1

Materials Engineering Department

Class: First

Date : 2009-2010

Subject: Engineering Mechanics

Lecturer: Dr. Emad AL-Hassani

Lecture # 1

1.1. WHAT IS MECHANICS?

Mechanics can be defined as that science which describes and

predicts the conditions of rest or motion of bodies under the action of

forces. It is divided into three parts: mechanics of rigid bodies, mechanics

of deformable bodies, and mechanics of fluids. The mechanics of rigid

bodies is subdivided into statics and dynamics, the former dealing with

bodies at rest, the latter with bodies in motion. In this part of the study of

mechanics, bodies are assumed to be perfectly rigid. Actual structures and

machines, however, are never absolutely rigid and deform under the loads

to which they are subjected. But these deformations are usually small and

do not appreciably affect the conditions of equilibrium or motion of the

structure under consideration. They are important, though, as far as the

resistance of the structure to failure is concerned and are studied in

mechanics of materials, which is a part of the mechanics of deformable

bodies. The third division of mechanics, the mechanics of fluids, is

subdivided into the study of incompressible fluids and of compressible

fluids. An important subdivision of the study of incompressible fluids is

hydraulics, which deals with problems involving water.

Mechanics is a physical science, since it deals with the study of

physical phenomena. However, some associate mechanics with

mathematics, while many consider it as an engineering subject. Both

these views are justified in part. Mechanics is the foundation of most

engineering sciences and is an indispensable prerequisite to their study.

2

However, it does not have the empiricism found in some engineering

sciences, i.e., it does not rely on experience or observation alone; by its

rigor and the emphasis it places on deductive reasoning it resembles

mathematics. But, again, it is not an abstract or even a pure science;

mechanics is an applied science. The purpose of mechanics is to explain

and predict physical phenomena and thus to lay the foundations for

engineering applications.

1.2. FUNDAMENTAL CONCEPTS AND PRINCIPLES

The basic concepts used in mechanics are space, time, mass, and

force. These concepts cannot be truly defined; they should be accepted on

the basis of our intuition and experience and used as a mental frame of

reference for our study of mechanics. The concept of space is associated

with the notion of the position of a point P. The position of P can be

defined by three lengths measured from a certain reference point, or

origin, in three given directions. These lengths are known as the

coordinates of P.

To define an event, it is not sufficient to indicate its position in space.

The time of the event should also be given.

The concept of mass is used to characterize and compare bodies on

the basis of certain fundamental mechanical experiments. Two bodies of

the same mass, for example, will be attracted by the earth in the same

manner; they will also offer the same resistance to a change in

translational motion.

A force represents the action of one body on another. It can be

exerted by actual contact or at a distance, as in the case of gravitational

forces and magnetic forces. A force is characterized by its point of

3

application, its magnitude, and its direction; a force is represented by a

vector .

In Newtonian mechanics, space, time, and mass are absolute

concepts, independent of each other. (This is not true in relativistic

mechanics, where the time of an event depends upon its position, and

where the mass of a body varies with its velocity.) On the other hand, the

concept of force is not independent of the other three. Indeed, one of the

fundamental principles of Newtonian mechanics listed below indicates

that the resultant force acting on a body is related to the mass of the body

and to the manner in which its velocity varies with time.

You will study the conditions of rest or motion of particles and

rigid bodies in terms of the four basic concepts we have introduced. By

particle we mean a very small amount of matter which may be assumed

to occupy a single point in space. A rigid body is a combination of a

large number of particles occupying fixed positions with respect to each

other. The study of the mechanics of particles is obviously a prerequisite

to that of rigid bodies. Besides, the results obtained for a particle can be

used directly in a large number of problems dealing with the conditions

of rest or motion of actual bodies. The study of elementary mechanics

rests on six fundamental principles based on experimental evidence.

The Parallelogram Law for the Addition of Forces

This states that two forces acting on a particle may be replaced by a

single force, called their resultant, obtained by drawing the diagonal of

the parallelogram which has sides equal to the given forces.

4

The Principle of Transmissibility

This states that the conditions of equilibrium or of motion of a rigid

body will remain unchanged if a force acting at a given point of the rigid

body is replaced by a force of the same magnitude and same direction,

but acting at a different point, provided that the two forces have the same

line of action.

Newton’s Three Fundamental Laws.

Formulated by Sir Isaac Newton in the latter part of the seventeenth

century, these laws can be stated as follows:

FIRST LAW

If the resultant force acting on a particle is zero, the particle will remain

at rest (if originally at rest) or will move with constant speed in a straight

line (if originally in motion) .

SECOND LAW

If the resultant force acting on a particle is not zero, the particle will have

an acceleration proportional to the magnitude of the resultant and in the

direction of this resultant force., this law can be stated as:

F=m a ……. (1.1)

Where F, m, and a represent, respectively, the resultant force acting on

the particle, the mass of the particle, and the acceleration of the particle,

expressed in a consistent system of units.

5

THIRD LAW

The forces of action and reaction between bodies in contact have the

same magnitude, same line of action, and opposite sense .

