5.2 Make the rubbing surfaces as smooth as possible.7.5.3 Make rubbing surfaces of different material so that the depressions or irregularities of

one would not fit into the other.7.5.4 Use proper lubricants to prevent direct contact between surfaces.

7.6 ADVANTAGES OF FRICTION. Friction is not always a disadvantage. Screw and nails hold their position in the objects in which they are driven by means of friction.

CHAPTER Ⅷ: WORK, ENERGY AND POWERCHAPTER OBJECTIVES

At the end of this chapter one should be able to: 1. Determine the work done by a force that moves an object through a certain displacement. 2. Determine the work done when an object is raise to a certain height.3. Determine the power output and give the units as watt, ft.lb/s or hp interchangeably.4. Determine potential, kinetic and resultants of energy.5. Compute and solve problems that involved moving objects using the principle of conservation of energy.8.1 CONCEPT OF WORK. In physics, work is done when a force act on an object moves from one place to another.8.2 COMPUTATION OF WORK. The value of work done by a force is the product of the force by distance through which its point of application moves in the direction of a force.

WORK=force x displacement8.3. ENERGY. The energy of a body is its ability to do work, and the work it can do is the measure of its energy. 8.4. KINDS OF ENERGY. 8.4.1 Potential Energy. Potential energy is the energy that a body possesses by virtue of its position with respect to other bodies.

PE=Weight x height8.4.1.1 Elastic Potential Energy. The energy stored in a stretch or compressed elastic material.

PE=(average force) x distance8.4.2 Kinetic Energy. Kinetic energy is the energy that a body possesses by virtue of its motion.By transforming work W=Fs from previous discussion,

F=ma

s=v2/2a

substituting, we have

KE=(ma) (v2/2a) or

KE=1/2 mv2

8.5 CONSERVATION OF ENERGY. Energy is never created nor destroy. The sum total of energy of the universe remains constant.8.6 POWER. Power is defined as the ratio of the work done W and the time t required to do that work.

Power=P=wt

8.6.1 Units of Power.8.6.1.1 Watt or Joule per second8.6.1.2 Horsepower which is equivalent to 746 watts.

EXAMPLE;A 5 kg block from rest at pt. A and slides along a plane that is inclined 30° to the horizontal. At the 3].0 m position the speed of the block is 3.0 m/s. What is the coefficient of friction between the block and the plane?

CHAPTER Ⅸ: SIMPLE MACHINECHAPTER OBJECTIVES

At the end of this chapter , one should be able to ;1. Discuss the principle of work in relation to machines.2. Explain the actual and ideal mechanical advantage.3. Enumerate and discuss each of the simple machines.4. Determine the efficiency of machines.

9.1 Simple Machines. A machine is a device by which magnitude, direction or method of application of a force is changed for the sake of gaining some practical advantage.9.2 Principle of Work. In an ideal or frictionless machine, the work done by machine (output) is equal t the work done upon the machine or the energy applied to it (input).

Work Input = useful work + friction work 9.3 Efficiency. Efficiency is the ratio of the output to the input.

Efficiency = work outputwork input

= power outputpower input

x 100

9.4 Actual Mechanical Advantage. The AMA of a machine is

AMA = force ratio =force exerted bymachine on load

force used ¿operatemachine¿

9.5 Ideal Mechanical Advantage. T he IMA of a machine is

IMA= distance ratio=distancemoved by input force

distancemoved by load9.6 Kinds of a Simple Machines.

9.6.1 The Lever. The Lever is a rigid bar, straight or curved, which rotates about the a fixed point called the fulcrum.

Class 1 Lever. The fulcrum lies between the effort and the resistance. I.e. scissors Class 2 Lever. The resistance lies between the fulcrum and the effort. i.e. the wheel barrow Class 3 Lever. The efforts is applied between the fulcrum and the resistance. i.e. the pincers.

9.6.2 The Wheel and Axle. It consists of a large and a small wheel rigidly joined to the same axle. The rope to which the effort is applied is wound around the large wheel while the rope that carries the weight is wound around the smaller, but in opposite sense, so that unwinding to the first results in the winding of the second.

CHAPTER 10;CHAPTER 11: CIRCULAR MOTIONCHAPTER OBJECTIVES;

At the end of this chapter, one should be able to;1. Explain how the mass, speed and radius of orbit affect the centripetal force.2. Convert radians to degree and vice versa.3. Find the maximum velocity that a car can attain without skidding or overturning.4. Compute for the period of the planets and satellites given their radius of orbits.5. Determine the required banking of road for given speed.

11.1 Uniform Circular Motion. Circular motion is the result of a central constant force and an instantaneous tangential force. The actions are at the every instant perpendicular to each other while their directions in space are changing continuously.11.2 Centripetal Acceleration. We can derive an expression for this acceleration by considering a circle with radius r and center at 0. On the circumference mark a point A and draw a tangent that will represent vertically the velocity due to the instantaneous force. Going over the arc of length mark a second point B and again draw a tangent which at velocity of the same magnitude as V but with different direction.