DocumentEM

St. PETER’S UNIVERSITY St. Peter’s Institute of Higher Education and Research

(Declared Under Section 3 of the UGC Act, 1956) AVADI, CHENNAI – 600 054

TAMIL NADU

STUDY MATERIAL

B. Tech. PROGRAMME (Code No. – 501 - 516)

(Effective From 2009 – 2010)

II SEMESTER

209BTT05 – ENGINEERING MECHANICS

St. PETER’S INSTITUTE OF DISTANCE EDUCATION (Recognized by Distance Education Council and Joint Committee of

UGC-AICTE-DEC, New Delhi. Ref. F. No. DEC/SPU/CHN/TN/Recog/09/14 dated 02.04.2009 and Ref.

F.No. DEC/Recog/2009/3169 dated 09.09.2009)

Copyright © Laxmi Publications (P) Ltd.

Published by: Laxmi Publications Pvt Ltd., 113, Golden House, Daryaganj, New Delhi-110 002. Tel: 43532500, E-mail: [email protected]

No part of this publication which is material protected by this copyright notice may be reproduced or transmitted or utilized or stored in any form or by any means now known or hereinafter invented, electronic, digital or mechanical, including photocopying, scanning, recording or by any information storage or retrieval system, without prior written permission from the publisher.

Information contained in this book has been published by Laxmi Publications (P) Ltd. and has been obtained by its authors from sources believed to be reliable and are correct to the best of their knowledge. The University has edited the study material to suit the curriculum and distance education mode. However, the publisher/university and its author shall in no event be liable for any errors, omissions or damages arising out of use of this information and specifically disclaim any implied warranties or merchantability or fitness for any particular use.

PREFACE

St. Peter’s University has been recognized by the Distance Education

Council and Joint Committee of UGC-AICTE-DEC, New Delhi, for offering

various programmes including B.Tech., D.Tech., MBA, MCA and other

programmes in Humanities and Sciences through Distance Education

mode.

The Methodology of distance education includes self-instructional

study materials in print form, face-to-face counselling, practical classes,

virtual classes in phased manner and end assessment.

The basic support for distance education students lies on the self-

instructional study materials. Keeping this in mind, the study materials

under distance mode are prepared. The main features of the study

materials are (1) learning objectives (2) self explanatory study

materials unitwise (3) self tests (4) list of references for further studies.

The material is prepared in simple English and graded in terms of

technical content. It is built upon the pre-requisite knowledge.

Students are advised to study the materials several times and get

benefitted. The face-to-face session in the counseling centre will help

them to clear their doubts and difficult concepts which they would have

faced during the learning process.

Students should remember that self study and sustained motivation

are the two important requirements for a successful learning under the

distance education mode.

We wish the students to put forth their best efforts to become

successful in their chosen field of learning.

Registrar

St. Peter’s University

CONTENTS

Page No.

Scheme of Examinations (vi)

Syllabus of Engineering Mechanics (vii)

Model Question Paper (ix)

Unit 1: Basics and Statics of Particles 1-43

Unit 2: Equilibrium of Rigid Bodies 44-66

Unit 3: Properties of Surfaces and Solids 67-124

Unit 4: Dynamics of Particles 125-194

Unit 5: Friction and Elements of Rigid Body Dynamics 195-217

Scheme of Examinations

I Semester

Code No. Course Title Credit Marks

Theory EA Total 109BTT01 Technical English – I 1 100 100 109BTT02 Mathematics – I 3 100 100 109BTT03 Engineering Physics – I 3 100 100 109BTT04 Engineering Chemistry – I 3 100 100 109BTT05 Engineering Graphics – I 3 100 100

109BTT06 Fundamentals of Computing and programming 3 100 100

109BTP01 Computer Practices Laboratory – I Record 1 90

10 100

109BTP02 Engineering Practices Laboratory – I Record 1 90

10 100

109BTP03 *Physics & Chemistry Laboratory - I - - - Total 18 800 800

*The Practical Examinations of First Semester will be held only along with those of the second semester.

II Semester

Code No. Course Title Credit Marks

Theory EA Total209BTT01 Technical English – II 2 100 100 209BTT02 Mathematics – II 3 100 100 209BTT03 Engineering Physics – II 2 100 100 209BTT04 Engineering Chemistry – II 2 100 100

209BTT05 209BTT06 209BTT07

Engineering Mechanics (For non-circuit branches) Circuit Theory (For branches under Electrical Faculty) Electric Circuits and Electron Devices (For branches under I & C Faculty)

3

100

100

209BTT08 209BTT09

Basic Electrical & Electronics Engineering (For non-circuit branches) Basic Civil & Mechanical Engineering (For circuit branches)

3

100 100

209BTP01 Computer Practices Laboratory – II Record 1 90

10 100

209BTP02 Physics & Chemistry Laboratory – II Record 1 90

10 100

209BTP03 209BTP04 209BTP05

Computer Aided Drafting and Modelling Laboratory(For non-circuit branches) Electrical Circuits Laboratory (For branches under Electrical Faculty) Circuits and Devices Laboratory (For branches under I & C Faculty) Record

1

90 90 90 10

100 100 100

Total 18 900 900

SYLLABUS 209BTT05 ENGINEERING MECHANICS UNIT I BASICS & STATICS OF PARTICLES Introduction – Units and Dimensions – Laws of Mechanics – Lame’s theorem, Parallelogram and triangular Law of forces – Vectors – Vectorial representation of forces and moments – Vector operations: additions, subtraction, dot product, cross product – Coplanar Forces – Resolution and Composition of forces – Equilibrium of a particle – Forces in space – Equilibrium of a particle in space – Equivalent systems of forces – Principle of transmissibility – Single equivalent force. UNIT II EQUILIBRIUM OF RIGID BODIES

Free body diagram – Types of supports and their reactions – requirements of stable equilibrium – Moments and Couples – Moment of a force about a point and about an axis – Vectorial representation of moments and couples – Scalar components of a moment – Varignon’s theorem – Equilibrium of Rigid bodies in two dimensions – Equilibrium of Rigid bodies in three dimensions – Examples UNIT III PROPERTIES OF SURFACES AND SOLIDS

Determination of Areas and Volumes – First moment of area and the Centroid of sections – Rectangle, circle, triangle from integration – T section, I section, - Angle section, Hollow section by using standard formula – second and product moments of plane area – Rectangle, triangle, circle from integration – T section, I section, Angle section, Hollow section by using standard formula – Parallel axis theorem and perpendicular axis theorem – Polar moment of inertia – Principal moments of inertia of plane areas – Principal axes of inertia – Mass moment of inertia – Derivation of mass moment of inertia for rectangular section, prism, sphere from first principle – Relation to area moments of inertia. UNIT IV DYNAMICS OF PARTICLES

Displacements, Velocity and acceleration, their relationship – Relative motion – Curvilinear motion – Newton’s law – Work Energy Equation of particles – Impulse and Momentum – Impact of elastic bodies. UNIT V FRICTION AND ELEMENTS OF RIGID BODY DYNAMICS

Frictional force – Laws of Coloumb friction – simple contact friction – Rolling resistance – Belt friction. Translation and Rotation of Rigid Bodies – Velocity and acceleration – General Plane motion.

TEXT BOOK:

1. Beer, F.P and Johnson Jr. E.R. “Vector Mechanics for Engineers”, Vol. 1 Statics and Vol. 2 Dynamics, McGraw-Hill International Edition, (1997).