Newton’s Law of Gravitation

This states that two particles of mass M and m are mutually attracted

with equal and opposite forces F and −F (Fig. 1.1) of magnitude F given

by the formula

F=G (Mm/r2)…….. (1.2)

Where

r = distance between the two particles

G = universal constant called the constant of gravitation

Newton’s law of gravitation introduces the idea of an action exerted at a

distance and extends the range of application of Newton’s third law: the

action F and the reaction −F in Fig. 1.1 are equal and opposite, and they

have the same line of action.

Fig: (1.1)

A particular case of great importance is that of the attraction of the

earth on a particle located on its surface. The force F exerted by the earth

6

on the particle is then defined as the weight W of the particle. Taking M

equal to the mass of the earth, m equal to the mass of the particle, and r

equal to the radius R of the earth, and introducing the constant

g = GM ……… (1.3)

The magnitude W of the weight of a particle of mass m may be

expressed as

W =mg ………. (1.4)

The value of R in formula (1.3) depends upon the elevation of the

point considered; it also depends upon its latitude, since the earth is not

truly spherical. The value of g therefore varies with the position of the

point considered. As long as the point actually remains on the surface of

the earth, it is sufficiently accurate in most engineering computations to

assume that g equals 9.81 m/s2 or 32.2 ft/s2.

1.3. SYSTEMS OF UNITS

With the four fundamental concepts introduced in the preceding

section are associated the so-called kinetic units, i.e., the units of length,

time, mass, and force. These units cannot be chosen independently if Eq.

(1.1) is to be satisfied. Three of the units may be defined arbitrarily; they

are then referred to as basic units. The fourth unit, however, must be

chosen in accordance with Eq. (1.1) and is referred to as a derived unit.

Kinetic units selected in this way are said to form a consistent system of

units.

7

International System of Units (SI Units)

In this system, which will be in universal use after the United

States has completed its conversion to SI units, the base units are the units

of length, mass, and time, and they are called, respectively, the meter (m),

the kilogram (kg), and the second (s). All three are arbitrarily defined.

The second, which was originally chosen to represent 1/86 400 of the

mean solar day, is now defined as the duration of 9 192 631 770 cycles of

the radiation corresponding to the transition between two levels of the

fundamental state of the cesium-133 atom. The meter, originally defined

as one ten-millionth of the distance from the equator to either pole, is now

defined as 1 650 763.73 wavelengths of the orange-red light

corresponding to a certain transition in an atom of krypton-86. The

kilogram, which is approximately equal to the mass of 0.001 m3 of water,

is defined as the mass of a platinum-indium standard kept at the

international Bureau of Weights and Measures at Sevres, near Paris,

France. The unit of force is a derived unit. It is called the Newton (N) and

is defined as the force which gives an acceleration of 1 m/s2 to a mass of

1 kg (Fig. 1.2). From Eq. (1.1) we write

1 N = (l kg) (l m/s2) = l kg.m/s2 …….. (1.5)

8

The SI units are said to form an absolute system of units. This

means that the three base units chosen are independent of the location

where measurements are made. The meter, the kilogram, and the second

may be used anywhere on the earth; they may even be used on another

planet. They will always have the same significance.

The weight of a body, or the force of gravity exerted on that body,

should, like any other force, be expressed in newtons. From Eq. (1.4) it

follows that the weight of a body of mass 1 kg (Fig. 1.3) is

W =mg

= (1 kg) (9.81 m/s2)

=9.81N

Multiples and submultiples of the fundamental SI units may be

obtained through the use of the prefixes defined in Table 1.1. The

multiples and submultiples of the units of length, mass, and force most

frequently used in engineering are, respectively, the kilometer (km) and

9

the millimeter (mm); the megagram (Mg) and the gram (g); and the

kilonewton (kN). According to Table 1.1, we have

1 km =1000 m 1 mm = 0.001 m

l Mg= 1000kg 1 g = 0.001 kg

1kN = 1000 N

The conversion of these units into meters, kilograms, and newtons,

respectively, can be effected by simply moving the decimal point three

places to the right or to the left. For example, to convert 3.82 km into

meters, one moves the decimal point three places to the right:

3.82 km = 3820 m

Similarly, 47.2 mm is converted into meters by moving the

decimal point three places to the left:

47.2 mm = 0.0472 m

10

Table (1.1): SI Prefixes

Multiplication Factor Pretixt Symbol

1000 000 000 000 =1012 tera T

1 000 000 000 = 109 giga G

1 000 000 = 106 mega M

1000 =103 kilo k

100 = 102 hecto h

10 = 101 deka da

0.1= 10−1 deci d

0.01 = 10−2 centi c

0.001 = 10−3 milli m

0.000 001 = 10−6 micro µ

0.000 000 001 = 10−9 nano n

0.000 000 000 001=10−12 pico p

0.000 000 000 000 001= 10−15 femto f

0.000 000 000 000 000 001 = 10−18 atto a

Using scientific notation, one may also write

3.82 km = 3.82 x 103 m

47.2 mm = 47.2 x 10−3

The multiples of the unit of time are the minute (min) and the hour (h).

Since 1 min = 60 s and 1 h 60 min = 3600 s, these multiples cannot be

converted as readily as the others.

11

By using the appropriate multiple or submultiple of a given unit,

one can avoid writing very large or very small numbers. For example, one

usually writes 427.2 km rather than 427 200 m, and 2.16 mm rather than

0.002 16 m.

Units of Area and Volume

The unit of area is the square meter (m2), which represents the area

of a square of side 1 m; the unit of volume is the cubic meter (m3), equal

to the volume of a cube of side 1 m. In order to avoid exceedingly small

or large numerical values in the computation of areas and volumes, one

uses systems of subunits obtained by respectively squaring and cubing

not only the millimeter but also two intermediate submultiples of the

meter, namely, the decimeter (dm) and the centimeter (cm). Since, by

definition,

1 dm = 0.1 m = 10−1m

1 cm = 0.01 m = 10−2 m

1 mm 0.001 m = 10−2 m

The submultiples of the unit of area are

1dm2 = (1dm) 2= (10−1) 2=10−2 m2

1 cm2 = (1 cm) 2 = (10−2m) 2 =10−4 m2

1 mm2 = (1 mm) 2 = (10−3m) 2 = 10−6m 2

and the submultiples of the unit of volume are

1dm3 = (1 dm) 3 = (10−1m) 3 = 10−3m 3

12

1 cm3 = (1 cm) 3 = (10−2m) 3= 10−6m 3

1 mm3 = (1 mm) 3 = (10−3m) 3 = 10−9m 3

It should be noted that when the volume of a liquid is being

measured, the cubic decimeter (dm3) is usually referred to as a liter (L).

Other derived SI units used to measure the moment of a force, the

work of a force, etc., are shown in Table 1.2. While these units will be

introduced in later chapters as they are needed, we should note an

important rule at this time: When a derived unit is obtained by dividing a

base unit by another base unit, a prefix may be used in the numerator of

the derived unit but not in its denominator. For example, the constant k of

a spring which stretches 20 mm under a load of 100 N will be expressed

as

k = 100 N/20 mm = l00N /0.020 m =5000 N/m

Or k = 5 kN /m

but never as k = 5 N/mm.

Table (1.2): Principal SI Units Used in Mechanics

Quantity Unit Symbol Formula

Acceleration Meter per second squared ……. m/s2

Angle Radian rad *

Angular acceleration Radian per second squared . rad/s2

Angular velocity Radian per second . Rad/s

13

Area Square meter . m2

Density Kilogram per cubic meter . kg/m3

Energy Joule J N. m

Force Newton N Kg. m/s2

Frequency Hertz Hz s−1

Impulse Newton-second . . Kg . m/s

Length Meter m

Mass Kilogram kg

Moment of a force Newton-meter . . . N. m

Power Watt W J/s

Pressure Pascal Pa N/m2

Stress Pascal Pa N/ m2

Time Second s

Velocity Meter per second . m/s

Volume

Solids Cubic meter . m3

Liquids Liter L 10−3m 3

Work Joule J N. m

*Supplementary unit (1 revolution = 2π rad = 360ο).

14

U.S. Customary Units

Most practicing American engineers still commonly use a system in

which the base units are the units of length, force, and time. These units

are, respectively, the foot (ft), the pound (lb), and the second (s). The

second is the same as the corresponding SI unit. The foot is defined as

0.3048 m. The pound is defined as the weight of a platinum standard,

called the standard pound, which is kept at the National Institute of

Standards and Technology outside Washington, the mass of which is

0.453 592 43 kg. Since the weight of a body depends upon the earth’s

gravitational attraction, which varies with location, it is specified that the

standard pound should be placed at sea level and at latitude of 45° to

properly define a force of 1 lb. Clearly the U.S. customary units do not

form an absolute system of units. Because of their dependence upon the

gravitational attraction of the earth, they form a gravitational system of

units.

While the standard pound also serves as the unit of mass in

commercial transactions in the United Sates, it cannot be so used in

engineering computations, since such a unit would not be consistent with

the base units defined in the preceding paragraph. Indeed, when acted

upon by a force of 1 lb, that is, when subjected to the force of gravity, the

standard pound receives the acceleration of gravity, g = 32.2 ft/s2 (Fig.

1.4), not the unit acceleration required by Eq. (1.1). The unit of mass

consistent with the foot, the pound, and the second is the mass which

receives an acceleration of 1 ft/s2 when a force of 1 lb is applied to it (Fig.

1.5). This unit, sometimes called a slug, can be derived from the equation

F = ma after substituting 1 lb and 1 ft/s2 for F and a, respectively. We

write

15

F=ma 1 lb= (1slug) (1ft/s2)

and obtain l slug = l lb / l (ft/s2) = 1 lb.s2/ft …….. (1.6)

Comparing Figs. 1.4 and 1.5, we conclude that the slug is a mass

32.2 times larger than the mass of the standard pound.

The fact that in the U.S. customary system of units bodies are

characterized by their weight in pounds rather than by their mass in slugs

will be a convenience in the study of statics, where one constantly deals

with weights and other forces and only seldom with masses. However, in

the study of dynamics, where forces, masses, and accelerations are

involved, the mass m of a body will be expressed in slugs when its weight

W is given in pounds. Recalling Eq. (1.4), we write

m= w/ g ………….. (1.7)

Where g is the acceleration of gravity (g = 32.2 ft/s2).

16

Other U.S. customary units frequently encountered in engineering

problems are the mile (mi), equal to 5280 ft; the inch (in.), equal to (1/12)

ft; and the kilo pound (kip), equal to a force of 1000 lb. The ton is often

used to represent a mass of 2000 lb but, like the pound, must be converted

into slugs in engineering computations.