REFERENCES:

1. Rajasekaran, S, Sankarasubramanian, G., “Fundamentals of Engineering Mechanics”, Vikas Publishing House Pvt. Ltd., (2000).

2. Hibbeller, R.C., “Engineering Mechanics”, Vol. 1 Statics, Vol. 2 Dynamics, Pearson Education Asia Pvt. Ltd., (2000).

3. Palanichamy, M.S., Nagam, S., “Engineering Mechanics – Statics & Dynamics”, Tata McGraw-Hill, (2001).

4. Irving H. Shames, “Engineering Mechanics – Statics and Dynamics”, IV Edition – Pearson Education Asia Pvt. Ltd., (2003).

5. Ashok Gupta, “Interactive Engineering Mechanics – Statics – A Virtual Tutor (CDROM)”, Pearson Education Asia Pvt., Ltd., (2002).

MODEL QUESTION PAPER

B.Tech. DEGREE EXAMINATIONS

Second Semester

209BTT05-ENGINEERING MECHANICS(Regulations 2009)

Time : 3 Hours Maximum Marks : 100

Answer All the Questions

PART–A (10 × 2 = 20 Marks)1. Define Lami’s Theorem.

2. Explain Principle of Transmissibility of Forces.

3. What are the Types of Loads?

4. Define Principle of Moments.

5. Define Centroid.

6. Write the Equation for Radius of Gyration.

7. Define D’Alemberts Principle.

8. Define Friction.

9. Define Velocity.

10. What is the equation of Power Transmitted by Belt Drive?

PART–B (5 × 16 = 80 Marks)

11. (a) Two forces P and Q are acting at a point O as shown in figure. The Force P = 240 N and Q = 200 N.If the resultant of the forces is equal to 400 N, then find the values of α, β and γ.

��

�

O P A

(180° – )��

B C

�

RQQ

Or(b) The figure shows the two vertical forces and a couple

of moment 2000 Nm acting on a horizontal rod whichis fixed at end A.

(i) Determine the resultant of the system.(ii) Determine an equivalent system through A. A

1 m 1.5 m

2000Nm

C B

4000 N 2500 N

0.8 m

D

12. (a) A beam AB of span 8 m overhanging on both sides, is loaded as shown in the figure. Calculate thereactions at both ends.

800 N 2000 N 1000 N

A

5 m

3 m 8 m 2 m

B

RA RB

Or(b) Determine the forces in all members of a cantilever truss shown in the figure.

2 m 2 m1000 N 1000 N

A B C

3 mD

E

13. (a) Figure shows the T section of dimensions 10 × 10 × 2 cm. Determine the moment of inertia of thesection about the horizontal and vertical axis passing through the centre of gravity of the section.

A B

2 cm

D CEH

1

10 cm

2

FG2 cm

10 cm

Or(b) Determine the polar moment of inertia of the I section shown in the figure. All dimensions are in mm.

12

12

10

1

2

3

128150

120

80

14. (a) A body is moving with a uniform acceleration and covers 15 m in fifth second and 25 m in the tenthsecond. Determine

(i) Initial velocity of the body.(ii) Acceleration of the body.

Or(b) A car moving on a straight level road skidded for a total distance of 60 m after the brake were

applied. Determine the speed of the car just before the brakes were applied, if the coefficient offriction between the car tyre and road is 0.4. Take g = 9.81 ms–2.

15. (a) A cord connects two bodies of weights 300 N and 800 N. The two bodies are placed on a inclinedplane and chord is parallel to inclined plane. The coefficient of friction for the weight of 400 N is 0.15and that for 800 N is 0.4. Determine the inclination of the plane to the horizontal and the tension in thechord when the motion is about to take place, down the inclination plane. The body weighing 400 N isbelow the body weighing 800 N.

Or(b) Determine the maximum power that can be transmitted using a belt of 100 mm × 10 mm with an

angle of lap of 160 degree. The density of belt is 10–3 gm/mm3 and coefficient of friction may betaken as 0.25. The tension in the belt not to exceed 1.5 N/mm2.

Self-Instructional Material 1

NOTES

Basics and Staticsof Particles

U N I T

1BASICS AND STATICS

OF PARTICLES

STRUCTURE

1.1 Introduction1.2 Lame’s Theorem1.3 Units and Dimensions1.4 Laws of Mechanics1.5 Parallelogram and Triangular Law of Forces1.6 Vectors1.7 Vector Operations1.8 Resolution and Composition of a Force1.9 Coplanar Forces1.10 Resultant of Coplanar Forces1.11 Equilibrium of a Particle1.12 Equilibrium of a Rigid Body1.13 Forces in Space1.14 Equilibrium of a Particle in Space1.15 Equivalent Systems of Foces1.16 Principle of Transmissibility1.17 Single Equivalent Force• Summary• Glossary• Review Questions• Further Readings

��

After going through this unit, you should be able to :• know about terms, generally used in Mechanics.• define unit and dimensions in all three types of system and laws and principles

considered as the foundation of Mechanics.

2 Self-Instructional Material

NOTES

Engineering Mechanics • define vectors, its representation and operations.• know about coplanar forces, its resolution and composition of forces.• know—equilibrium of particle, forces in space, principle of transmissibility

and single equivalent force.

��

Engineering mechanics is that branch of science which deals with the behaviourof a body when the body is at rest or in motion. The engineering mechanics may bedivided into statics and dynamics. The branch of science, which deals with the studyof a body when the body is at rest, is known as statics while the branch of sciencewhich deals with the study of a body when the body is in motion, is known asdynamics. Dynamics is further divided into kinematics and kinetics. The study of abody in motion, when the forces which cause the motion are not considered, is calledkinematics and if the forces are also considered for the body in motion, that branchof science is called kinetics. The classification of Engineering Mechanics are shownin Fig. 1.1 below.

ENGINEERING MECHANICS

1. Statics(Body is at rest)

2. Dynamics(Body is in motion)

(i) Kinematics (ii) Kinetics

Fig. 1.1

Note. Statics deals with equilibrium of bodies at rest, whereas dynamics deals withthe motion of bodies and the forces that cause them.

The following terms are generally used in Mechanics :1. Vector quantity, 2. Scalar quantity, 3. Particle, 4. Law of parallelogram of

forces, 5. Triangle law and 6. Lame’s theorem.1. Vector Quantity. A quantity which is completely specified by magnitude

and direction, is known as a vector quantity. Some examples of vector quantities are :velocity, acceleration, force and momentum. A vector quantity is represented by meansof a straight line with an arrow as shown in Fig. 1.2. The length of the straight line(i.e., AB) represents the magnitude and arrow represents the direction of the vector.

The symbol AB→

also represents this vector, which means it is acting from A to B.

A B

Fig. 1.2. Vector quantity

2. Scalar Quantity. A quantity, which is completely specified by magnitudeonly, is known as a scalar quantity. Some examples of scalar quantity are : mass,length, time and temperature.

3. A Particle. A particle is a body of infinitely small volume (or a particle is abody of negligible dimensions) and the mass of the particle is considered to be con-centrated at a point. Hence a particle is assumed to a point and the mass of theparticle is concentrated at this point.


NOTES


4. Law of Parallelogram of Forces. The law of parallelogram of forces isused to determine the resultant of two forces acting at a point in a plane. It states,“If two forces, acting at a point be represented in magnitude and direction by thetwo adjacent sides of a parallelogram, then their resultant is represented in magnitudeand direction by the diagonal of the parallelogram passing through that point.”