The conversion into feet, pounds, and seconds of quantities

expressed in other U.S. customary units is generally more involved and

requires greater attention than the corresponding operation in SI units. If,

for example, the magnitude of a velocity is given as v = 30 mi/h, we

convert it to ft/s as follows. First we write

v = 30 mi/h

Since we want to get rid of the unit miles and introduce instead the

unit feet, we should multiply the right-hand member of the equation by an

expression containing miles in the denominator and feet in the numerator.

But, since we do not want to change the value of the right-hand member,

the expression used should have a value equal to unity. The quotient

(5280 ft)/ (1 mi) is such an expression. Operating in a similar way to

transform the unit hour into seconds, we write

v = (30 mi/h) (5280ft/1mi) (1 h/3600s)

Carrying out the numerical computations and canceling out units

which appear in both the numerator and the denominator, we obtain

v = 44 ft /s

17

1.4. CONVERSION FROM ONE SYSTEM OF UNITS TO ANOTHER

There are many instances when an engineer wishes to convert into

SI units a numerical result obtained in U.S. customary units or vice versa.

Because the unit of time is the same in both systems, only two kinetic

base units need be converted. Thus, since all other kinetic units can be

derived from these base units, only two conversion factors need be

remembered.

Units of Length

By definition the U.S. customary unit of length is

1 ft =0.3048 m ……. (1.8)

It follows that

1 mi = 5280 ft = 5280(0.3048m) = 1609 m

or 1 mi = 1.609 km ………… (1.9)

also 1 in. = 1/12 ft = 1/12 (0.3048 m) = 0.0254 m

Or 1 in. = 25.4 mm …………. (1.10)

Units of Force

Recalling that the U.S. customary unit of force (pound) is defined

as the weight of the standard pound (of mass 0.4536 kg) at sea level and

at a latitude of 45° (where g = 9.807 m /s2) and using Eq. (1.4), we write

W = mg

1 lb = (0.4536 kg)(9.807 m/s2) = 4.448 kg• m/s2

18

or, recalling Eq. (1.5),

l Ib = 4.448N …………… (1.11)

Units of Mass

The U.S. customary unit of mass (slug) is a derived unit. Thus,

using Eqs. (1.6), (1.8), and (1.11), we write

l slug =1lb.s2/ft =1 Ib/ 1 ft/s2 = 4.448N/ 0.3048 m/s2

and, recalling Eq. (1.5),

1 slug = 1 Ib. s2/ft = 14.59 kg ………….. (1.12)

Although it cannot be used as a consistent unit of mass, we recall

that the mass of the standard pound is, by definition,

1 pound mass = 0.4536 kg ……………… (1.13)

This constant may be used to determine the mass in SI units

(kilograms) of a body which has been characterized by its weight in U.S.

customary units (pounds).

To convert a derived U.S. customary unit into SI units, one simply

multiplies or divides by the appropriate conversion factors. For example,

to convert the moment of a force which was found to be M = 47 lb. in.

into SI units, we use formulas (1.10) and (1.11) and write

M = 47 lb. in. = 47(4.448 N)(25.4 mm)

= 5310 N. mm =5.31 N. m

19

The conversion factors given in this section may also be used to

convert a numerical result obtained in SI units into U.S. customary units.

For example, if the moment of a force was found to be M = 40 N. m, we

write, following the procedure used in the last paragraph of Sec. 1.3,

M= 40N.m = (40N.m)( 1 Ib/4.448 N) ( 1 ft /0.3048 m )

Carrying out the numerical computations and canceling out units

which appear in both the numerator and the denominator, we obtain

M = 29.5 lb.ft

The U.S. customary units most frequently used in mechanics are

listed in Table 1.3 with their SI equivalents.

20

1

Materials Engineering Department Class: First Date : 2009-2010

Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani Lecture # 2

Introduction • The objective for the current chapter is to investigate the

effects of forces on particles: - replacing multiple forces acting on a particle with a single equivalent or resultant force, Relations between forces acting on a particle that is in a state of equilibrium

• The focus on particles does not imply a restriction to miniscule bodies. Rather, the study is restricted to analyses in which the size and shape of the bodies is not significant so that all forces may be assumed to be applied at a single point.

Resultant of Two Forces

• .Force: action of one body on another; characterized by its point of

application, magnitude, line of action, and sense • Experimental evidence shows that the combined effect of two forces

may be represented by a single resultant force. • The resultant is equivalent to the diagonal of a parallelogram which

contains the two forces in adjacent legs.

• Force is a vector quantity

2

Vectors

• Vector: parameter possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations.

• Scalar: parameter possessing magnitude but not direction. Examples: mass, volume, temperature

• Vector classifications: - Fixed or bound vectors have well defined points of

application that cannot be changed without affecting an analysis.

- Free vectors may be freely moved in space without changing their effect on an analysis.

- Sliding vectors may be applied anywhere along their line of action without affecting an analysis.

• Equal vectors have the same magnitude and direction.

• Negative vector of a given vector has the same magnitude and the opposite direction

3

Addition of Vectors

• Trapezoid rule for vector addition

• Triangle rule for vector addition • Law of cosines, • Law of sines,

• Vector addition is commutative

• Vector subtraction

• Addition of three or more vectors through repeated application of

the triangle rule

QPR

BPQQPRrrr

+=

−+= cos2222

AC

RB

QA sinsinsin

==

PQQPrrrr

+=+

B

B

C

C

4

• The polygon rule for the addition of three or more vectors.

• Vector addition is associative,

• Multiplication of a vector by a scalar Resultant of Several Concurrent Forces

• Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.

• Vector force components: two or more force vectors

which, together, have the same effect as a single force vector.

( ) ( )SQPSQPSQPrrrrrrrrr

++=++=++

5

Sample Problem The two forces act on a bolt at A. Determine their resultant.

SOLUTION:

• Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal.

• Trigonometric solution - use the triangle rule for vector addition in

conjunction with the law of cosines and law of sines to find the resultant.

• Graphical solution - A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,

°== 35N 98R

6

• Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured,

• Trigonometric solution - Apply the triangle rule. From the Law of Cosines,

From the Law of Sines,

°== 35N 98 αR

( ) ( ) ( )( ) °−+=

−+=

155cosN60N402N60N40

cos222

222 BPQQPR

N73.97=R

AA

RQBA

RB

QA

+°=°=

°=

=

=

2004.15

N73.97N60155sin

sinsin

sinsin

α

°= 04.35α

7

Rectangular Components of a Force: Unit Vectors • May resolve a force vector into perpendicular components so

that the resulting parallelogram is a rectangle. are referred to as rectangular vector components and

• Define perpendicular unit vectors i and j which are parallel to the x and y axes.

Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. Fx and Fy are referred to as the scalar components of

yx FFrr

and

yx FFFrrr

+=

Fr

8

Addition of Forces by Summing Components

• Wish to find the resultant of 3 or more concurrent forces,

Resolve each force into rectangular components

The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces and to find the resultant magnitude and direction,

SQPRrrrr

++=

( ) ( ) jSQPiSQP

jSiSjQiQjPiPjRiR

yyyxxx

yxyxyxyxrr

rrrrrrrr

+++++=

+++++=+

∑=++=

y

yyyy

F

SQPR

∑=++=

x

xxxx

FSQPR

x

yyx R

RRRR 122 tan−=+= θ

9

Sample Problem Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

SOLUTION:

• Resolve each force into rectangular components.

• Determine the components of the resultant by adding the corresponding force components.

• Calculate the magnitude and direction of the resultant.

SOLUTION:

• Resolve each force into rectangular components Determine the components of the resultant by adding the corresponding force components. Calculate the magnitude and direction.

9.256.961000.11001102.754.27800.759.129150

4

3

2

1

−+−+−++−−

FFFF

compycompxmagforc

r

r

r

r

°=== 1.4N1199N314tan αα

..

RR

x

y °= 1.4α

10

Equilibrium of a Particle

• When the resultant of all forces acting on a particle is zero, the particle is in equilibrium.

• Newton’s First Law: If the resultant force on a particle is zero, the

particle will remain at rest or will continue at constant speed in a straight line.

• Particle acted upon by two forces: - equal magnitude - same line of action - opposite sense

• Particle acted upon by three or more forces:

- graphical solution yields a closed polygon - algebraic solution

N6.199sin

N3.14==R

000

==

==

∑∑∑

yx FFFRrr

11

Free-Body Diagrams

Space Diagram: A sketch showing the physical conditions of the problem.

Free-Body Diagram: A sketch showing only the forces on the selected particle

Rectangular Components in Space

The vector F is contained in the plane OBAC.

• Resolve into F horizontal and vertical components.

Resolve Fh into rectangular components

yy FF θcos=

yh FF θsin= φθ

φ

φθφ

sinsin

sin

cossincos

y

hy

y

hx

FFFFFF

=

=

==

12

With the angles between F and the axes,

λ is a unit vector along the line of action of F and cosθx, cosθy, and cosθz are the direction cosines for F Direction of the force is defined by the location of two points

Fr

( )

kji

F

kjiF

kFjFiFF

FFFFFF

zyx

zyx

zyx

zzyyxx

rrrr

r

rrr

rrrr

θθθλ

λ

θθθ

θθθ

coscoscos

coscoscos

coscoscos

++=

=

++=

++=

===

( ) ( )222111 ,, and ,, zyxNzyxM

( )

dFdF

dFd

Fd

FdF

kdjdidd

FF

zzdyydxxdkdjdid

NMd

zz

yy

xx

zyx

zyx

zyx

===

++=

=

−=−=−=

++=

=

rrrr

rr

rrr

r

1

and joining vector

121212

λ

λ

1

Materials Engineering Department Class: First Date : 2009-2010

Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani Lecture # 3

Rigid Bodies: Equivalent Systems of Forces Introduction • Treatment of a body as a single particle is not always possible. In

general, the size of the body and the specific points of application of the forces must be considered.

• Most bodies in elementary mechanics are assumed to be rigid, i.e.,

the actual deformations are small and do not affect the conditions of equilibrium or motion of the body.

• Current chapter describes the effect of forces exerted on a rigid body

and how to replace a given system of forces with a simpler equivalent system.

- Moment of a force about a point - Moment of a force about an axis - Moment due to a couple

• Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple.

External and Internal Forces • Forces acting on rigid bodies are divided into two groups:

- External forces - Internal forces

• External forces are shown in a free-body diagram.