Let two forces P and Q act at a point O as shown in Fig. 1.3. The force P isrepresented in magnitude and direction by OA whereas the force Q is presented inmagnitude and direction by OB. Let the angle between the two forces be ‘α’. Theresultant of these two forces will be obtained in magnitude and direction by thediagonal (passing through O) of the parallelogram of which OA and OB are twoadjacent sides. Hence draw the parallelogram with OA and OB as adjacent sides asshown in Fig. 1.4. The resultant R is represented by OC in magnitude and direction.

B

Q

O

α αα

θ

P A

B C

Q

O P A D

R

Fig. 1.3 Fig. 1.4

Magnitude of Resultant (R)From C draw CD perpendicular to OA produced.Let α = Angle between two forces P and Q = ∠AOBNow ∠DAC = ∠AOB (Corresponding angles)

= αIn parallelogram OACB, AC is parallel and equal to OB.∴ AC = Q.In triangle ACD,

AD = AC cos α = Q cos αand CD = AC sin α = Q sin α.

In triangle OCD,OC2 = OD2 + DC2.

But OC = R, OD = OA + AD = P + Q cos αand DC = Q sin α.

∴ R2 = (P + Q cos α)2 + (Q sin α)2

= P2 + Q2 cos2 α + 2PQ cos α + Q2 sin2 α= P2 + Q2 (cos2 α + sin2 α) + 2PQ cos α= P2 + Q2 + 2PQ cos α (� cos2 α + sin2 α = 1)

∴ R = P Q PQ2 2 2+ + cos α ...(1.1)

Equation (1.1) gives the magnitude of resultant force R.


NOTES

Engineering Mechanics Direction of Resultant

Let θ = Angle made by resultant with OA.Then from triangle OCD,

tan θ = CDOD

QP Q

=+

sincos

αα

∴ θ = tan–1 Q

P Qsin

cosα

α+��

��

...(1.2)

Equation (1.2) gives the direction of resultant (R).

αθ

QR

C

(180– )α

AO

( – )α θ

P

Fig. 1.5

The direction of resultant can also be obtained by using sine rule [In triangleOAC, OA = P, AC = Q, OC = R, angle OAC = (180 – α), angle ACO = 180 – [θ + 180– α] = (α – θ)]

sin sin ( ) sin ( )θ α α θAC OC OA

= − = −180

sin sin ( ) sin ( )θ α α θQ R P

= − = −180

Two cases are important.1st Case. If the two forces P and Q act at right angles, then

α = 90°From equation (1.1), we get the magnitude of resultant as

R = P Q PQ P Q PQ2 2 2 22 2 90+ + = + + °cos cosα

= P Q2 2+ (� cos 90° = 0) ...(1.2A)

From equation (1.2), the direction of resultant is obtained as

θ = tan–1 Q

P Qsin

cosα

α+��

��

= tan–1 Q

P QQP

sincos

tan90

901°

+ °��

��

= −

(� sin 90° = 1 and cos 90° = 0)2nd Case. The two forces P and Q are equal and are acting at an angle α

between them. Then the magnitude and direction of resultant is given as

R = P Q PQ P P P P2 2 2 22 2+ + = + + × ×cos cosα α(� P = Q)

= 2 2 2 12 2 2P P P+ = +cos ( cos )α α

= 2 22

2 2P × cosα

� 1 22

2+ =��

��

cos cosα α


NOTES

Basics and Staticsof Particles= 4

22

22 2P Pcos cos

α α= ...(1.3)

and θ = tan–1 QP Q

PP P

sincos

tansin

cosα

αα

α+��

��

=+

−1 (� P = Q)

= tan–1 PP

sin( cos )

tansin

cosα

αα

α1 11

+=

+−

= tan–1 2

2 2

22

2

sin cos

cos

α α

α� sin sin cosα α α=��

��

22 2

= tan–1 sin

costan tan

α

αα α2

22 2

1= ��

��

=− ...(1.4)

It is not necessary that one of two forces,should be along x-axis. The forces P and Q maybe in any direction as shown in Fig. 1.6. If theangle between the two forces is ‘α’, then theirresultant will be given by equation (1.1). Thedirection of the resultant would be obtainedfrom equation (1.2). But angle θ will be theangle made by resultant with the direction ofP.

5. Law of Triangle of Forces. It statesthat, “if three forces acting at a point berepresented in magnitude and direction by thethree sides of a triangle, taken in order, they will be in equilibrium.”

��

It states that, “If three forces actingat a point are in equilibrium, each force willbe proportional to the sine of the anglebetween the other two forces.”

Suppose the three forces P, Q and Rare acting at a point O and they are in equi-librium as shown in Fig. 1.7.

Let α = Angle between force P and Q.β = Angle between force Q and R.γ = Angle between force R and P.

Then according to Lame’s theorem,P is proportional sine of angle between Q and R α sin β.

∴P

sin β = constant

Fig. 1.6

Fig. 1.7

α θ

Q

R

P

O

Q

β α

γ

O

R

P


NOTES

Engineering MechanicsSimilarly

Qsin γ

= constant and R

sin α = constant

or P Q R

sin sin sinβ γ α= = ...(1.5)

Proof of Lame’s Theorem. The three forces acting on a point, are inequilibrium and hence they can be represented by the three sides of the triangle takenin the same order. Now draw the force triangle as shown in Fig. 1.8.

P

R

O

Q

α(180 – )α

β

(180 –)β

(180 – )γ

γ

Fig. 1.8

Now applying sine rule, we getP Q R

sin ( ) sin ( ) sin ( )180 180 180−=

−=

−β γ αThis can also be written

P Q Rsin sin sinβ γ α

= =

This is same equation as equation (1.5).Note. All the three forces should be acting either towards the point or away from the

point.

��

The following units of different systems are mostly used :1. C.G.S. (i.e., Centimetre-Gram Second) system of units.2. M.K.S. (i.e., Metre-Kilogram-Second) system of units.3. S.I. (i.e., International) system of units.C.G.S. System of Units. In this system, length is expressed in centimetre,

mass in gram and time in second. The unit of force in this system is dyne, which isdefined as the force acting on a mass of one gram and producing an acceleration ofone centimetre per second square.

M.K.S. System of Units. In this system, length is expressed in metre, massin kilogram and time in second. The unit of force in this system is expressed askilogram force and is represented as kgf.

S.I. System of Units. S.I. is abbreviation for ‘The System International Units’.It is also called the International System of Units. In this system length is expressedin metre mass in kilogram and time in second. The unit of force in this system isNewton and is represented N. Newton is the force acting on a mass of one kilogram


NOTES


and producing an acceleration of one metre per second square. The relation betweennewton (N) and dyne is obtained as

One Newton = One kilogram mass × One metre

s2

= 1000 gm × 100 cm

s2 (� one kg = 1000 gm)

= 1000 × 100 × gm cm

s2×

= 105 dyne � dynegm cm

s2= ×��

��

When the magnitude of forces is very large, then the unit of force like kilo-newton and mega-newton is used. Kilo-newton is represented by kN.

One kilo-newton = 103 newtonor 1 kN = 103 Nand One mega newton = 106 Newton

The large quantities are represented by kilo, mega, giga and tera. They standfor :

Kilo = 103 and represented by .......kMega = 106 and represented by .......M Giga = 109 and represented by .......G Tera = 1012 and represented by ........T

Thus mega newton means 106 newton and is represented by MN. Similarly,giga newton means 109 N and is represented by GN. The symbol TN stands for 1012

N.The small quantities are represented by milli, micro, nano and pico. They are

equal to Milli = 10–3 and represented by .......m

Micro = 10–6 and represented by .......µNano = 10–9 and represented by ........nPico = 10–12 and represented by .......p.