2

• If unopposed, each external force can impart a motion of translation or

rotation, or both. Principle of Transmissibility: Equivalent Forces

• Principle of Transmissibility -

Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are equivalent forces.

• Moving the point of application of the force F to the rear bumper does

not affect the motion or the other forces acting on the truck.

• Principle of transmissibility may not always apply in determining internal

forces and deformations.

3

Vector Product of Two Vectors • Concept of the moment of a force about a point is more easily understood

through applications of the vector product or cross product. • Vector product of two vectors P and Q is defined as the vector V which

satisfies the following conditions: 1. Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is V = PQ sin θ 3. Direction of V is obtained from the right-

hand rule

• Vector products:

- are not commutative, ( )QPPQ ×−=×

- are distributive, ( ) 2121 QPQPQQP ×+×=+×

-are not associative, ( ) ( )SQPSQP ××≠×× Vector Products: Rectangular Components • Vector products of Cartesian unit vectors,

00

0

=×=×−=×−=×=×=×

=×−=×=×

kkikjjkiijkjjkji

jikkijii

rrrrvrrr

rrrrrrrr

rrrrrrrr

4

• Vector products in terms of rectangular coordinates

( ) ( )kQjQiQkPjPiPV zyxzyx

rrrrrrr++×++=

( ) ( )( )kQPQP

jQPQPiQPQP

xyyx

zxxzyzzyr

rr

−+

−+−=

zyx

zyx

QQQPPPkjirrr

=

Moment of a Force About a Point • A force vector is defined by its magnitude and direction. Its effect on the

rigid body also depends on it point of application. • The moment of F about O is defined as FrMO ×= • The moment vector MO is perpendicular to the plane containing O and

the force F. • Magnitude of MO measures the tendency of the force to cause rotation of

the body about an axis along MO. FdrFM O == θsin • The sense of the moment may be determined by the right-hand rule.

5

• Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment.

• Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure.

• The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane

• If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive.

• If the force tends to rotate the structure

counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative

Varignon’s Theorem • The moment about a give point O of the resultant of several concurrent

forces is equal to the sum of the moments of the various moments about the same point O.

( ) L

rrrrL

rrr+×+×=++× 2121 FrFrFFr

• Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F.

6

Rectangular Components of the Moment of a Force • The moment of F about O,

kFjFiFFkzjyixrFrM

zyx

O rrrr

rrrrrrr

++=++=×= ,

( ) ( ) ( )kyFxFjxFzFizFyF

FFFzyxkji

kMjMiMM

xyzxyz

zyx

zyxO

rrr

rrr

rrrr

−+−+−=

=

++=

• The moment of F about B,

FrM BAB

rrr×= /

( ) ( ) ( )kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

rrrr

rrr

rrr

++=

−+−+−=

−=/

( ) ( ) ( )zyx

BABABAB

FFFzzyyxx

kjiM −−−=

rrr

r

7

• For two-dimensional structures

( )

zy

ZO

zyO

yFxFMM

kyFxFM

−==

−=rr

( ) ( )[ ]

( ) ( ) zBAyBA

ZO

zBAyBAO

FyyFxxMM

kFyyFxxM

−−−==

−−−=rr

Sample Problem A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O. Determine:

a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-lb vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original

force

8

a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper

( )( )( )in. 12lb 100

in. 1260cosin.24=

=°==

O

O

Md

FdM

in lb 1200 ⋅=OM

b) Horizontal force at A that produces the same moment,

( )

( )

in. 8.20in. lb 1200

in. 8.20in. lb 1200

in. 8.2060sinin. 24

⋅=

=⋅=

=°=

F

FFdM

d

O

lb 7.57=F

c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.

( )

in. 42in. lb 1200

in. 42in. lb 1200⋅

=

=⋅=

F

FFdM O

lb 50=F

d) To determine the point of application of a 240 lb force to produce the same moment,

in. 10=OB

( )

in. 5cos60

in. 5lb 402

in. lb 1200lb 240in. lb 1200

=°

=⋅

=

=⋅=

OB

d

dFdMO

9

e)Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force.

Sample Problem The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C.

SOLUTION: The moment MA of the force F exerted by the wire is obtained by evaluating the vector product,

( ) ( ) jirrr ACAC

rrrrr m 08.0m 3.0 +=−=

FrM ACA

rrr×=

10

( )

( ) ( ) ( ) ( )

( ) ( ) ( )kji

kji

rr

FFDC

DC

rrr

rrr

rrr

N 128N 69N 120m 5.0

m 32.0m 0.24m 3.0N 200

N 200

−+−=

−+−=

== λ

( ) ( ) ( )kjiM A

rrrvmN 8.82mN 8.82mN 68.7 ⋅+⋅+⋅−=

Scalar Product of Two Vectors • The scalar product or dot product between two vectors P and Q is

defined as ( )resultscalarcosθPQQP =•

rr

• Scalar products:

-are commutative, PQQPrrrr

•=•

-are distributive, ( ) 2121 QPQPQQPrrrrrrr

•+•=+•

-are not associative ( ) undefined =•• SQPrrr

11

• Scalar products with Cartesian unit components

( ) ( )kQjQiQkPjPiPQP zyxzyx

rrrrrrrr++•++=•

000111 =•=•=•=•=•=• ikkjjikkjjiirrrvrrrrrrrr

2222 PPPPPP

QPQPQPQP

zyx

zzyyxx

=++=•

++=•rr

rr

1

Materials Engineering Department Class: First Date : 2009-2010

Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani Lecture # 4

Equilibrium of Rigid Bodies Introduction • For a rigid body in static equilibrium, the external forces and moments

are balanced and will impart no translational or rotational motion to the body.

• The necessary and sufficient condition for the static equilibrium of a

body are that the resultant force and couple from all external forces form a system equivalent to zero,

• Resolving each force and moment into its rectangular components

leads to 6 scalar equations which also express the conditions for static equilibrium,

Free-Body Diagram • First step in the static equilibrium analysis of a rigid body is

identification of all forces acting on the body with a free-body

diagram.

• Select the extent of the free-body and detach it from the ground and all

other bodies

( )∑ ∑ ∑ =×== 00 FrMF O

rrrr

∑ =∑ =∑ =∑ =∑ =∑ =

000000

zyx

zyxMMMFFF

2

• Indicate point of application, magnitude, and direction of external

forces, including the rigid body weight

• Indicate point of application and assumed direction of unknown

applied forces. These usually consist of reactions through which the

ground and other bodies oppose the possible motion of the rigid body

• Include the dimensions necessary to compute the moments of the

forces Reactions at Supports and Connections for a Two-Dimensional Structure

• Reactions equivalent to a force with known line of action.

3

• Reactions equivalent to a force of unknown direction and magnitude. • Reactions equivalent to a force of unknown direction and magnitude

and a couple.of unknown magnitude Equilibrium of a Rigid Body in Two Dimensions • For all forces and moments acting on a two-dimensional structure

Ozyxz MMMMF ==== 00

4

• Equations of equilibrium become where A is any point in the plane of the structure • The 3 equations can be solved for no more than 3 unknowns

• The 3 equations can not be augmented with additional equations, but

they can be replaced

Statically Indeterminate Reactions

More unknowns than equations

Fewer unknowns than equations, partially

constrained

Equal number unknowns and equations but improperly constrained

∑ ∑ ∑ === 000 Ayx MFF

∑ ∑ ∑ === 000 BAx MMF

5

Sample Problem A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B.

SOLUTION:

• Create a free-body diagram for the crane • Determine B by solving the equation for the sum of the moments of

all forces about A. Note there will be no contribution from the unknown reactions at A.

• Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components

• Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.

• Create the free-body diagram

• Determine B by solving the equation for the sum of the moments of all forces about A.

( ) ( )( ) 0m6kN5.23

m2kN81.9m5.1:0=−

−+=∑ BM A

kN1.107+=B

6

• Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces

• Check the values obtained Equilibrium of a Rigid Body in Three Dimensions • Six scalar equations are required to express the conditions for the

equilibrium of a rigid body in the general three dimensional case. • These equations can be solved for no more than 6 unknowns which

generally represent reactions at supports or connections • The scalar equations are conveniently obtained by applying the vector

forms of the conditions for equilibrium

0:0 =+=∑ BAF xx

kN1.107−=xA

0kN5.23kN81.9:0 =−−=∑ yy AF

kN 3.33+=yA

∑ =∑ =∑ =∑ =∑ =∑ =

000000

zyx

zyxMMMFFF

( )∑ ∑ =∑ ×== 00 FrMF Orrrr

7

Reactions at Supports and Connections for a Three-Dimensional Structure

8

Sample Problem A sign of uniform density weighs 270 lb and is supported by a ball-and-socket joint at A and by two cables. Determine the tension in each cable and the reaction at A.

SOLUTION:

• Create a free-body diagram for the sign • Apply the conditions for static equilibrium to develop equations for

the unknown reactions • Create a free-body diagram for the sign.

Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading.

( )

( )kjiT

kjiT

rrrrTT

kjiT

kjiT

rrrrTT

EC

EC

EC

ECECEC

BD

BD

BD

BDBDBD

rrr

rrr

rr

rrr

rrr

rrr

rr

rrr

72

73

76

32

31

32

7236

12848

++−=

++−=

−−

=

−+−=

−+−=

−−

=

9

• Apply the conditions for static equilibrium to develop equations for the unknown reactions.

Solve the 5 equations for the 5 unknowns,

( )

( ) ( )

0lb1080571.2667.2:

0714.1333.5:

0lb 270ft 4

0:

0lb 270:

0:

0lb 270

72

32

73

31

76

32

=−+

=−

=−×+×+×=

=+−

=−++

=−−

=−++=

∑

∑

ECBD

ECBD

ECEBDBA

ECBDz

ECBDy

ECBDx

ECBD

TTk

TTj

jiTrTrM

TTAk

TTAj

TTAi

jTTAF

r

r

rrrrrrr

r

r

r

rrrrr

( ) ( ) ( )kjiA

TT ECBDrrvr

lb 22.5lb 101.2lb 338

lb 315lb 3.101

−+=

==

1

Materials Engineering Department Class: First

Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani

Centroids and Centers of Gravity Introduction: • The earth exerts a gravitational force on each of the particles forming

a body. These forces can be replace by a single equivalent force equal

to the weight of the body and applied at the center of gravity for the

body

• The centroid of an area is analogous to the center of gravity of a body.