Thus milli newton means 10–3 newton and is represented by mN. Micro newtonmeans 10–6 N and is represented by µN.

The relation between kilogram force (kgf) and newton (N) is given by One kgf= 9.81 N

Weight of a body is the force with which the body is attracted towardsearth. If W = weight of a body, m = mass in kg, then W = m × g Newtons

If mass, m of the body is 1 kg, then its weight will be,

W = 1 (kg) × 9.81 ms2 = 9.81 N. � N = kg

ms2

��

��


NOTES

Engineering Mechanics Table 1.1 shows the multiples and sub-multiples of the S.I. units prefixes.

Table 1.1. S.I. Prefixes

Prefix Symbol Multiplying factor

Tera T 1012 = 1 000 000 000 000

Giga G 109 = 1 000 000 000

Mega M 106 = 1 000 000

Kilo k 103 = 1 000

Milli m 10–3 = 0.001

Micro µ 10–6 = 0.000 001

Nano n 10–9 = 0.000 000 001

Pico p 10–12 = 0.000 000 000 001

The basic, supplementary and some derived units with dimensions are given as :

(A) Basic Units

Physical quantity Notation or Unit Dimension or Symbol

Length metre m

Mass kilogram kg

Time Second s

Electric Current Ampere A

Temperature Kelvin K

Luminous Intensity Candella cd

(B) Supplementary Units

Plane angle Radian rad

Solid angle Steradian sr

(C) Derived Units

Acceleration metre/second2 m/s2

Angualr velocity radian/second rad/s

Angular acceleration radian/second2 rad/s2

Force Newton N = kg m/s2

Moment of force Newton metre Nm

Work, Energy Joule J = Nm = kg m2/s2

Torque Newton metre Nm

Power Watt W = J/s

Pressure Pascal Pa = N/m2

Frequency Hertz Hz = s–1

Problem 1. Two forces of magnitude 10 N and 8 N are acting at a point. If theangle between the two forces is 60°, determine the magnitude of the resultant force.

Sol. Given :Force P = 10 NForce Q = 8 N


NOTES


Angle between the two forces, α = 60°The magnitude of the resultant force (R) is given by equation (1.1)

R = P Q PQ2 2 2 22 10 8 2 10 8 60+ + = + + × × × °cos cosα

= 100 64 2 10 8 12+ + × × × (� cos 60° = 1

2 )

= 100 64 80 244+ + = = 15.62 N. Ans.Problem 2. An electric light fixture weighing

15 N hangs from a point C, by two strings AC andBC. AC is inclined at 60° to the horizontal and BCat 45° to the vertical as shown in Fig. 1.9. UsingLame’s theorem or otherwise determine the forces inthe strings AC and BC.

Sol. Given : Weight at C = 15 N∠OAC = 60°∠CBD = 45°

Let T1 = Force in string BC T2 = Force in string AC

1st MethodUsing Lame’s theorem at C

15 1 2

sin sin sinof of of∠=

∠=

∠BCAT

ACET

ACEBut ∠BCA = 45° + 30° = 75°

∠ACE = 180° – 30° = 150° ∠BCE = 180° – 45° = 135°

∴ 15

75 150 1351 2

sin sin sin°=

°=

°T T

∴ T1 = 15 150

75× °

°sin

sin = 7.76 N. Ans.

and T2 = 15 135

75× °

°sin

sin = 10.98 N. Ans.

2nd MethodThe point C is in the equilibrium. The forces acting at C are 15 N, T1 and T2.Resolving all forces at C in the horizontal direction

T1 sin 45° = T2 sin 30° or T1 × 1

2

122= ×T

∴ T2 = T1 × 2

22 1= × T ...(i)

Resolving all forces at C in the vertical direction, T1 cos 45° + T2 cos 30° = 15

or T1 × 1

2

32

152+ × =T ...(ii)

Substituting the value of T2 from equation (i) into equation (ii),

T1 × 1

22

321+ × ×T = 15

45°

60°

30°45°

F

T1

T2

CE

15 N

B

OA

D

Fig. 1.9


NOTES

Engineering Mechanics

or T T1 1

2

3

215+ = �

22

22 2

12

=×

=�

��

��

or T1 + 3 T1 = 15 × 2 or T1(1 + 3 ) = 15 × 2

∴ T1 = 15 21 3

×× = 7.76 N. Ans.

Substituting this value of T1 in equation (i), we get

T2 = 2 2 7 761× = ×T . = 10.98 N. Ans.

��

The following basic laws and principles are considered to be the foundation ofmechanics :

1. Newton’s first and second laws of motion2. Newton’s third law3. The gravitational law of attraction4. The parallelogram law5. The principle of transmissibility of forces.

Newton’s First and Second Laws of Motion

Newton’s first law states, “Every body continues in a state of rest or uniformmotion in a straight line unless it is compelled to change that state by some externalforce acting on it.”

Newton’s second law states, “The net external force acting on a body in adirection is directly proportional to the rate of change of momentum in that direction.”

Newton’s Third Law

Newton’s third law states, “To every action there is always equal and oppositereaction.”

Fig. 1.10 shows two bodies A and B which are placed one above the other ona horizontal surface.

Here F1 = Force exerted by horizontal surface on body A (action)– F1 = Force exerted by body A on horizontal surface (reaction)

F2 = Force exerted by body A on body B (action)– F2 = Force exerted by body B on body A (reaction)

B

A

B

F2

F1

–F2

–F1

A

Fig. 1.10


NOTES


The Gravitational Law of Attraction

It states that two bodies will be attracted towards each other along theirconnecting line with a force which is directly proportional to the product of theirmasses and inversely proportional to the square of the distance between their centres.

Refer to Fig. 1.11.

m1m2

F F

r

Fig. 1.11

Let m1 = Mass of first bodym2 = Mass of second body

r = Distance between the centre of bodiesF = Force of attraction between the bodies.

Then according to the law of gravitational attraction.F ∝ m1 . m2

∝ 12r

or F ∝ m m

r1 2

2

or F = G m m

r2 2

2 ...(1.6)

where G = Universal gravitational constant of proportionality.In equation (1.6), F is expressed in N, m1 and m2 in kg and r in m. Hence di-

mensionally, equation (1.6) becomes as

N = G kg kg

m2×

or G = Nmkg

2

2 (i)

But 1 N = 1 kg × 1 ms2 or N = kg ×

ms2

Substituting the value of N is equation (i),

G = kgms

mkg

mkg s2

2

2

3

2×��

��

× = ...(ii)

Hence from equations (i) and (ii), dimension of Universal GravitationalConstant G is N m2/kg2 or m3/kg s2.

The value of G is 6.67 × 10–11 N m2/kg2 or m3/kg s2.In equation (1.6), if m1 = 1 kg, m2 = 1 kg and r = 1 m, then F = G. This means

that the force of attraction between two bodies of mass 1 kg each when they are at adistance of 1 m apart, will be 6.67 × 10–11 N i.e., 0.0000000000667 N. This force isvery very small.

Weight

The weight of a body is defined with the help of law of gravitation. Weight isdefined as the force with which a body is attracted towards the centre of earth.