The concept of the first moment of an area is used to locate the

centroid.

• Determination of the area of a surface of revolution and the volume of

a body of revolution are accomplished with the Theorems of Pappus-

Guldinus

Center of Gravity of a 2D Body

• Center of gravity of a plate

• Center of gravity of a wire

∫

∑∑∫

∑∑

=

∆=

=

∆=

dWy

WyWyMdWx

WxWxM

y

y

2

Centroids and First Moments of Areas and Lines • Centroid of an area

•

• Centroid of a line First Moments of Areas and Lines • An area is symmetric with respect to an axis BB’ if for every point P

there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’.

( ) ( )

x

QdAyAy

y

QdAxAx

dAtxAtx

dWxWx

x

y

respect toh moment witfirst

respect toh moment witfirst

=

==

=

==

=

=

∫

∫∫∫

γγ

( ) ( )

∫∫∫∫

=

=

=

=

dLyLy

dLxLx

dLaxLax

dWxWx

γγ

3

• The first moment of an area with respect to a line of symmetry is zero.

• If an area possesses a line of symmetry, its centroid lies on that axis

• If an area possesses two lines of symmetry, its centroid lies at their

intersection

• An area is symmetric with respect to a center O if for every element

dA at (x,y) there exists an area dA’ of equal area at (-x,-y).

• The centroid of the area coincides with the center of symmetry

4

Centroids of Common Shapes of Areas

Centroids of Common Shapes of Lines

5

Composite Plates and Areas

• Composite plates

• Composite area Sample Problem For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid.

SOLUTION:

• Divide the area into a triangle, rectangle, and semicircle with a circular cutout.

• Calculate the first moments of each area with respect to the axes • Find the total area and first moments of the triangle, rectangle, and

semicircle. Subtract the area and first moment of the circular cutout

∑∑∑∑

==

WyWYWxWX

∑∑∑∑

==

AyAYAxAX

6

• Compute the coordinates of the area centroid by dividing the first moments by the total area.

• Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout.

• Compute the coordinates of the area centroid by dividing the first moments by the total area

33

33

mm107.757

mm102.506

×+=

×+=

y

x

QQ

23

33

mm1013.828mm107.757

××+

==∑∑

AAx

X

mm 8.54=X

23

33

mm1013.828mm102.506

××+

==∑∑

AAy

Y

mm 6.36=Y

1

Materials Engineering Department

Class: First

Subject: Engineering Mechanics

Lecturer: Dr. Emad AL-Hassani

Friction Introduction • In preceding chapters, it was assumed that surfaces in contact were

either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent relative motion between surfaces).

• Actually, no perfectly frictionless surface exists. For two surfaces in

contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other.

• However, the friction forces are limited in magnitude and will not

prevent motion if sufficiently large forces are applied. • The distinction between frictionless and rough is, therefore, a matter

of degree. • There are two types of friction: dry or Coulomb friction and fluid

friction. Fluid friction applies to lubricated mechanisms. The present discussion is limited to dry friction between nonlubricated surfaces.

The Laws of Dry Friction. Coefficients of Friction • Block of weight W placed on horizontal surface. Forces acting on

block are its weight and reaction of surface N. • Small horizontal force P applied to block. For block to remain

stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force.

2

• As P increases, the static-friction force F increases as well until it reaches a maximum value Fm.

NF sm µ=

• Further increase in P causes the block to begin to move as F drops to a

smaller kinetic-friction force Fk.

NF kk µ= • Maximum static-friction force: • Kinetic-friction force:

NF sm µ=

sk

kk NFµµ

µ75.0≅

=

3

• Four situations can occur when a rigid body is in contact with a horizontal surface:

No friction, No motion (Px = 0) ,(Px < Fm)

• Motion impending, Motion, (Px > Fm) (Px = Fm)

4

Angles of Friction • It is sometimes convenient to replace normal force N and friction force

F by their resultant R:

No friction

• Motion impending

• No motion

ss

sms N

NN

F

µφ

µφ

=

==

tan

tan

• Motion

kk

kkk N

NNF

µφ

µφ

=

==

tan

tan

5

• Consider block of weight W resting on board with variable inclination angle θ.

• No friction No motion

• Motion impending Motion

6

Problems Involving Dry Friction

• All applied forces known • Coefficient of static friction is known • Determine whether body will remain at rest or slide

• All applied forces known • Motion is impending • Determine value of coefficient of static friction.

• Coefficient of static friction is known • Motion is impending • Determine magnitude or direction of one of the applied forces

7

Sample Problem A 450 N force acts as shown on a 1350 N block placed on an inclined plane. The coefficients of friction between the block and plane are ms = 0.25 and mk = 0.20. Determine whether the block is in equilibrium and find the value of the friction force.

SOLUTION:

• Determine values of friction force and normal reaction force from plane required to maintain equilibrium.

• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide.

• If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force.

The block will slide down the plane

:0=∑ xF ( ) 0N 1350 N 504 53 =−− F

N 360−=F

:0=∑ yF ( ) 0N 1350 54 =−N

N 1080=N

( ) N 270N 080125.0 === msm FNF µ

( )N 080120.0=== NFF kkactual µ

N 216=actualF