NOTES

Engineering Mechanics Let M = Mass of the bodyME = Mass of the earth = 5.9761 × 1024 kg

r = Distance between the centres of the earth and the body= 6.371 × 106 m (i.e., radius of earth)

G = Universal gravitational constant = 6.67 × 10–11 N m2/kg2

F = Force of attraction which is equal to weight (W)Substituting these values in equation (1.6), we get

W = G M M

rE ×

2 (� F = W, m1 = M, m2 = ME)

=

6 67 10 5 9761 10

6 371 10

11 24

6 2

. . ( ) ( )

( . )

×�

��

��× × ×

×

− N mkg

kg kg

m

2

2 M

= 6 67 10 5 9761 10

6 371 10

11 24

2 12 2. .

.× × ×

× ×

− Mm

N mkg

2

2 × kg × kg = (9.81 × M) N

where 9.81 is acceleration due to gravity and is denoted by ‘g’.∴ W = g × M or M × g

Actually the term GM

rE

2 is equal to 9.81 ms2

, which is represented by ‘g’.

The Parallelogram Law

This law has been already defined. It states that if two forces acting at a pointbe represented in magnitude and direction by the two adjacent sides of aparallelogram, then their resultant is represented in magnitude and direction by thediagonal of the parallelogram passing through that point.

The Principle of Transmissibility of Forces

It states that if a force, acting at a point on a rigid body, is shifted to any otherpoint which is on the line of action of the force, the external effect of the force on thebody remains unchanged.

F = F1F = F1

F = F2

O′ O′O′

OO

O

FF

(a) (b) (c)

Fig. 1.12

For example, consider a force F acting at point O on a rigid body as shown inFig. 1.12 (a). On this rigid body, “there is another point O′ in the line of action of theforce F. Suppose at this point O′, two equal and opposite forces F1 and F2 (each equalto F and collinear with F) are applied as shown in Fig. 1.12 (b). The force F and F2,


NOTES


being equal and opposite, will cancel each other, leaving a force F1 at point O′ asshown in Fig. 1.12 (c). But force F1 is equal to force F.

The original force F acting at point O, has been transferred to point O′ whichis along the line of action of F without changing the effect of the force on the rigidbody. Hence any force acting at a point on a rigid body can be transmitted to act atany other point along its line of action without changing its effect on the rigid body.This proves the principle of transmissibility of a force.

��

Parallelogram law of forces states that, “If two forces acting at a point be rep-resented in magnitude and direction by the two adjacent sides of a parallelogram, thentheir resultant is represented in magnitude and direction by the diagonal of theparallelogram passing through that point.’’

Triangle law of forces states that, “if three forces acting at a point berepresented in magnitude and direction be the three sides of the triangle taken in order,they will be in equilibrium”

��

All quantities that have magnitude and direction are known as vectors. A vectorwill be represented by a bold face letter (i.e., with thick capital letter). The force Fwhich is a vector will be represented by F. The alternate way of representing the

force vector is by putting an arrow as F→

.

Magnitude of a Vector. The magnitude of a quantity is always a positivenumber. Thus the magnitude of a quantity (– 40) is + 40 units. The magnitude of aquantity mathematically is represented by a set of vertical lines enclosing thequantity. Hence mathematically the magnitude of a quantity (– 40) is representedas :

| – 40 units | = + 40 unitsSimilarly, the magnitude of a vector is a positive number of units corresponding

to the length of the vector in those units. Hence magnitude of a vector A is representedas :

Magnitude of vector A = | A |where | A | is a positive scalar quantity. The magnitude is also represented by A(i.e., capital letter A).

∴ Magnitude of a vector A = | A | = A.

Multiplication of a Vector by a ScalarLet A = a given vector

m = a given scalar

∴ Vector A multiplied by scalar m = mAwhere mA = a vector having same direction as A and a magnitude equal to theordinary scalar product between magnitudes of m and A.


NOTES

Engineering Mechanics Vectorial Representation of Forces and Moments

(A) Vectorial representation of a force.Fig. 1.13 shows a force F acting at the origin O. Letthe magnitude of this force is equal to length OA.Then this force vector F is represented by vector OA.Through point A, draw planes parallel to co-ordinateplanes.

These planes along with co-ordinate planesmake a rectangular box. The force F is then repre-sented by the diagonal of the box and its threecomponents Fx , Fy and Fz by its edges.

Let θx = Angle made by force F with x-axisθy = Angle made by force F with y-axisθz = Angle made by force F with z-axis

Now Fx = F cos θxFy = F cos θy ...(1.7)Fz = F cos θz

Also F = F F Fx y z2 2 2+ + ...(1.7A)

The cosines of θx, θy and θz are known as the direction cosines of the force Fand denoted by

l = cos θx, m = cos θy and n = cos θz ...(1.7B)The three angles are related by

cos2 θx + cos2 θy + cos2 θz = 1 ...(1.8)or l2 + m2 + n2 = 1

Let i = vector of unit length in the positive x-direction,

j = vector of unit length in the positive y-direction,

and k = vector of unit length in the positive z-direction.

Then force vector F is represented by

F = Fx i + Fy j + Fz k ...(1.9)

Magnitude of the vector F is given by

F = | F | = F F Fx y z2 2 2+ +

But Fx = F cos θx, Fy = F cos θy and Fz = F cos θz. Substituting these values inequation (1.9), we get

F = (F cos θx) i + (F cos θy) j + (F cos θz) k ...(1.10)Unit Vector. The vector having a unit length (or unit magnitude) is known

as unit vector. Hence unit vector corresponding to force vector F is equal to F ÷Magnitude of vector F.

∴ Unit vector = Force vector

Magnitude of vectorF F

F F=

= F

F F Fx y z2 2 2+ +

...(1.11)

=+ +

+ +

F F F

F F F

x y z

x y z

i j k2 2 2 (� F = Fxi + Fy j + Fzk)

θyθx

θz

Y

Z

OX

FA

Fig. 1.13�

�


NOTES


Note. Force vector is represented by THICK, BOLD FACE and magnitude is repre-sented by capital letters only.

(B) Vectorial representation of moments.The moment of a force is a vector which is the productof distance and force. Hence in case of moment of a forcethe cross-product of distance and force would be taken.Consider the Fig. 1.14.

Let F = Force vector (Fxi + Fy j + Fzk)

r = Distance (or position) vector with respect

to O

= xi + yj + zk

M = Moment of force about point O

Then M = r × F

or M = r × F = i j kx y z

F F Fx y z

∴ M = (yFz – zFy) i + (zFx – xFz) j + (xFy – yFx) kThe moment of the given force about x, y and z-axis are equal to

or Mx = yFz – zFy , My = zFx – xFz , Mz = xFy – yFx

where Mx = Moment of F about x-axisMy = Moment of F about y-axis, andMz = Moment of F about z-axis.

Problem 3. A force F = 2i + 4j – 3k is applied ata point P(1, 1, – 2). Find the moment of the force F aboutthe point (2, – 1, 2).

Sol. Given :Force F = 2i + 4j – 3kThe position vector r of the point P w.r.t. O.

= Position vector of point P– Position vector of point O.

= (i + j – 2k) – (2i – j + 2k)∴ r = (1 – 2)i + [1 + (1)] j + [– 2 – 2]k = – i + 2j – 4kThe moment M is given by

M = r × F = i j k

− −−

1 2 42 4 3

= [(2)(– 3) – (– 4)(4)] i + [(– 4)(2) – (– 1) (– 3)] j + [(– 1)(4) – (2)(2)]k= (– 6 + 16) i + (– 8 – 3) j + (– 4 – 4) k = 10i – 11j – 8k. Ans.

��

The followings are the important vector operations :(i) Vector addition, (ii) Vector subtraction

(iii) Dot product, and (iv) Cross product or Vector product.

Fz

Fx

FFyY

My

r

Z

MzMx

OX

Fig. 1.14

Fig. 1.15

r

O (2, –1, 2)

F = 2 + 4 – 3ki j

P(1, 1, –2)


NOTES

Engineering Mechanics Vector Addition. The sum of two or more vectors can be obtained by addingthe respective components of the vectors. The two vectors A and B will be added asgiven below :

Vector A = Axi + Ay j + AzkVector B = Bxi + By j + Bzk∴ A + B = (Ax + Bx) i + (Ay + By) j + (Az + Bz) kAlso A + B = B + A (Commutative law)Vector Subtraction. The subtraction of a vector

from the other vector is obtained by subtracting therespective components. The vector B can be subtracted fromvector A as given below :

Vector A = Axi + Ayj + AzkVector B = Bxi + Byj + Bzk∴ A – B = (Ax – Bx) i + (Ay – By) j + (Az – Bz) k.Dot Product. Dot product is also known as a scalar

product. The dot product of two vectors A and B isdenoted by A . B (read as A dot B) and is a scalar equal tomagnitude of A times the magnitude of B times the cosine of the smaller angle θbetween them.

∴ A . B = | A | . | B | cos θ = AB cos θBut from Fig. 1.16,B cos θ = Projection of B on A

∴ A . B = | A | times [Projection of B on A]

Let A = Axi + Ay j + Azk

B = Bxi + By j + Bzk

A . B = (Axi + Ay j + Azk) . (Bxi + By j + Bzk)

= AxBxi . i + Ax By i . j + AxBzi . k + AyBx j . i + AyBy j . j

+ AyBz j . k + AzBx k . i + AzBy k . j + AzBz k . k

= AxBx + 0 + 0 + 0 + AyBy + 0 + 0 + 0 + AzBz

(� i . i = 1, j . j = 1 and k . k = 1, as the angle between unit vector and itself is zeroand cosine of angle zero is equal to 1. Also the magnitude of unit vector is unity.

And i . j = j . k = k . i = i . k = j . i = k . j = 0 as the angle between them is 90°and cosine of 90° is zero.)

∴ A . B = AxBx + AyBy + AzBz

Hence the dot product of two vectors is equal to the sum of the products oftheir respective scalar components.

Then scalar product in terms of components

AB = AxBx + AyBy + AzBz ...(1.12)

The angle between vectors A and B is given by

cos θ = A . BA B

A . B| || |

=AB

. ..(1.13)

θ

AB cos θProjectionof B on A

B

Fig. 1.16


NOTES


Note. (i) If θ = 0, the vectors A and B are acting in the same direction and hence

A . B = AB

(ii) If θ = 90° (cos θ = 0), the vectors A and B are perpendicular and hence

A . B = 0.

(iii) A . B = B . A (Commutative law)

(iv) The dot product of orthogonal unit vectors i, j, k is given as : i . i = 1, j . j = 1, k . k =1 because the dot product of a unit vector with itself is unity as it is the product ofunit magnitude by cosine of zero angle between their directions and i . j = j . k = k . i= 0 as now the angle between their direction is 90° and cosine of 90° is zero.

Cross Product or Vector Product. Cross product is also known as vectorproduct. The cross product of two vectors A and B is denoted by A × B (read as Across B) and is a vector :

(i) whose magnitude is equal to the magnitude of A times the magnitude of Btimes the sine of the smaller angle between the two vectors ;

(ii) whose direction is perpendicular to the plane containing the vectors A and B; and

(iii) whose sense is given by right-hand screw rule.

Hence cross product of vector A and vector B inclined to each other at an angleθ is a vector whose magnitude is AB sin θ and direction is perpendicular to theplane containing vector A and vector B as shown in Fig. 1.17.

A

B

A × B (= C)

Normal to theplane of A and B

θ

Fig. 1.17

Problem 4. Three vectors A, B and C are given as :

A = i + j + 2k, B = 4i – 3j + 2k and C = 2i – j – k.

Find the resultant vector and a unit vector in the direction of the resultant.

Sol. Given :

A = i + j + 2k B = 4i – 3j + 2k C = 2i – j – k

Resultant vector, R = A + B + C

= (i + j + 2k) + (4i – 3j + 2k) + (2i – j – k)

= (1 + 4 + 2) i + (1 – 3 – 1) j + (2 + 2 – 1) k

= 7i – 3j + 3k. Ans.

Unit vector in the direction of the resultant R is

= 7 3 3

7 3 32 2 2

i j k− +

+ − +( ) ( ) = 7 3 3

49 9 9

i j k− ++ +

= 7 3 3

67

i j k− +


NOTES

Engineering Mechanics=

7 3 38 185

i j k− +.

= 7

8 1853

8 1853

8 185. . .i j k− +

= 0.855 i – 0.366 j + 0.366 k. Ans.Problem 5. Two vectors A and B are given as :

A = 4i – 3j – 2k and B = 3i + 2j – k.Determine : (i) their dot product and (ii) angle between them.

Sol. Given :Vector A = 4i – 3j – 2k

Vector B = 3i + 2j – k(i) Dot productThe dot product of two vectors is given by equation (1.12) as

A.B = AxBx + AyBy + AzBz

∴ A.B = 4 × 3 + [(– 3) × (2)] + [(– 2) × (– 1)]= 12 + (– 6) + 2 = 8. Ans.

(ii) Angle between the vectorsLet θ = Angle between the two vectors.Using equation (1.13),

cos θ = A .BA B

A .B| || |

=AB

...(i)

But | A | = A = 4 3 22 2 2+ − + −( ) ( ) = 16 9 4 29 5 385+ + = = .

and | B | = B = 3 2 12 2 2+ + −( ) = 9 4 1 14 3 74+ + = = .

Substituting these values in equation (i),

cos θ = 8

5 385 3 74. .× = 0.3972 (� A.B = 8)

∴ θ = cos–1 0.3972 = 66.59°. Ans.Problem 6. A force vector F is equal to 10i + 5j – 8k. The point of application

of this force moves from the point 2i + k to the point 4i – j – 4k. Find the work done bythe force.

Sol. Given :Force vector, F = 10i+ 5j – 8k

Initial point = 2i + kFinal point = 4i – j – 4k.∴ Distance vector d = Final point – Initial point

= (4i – j – 4k) – (2i + k) = (4 – 2)i + (– 1)j + (– 4 – 1)k= 2i – j – 5k

Work is a product of force and distance. Work is a scalar quantity. As work isscalar, hence dot product of force and distance will give work done.

∴ Work done = Force vector . Distance vector= (10i + 5j – 8k) . (2i – j – 5k) [See equation (1.12)]= 10 × 2 + [(5) × (– 1) + (– 8) (– 5)]= 20 + (– 5) + 40 = 55. Ans.


NOTES


Problem 7. Determine the cross product of vectors A = 3i – 5j – 4kand B = 3i + 4j – 2k and the angle between them.

Sol. Given :Vector A = 3i – 5j – 4kVector B = 3i + 4j – 2k

The cross product of vectors A and B is given by

A × B = i j k3 5 43 4 2

− −−

= [(– 5)(– 2) – (– 4)(4)]i + [(– 4) (3) – (3) (– 2)]j+ [(3)(4) – (– 5)(3)] k

or A × B = [10 – (– 16)] i + [– 12 – (– 6)] j + [12 – (– 15)] k= 26 i – 6 j + 27 k. Ans.

The angle between the vector A and B is given by

sin θ = | A B || A | | B |

×...(i)

Now | A × B | = 26 6 272 2 2+ − +( ) ( ) = 676 36 729+ + = 1441 37 96= .

| A | = 3 5 42 2 2+ − + −( ) ( ) = 9 25 16 7 071+ + = .

| B | = 3 4 22 2 2+ + −( ) = 9 16 4+ + = 5.385

Substituting these values in equation (i),

sin θ = 37 96

7 071 5 385.

. .× = 0.9969

∴ θ = sin–1 0.9969 = 85.5°. Ans.

��

Resolution of a force means “finding the com-ponents of a given force in two given directions.”

Let a given force be R which makes an angleθ with X-axis as shown in Fig. 1.18. It is required tofind the components of the force R along X-axis andY-axis.

Components of R along X-axis = R cos θ.Components of R along Y-axis = R sin θ.Hence, the resolution of forces is the process of

finding components of forces in specified directions.Resolution of a Number of Coplanar Forces. Let a number of coplanar forces

(forces acting in one plane are called co-planar forces) R1, R2, R3, .... are acting at apoint as shown in Fig. 1.19.

Let θ1 = Angle made by R1 with X-axisθ2 = Angle made by R2 with X-axisθ3 = Angle made by R3 with X-axisH = Resultant component of all forces along X-axis

Y

BC

RR sin θ

θO A XR cos θ

Fig. 1.18


NOTES

Engineering Mechanics V = Resultant component of all forcesalong Y-axis

R = Resultant of all forcesθ = Angle made by resultant with X-axis.

Each force can be resolved into twocomponents, one along X-axis and other along Y-axis.

Component of R1 along X-axis = R1 cos θ1

Component of R1 along Y-axis = R1 sin θ1.Similarly, the components of R2 and R3 along

X-axis and Y-axis are (R2 cos θ2, R2 sin θ2) and (R3 cos θ3, R3 sin θ3) respectively.Resultant components along X-axis

= Sum of components of all forces along X-axis.∴ H = R1 cos θ1 + R2 cos θ2 + R3 cos θ3 + ... ...(1.14)Resultant component along Y-axis.

= Sum of components of all forces along Y-axis.∴ V = R1 sin θ1 + R2 sin θ2 + R3 sin θ3 + ... ...(1.15)

Then resultant of all the forces, R = H V2 2+ ...(1.16)

The angle made by R with X-axis is given by, tan θ = VH

...(1.17)

Problem 8. Two forces are acting at a point O asshown in Fig. 1.20. Determine the resultant in magnitudeand direction.

Sol. The above problem has been solved earlier. Henceit will be solved by resolution of forces.

Force P = 50 N and force Q = 100 N.Let us first find the angles made by each force with

X-axis.Angle made by P with x-axis = 15°Angle made by Q with x-axis = 15 + 30 = 45°Let H = Sum of components of all forces along X-axis.

V = Sum of components of all forces along Y-axis.The sum of components of all forces along X-axis is given by,

H = P cos 15° + Q cos 45° = 50 × cos 15° + 100 cos 45° = 119 N

The sum of components of all forces along Y-axis is given by,V = P sin 15° + Q sin 45°

= 50 sin 15° + 100 sin 45° = 83.64 NThe magnitude of the resultant force is given by equation (1.16),

R = H V2 2 2 2119 83 64+ = + . = 145.46 N. Ans.

The direction of the resultant force is given by equation (1.17), tan θ = VH

= 83 64119

.

∴ θ = tan–1 83 64119

. = 35.10°. Ans.

Here θ is the angle made by resultant R with x-axis.

Y

R2

R3

R1

θ2θ3 θ1

XO

Fig. 1.19

15°30°

Q100 N

P50 N

O

α

Fig. 1.20


NOTES

Basics and Staticsof Particles��

Coplanar forces means the forces in a plane. When several forces act on a body,then they are called a force system or a system of forces. In a system in which all theforces lie in the same plane, it is known as coplanar force system. Hence this articledeals with a system of forces which are acting in the same plane and the forces areeither having a common line of action or intersecting at a common point. If the forcesare having common line of action, then they are known as collinear whereas if the forcesintersect at a common point, then they are known as concurrent.

A force system may be coplanar or non-coplanar. If in a system all the forces liein the same plane then the force system is known as coplanar. But if in a system allthe forces lie in different planes, then the force system is known as non-coplanar. Hencea force system is classified as shown in Fig. 1.21.

Force System

1. Coplanar 2. Non-coplanar

Collinear Concurrent Parallel Non-concurrentNon-parallel

Non-concurrentNon-parallel

ParallelConcurrent

Fig. 1.21

Here we shall discuss only coplanar force system, in which the forces may be :1. Collinear2. Concurrent3. Parallel4. Non-concurrent, non-parallel (or General system of forces).

1. Coplanar Collinear. Fig. 1.22 showsthree forces F1, F2 and F3 acting in a plane. Thesethree forces are in the same line i.e., these threeforces are having a common line of action. Thissystem of forces is known as coplanar collinear forcesystem. Hence in coplanar collinear system of forces,all the forces act in the same plane and have acommon line of action.

2. Coplanar Concurrent. Fig. 1.23 showsthree forces F1, F2 and F3 acting in a plane andthese forces intersect or meet at a common point O.This system of forces is known as coplanarconcurrent force system. Hence in coplanar concur-rent system of forces, all the forces act in the sameplane and they intersect at a common point.

A Plane

F1 F2 F3

F2

F1

F3

APlane

O

Fig. 1.22. Coplanar collinear forces

Fig. 1.23. Concurrent coplanar forces


NOTES

Engineering Mechanics 3. Coplanar Parallel. Fig. 1.24 showsthree forces F1, F2 and F3 acting in a plane andthese forces are parallel. This system of forces isknown as coplanar parallel force system. Hence incoplanar parallel system of forces, all the forces actin the same plane and are parallel.

4. Coplanar Non-concurrent Non-par-allel. Fig. 1.25 shows four forces F1, F2, F3 andF4 acting in a plane. The lines of action of theseforces lie in the same plane but they are neitherparallel nor meet or intersect at a common point.This system of forces is known as coplanar non-concurrent non-parallel force system. Hence incoplanar non-concurrent non-parallel system offorces, all the forces act in the same plane butthe forces are neither parallel nor meet at acommon point. This force system is also knownas general system of forces.

��

When a number of coplanar forces are acting on a rigid body, then these forcescan be replaced by a single force which has the same effect on the rigid body as that ofall the forces acting together, then this single force is known as the resultant of severalforces. Hence a single force which can replace a number of forces acting on a rigid body,without causing any change in the external effects on the body, is known as the resultantforce.

The resultant of coplanar forces may be determined by the following two methods :1. Graphical method2. Analytical method.The resultant of the following coplanar forces will be determined by the above

two methods :(i) Resultant of collinear coplanar forces

(ii) Resultant of concurrent coplanar forces.

Resultant of Collinear Coplanar Forces

As defined above, collinear coplanar forces are those forces which act in the sameplane and have a common line of action. The resultant of these forces are obtained byanalytical method or graphical method.

1. Analytical method. The resultant is obtained by adding all the forces if theyare acting in the same direction. If any one of the forces is acting in the opposite direction,then resultant is obtained by subtracting that force.

Fig. 1.26 shows three collinear coplanar forces F1, F2 and F3 acting on a rigidbody in the same direction. Their resultant R will be sum of these forces.

∴ R = F1 + F2 + F3 ...(1.18)

F2

F1

F3

A Plane

Fig. 1.24. Coplanar parallel forces.

F2

F1

F4

F3

A Plane

Fig. 1.25. Non-concurrent non-parallel


NOTES


F3F2F1 F1 F2 F3

Fig. 1.26 Fig. 1.27

If any one of these forces (say force F2) is acting in the opposite direction, as shownin Fig. 1.27, then their resultant will be given by

R = F1 – F2 + F3 ...(1.19)2. Graphical method. Some suitable scale is chosen and vectors are drawn to

the chosen scale. These vectors are added/or subtracted to find the resultant. Theresultant of the three collinear forces F1, F2 and F3acting in the same direction will be obtained by addingall the vectors. In Fig. 1.28, the force F1 = ab to somescale, force F2 = bc and force F3 = cd. Then the lengthad represents the magnitude of the resultant on the scalechosen.

The resultant of the forces F1, F2 and F3 acting on a body shown in Fig. 1.27 willbe obtained by subtracting the vector F2. This resultantis shown in Fig. 1.29, in which the force F1 = ab to somesuitable scale. This force is acting from a to b. The forceF2 is taken equal to bc on the same scale in opposite di-rection. This force is acting from b to c. The force F3 istaken equal to cd. This force is acting from c to d. Theresultant force is represented in magnitude by ad on the chosen scale.

Problem 9. Three collinear horizontal forces of magnitude 200 N, 100 N and300 N are acting on a rigid body. Determine the resultant of the forces analyticallyand graphically when

(i) all the forces are acting in the same direction,(ii) the force 100 N acts in the opposite direction.Sol. Given : F1 = 200 N, F2 = 100 N and F3 = 300 N(a) Analytical method(i) When all the forces are acting in the same direction, then resultant is given by

equation (1.18) asR = F1 + F2 + F3 = 200 + 100 + 300 = 600 N. Ans.

(ii) When the force 100 N acts in the opposite direction, then resultant is given byequation (1.19) as

R = F1 – F2 + F3 = 200 – 100 + 300 = 400 N. Ans.(b) Graphical methodSelect a suitable scale. Suppose 100 N = 1 cm. Then to this scale, we have

F1 = 200100

= 2 cm,

F2 = 100100

= 1 cm,

and F3 = 300100

= 3 cm.

F1 F2 F3

b c da

R = F + F + F1 2 3

Fig. 1.28

Fig. 1.29

F1F2 F3

bc da

R = F – F + F1 2 3

F1 F2 F3

b c da

Fig. 1.30


NOTES

Engineering Mechanics (i) When all the forces act in the same direction.Draw vector ab = 2 cm to represent F1,

vector bc = 1 cm to represent F2 andvector cd = 3 cm to represent F3 as shown in Fig. 1.30.

Measure vector ad which represents the resultant.By measurement length ad = 6 cm∴ Resultant = Length ad × chosen scale

(� Chosen scale is 1 cm = 100 N)= 6 × 100 = 600 N. Ans.

(ii) When force 100 N = F2, acts in the opposite directionDraw length ab = 2 cm to represent force F1.From b, draw bc = 1 cm in the opposite direction

to represent F2. From c draw cd = 3 cm to represent F3as shown in Fig. 1.31.

Measure length ad. This gives the resultant.By measurement, length ad = 4 cm∴ Resultant = Length ad × chosen scale

= 4 × 100 = 400 N. Ans.

Resultant of Concurrent Coplanar Forces

As defined earlier, concurrent coplanar forces are those forces which act in thesame plane and they intersect or meet at a common point. We will consider the followingtwo cases:

(i) When two forces act at a point(ii) When more than two forces act at a point.

(i) When two forces act at a point(a) Analytical methodWe have mentioned earlier that when two forces act at a point, their resultant is

found by the law of parallelogram of forces. Themagnitude of resultant is obtained from equation (1.1)and the direction of resultant with one of the forces isobtained from equation (1.2).

Suppose two forces P and Q act at point O asshown in Fig. 1.32 and α is the angle between them. Letθ is the angle made by the resultant R with the directionof force P.

Forces P and Q form two sides of a parallelogram and according to the law, thediagonal through the point O gives the resultant R as shown.

The magnitude of resultant is given by

R = P Q PQ2 2 2+ + cos αThe above method of determining the resultant is also known as the cosine law

method.The direction of the resultant with the force P is given by

θ = tan–1 Q

P Qsin

cosα

α+��

��

Fig. 1.31

B C

AO Pθ

α

RQ

Fig. 1.32

F1F2 F3

bc da


NOTES


(b) Graphical method(i) Choose a convenient scale to represent the forces

P and Q.(ii) From point O, draw a vector Oa = P.

(iii) Now from point O, draw another vector Ob = Qand at an angle of α as shown in Fig. 1.33.

(iv) Complete the parallelogram by drawing lines ac|| to Ob and bc || to Oa.

(v) Measure the length Oc.Then resultant R will be equal to length Oc × chosen scale.(vi) Also measure the angle θ, which will give the direction of resultant.The resultant can also be determine graphically by drawing a triangle oac as

explained below and shown in Fig. 1.44.(i) Draw a line oa parallel to P and equal to P.

(ii) From a, draw a vector ac at an angle α with thehorizontal and cut ac equal to Q.

(iii) Join oc. Then oc represents the magnitude anddirection of resultant R.

Magnitude of resultant R = Length Oc × chosenscale. The direction of resultant is given by angle θ. Hencemeasure the angle θ.

(ii) When more than two forces act at a point(a) Analytical method. The resultant of three or more forces acting at a point

is found analytically by a method which is known as rectangular components methods.According to this method all the forces acting at a point are resolved into horizontaland vertical components and then algebraic summation of horizontal and verticalcomponents is done separately. The summation of horizontal component is written asΣH and that of vertical as ΣV. Then resultant R is given by

R = ( ) ( )Σ ΣH V2 2+ .

The angle made by the resultant with horizontal is given by

tan θ = ( )( )ΣΣ

VH

∴ Let four forces F1, F2, F3 and F4 act at a point O as shown in Fig. 1.35.

Y

F1

F3

F2

F4

θ1

θ4θ3

θ2

X′

Y′

XO

Y

F1

XF cos1 1θ

θ1

Fsi

n1

1θ

OX′

Y′

Fig. 1.35 Fig. 1.35(a)

b c

aO

Q

αθ

R

P

Fig. 1.33

Fig. 1.34

O

c

R

θP a

α

Q


NOTES


XOX′ F cos2 2θ

θ2 Fsi

n2

2θ

F2

Y

X

OX′

F cos3 3θ

θ3

Fsi

n3

3θ

F3

Y

X

OX′

F cos4 4θ

θ4

Fsi

n4

4θ

F4

Y

Fig. 1.35(b) Fig. 1.35(c) Fig. 1.35(d)

The inclination of the forces is indicated with respect to horizontal direction. Letθ1 = Inclination of force F1 with OXθ2 = Inclination of force F2 with OX′θ3 = Inclination of force F3 with OX′θ4 = Inclination of force F4 with OX.

The force F1 is resolved into horizontal and vertical components and thesecomponents are shown in Fig. 1.35(a). Similarly, Figs. 1.35b), (c) and (d) show thehorizontal and vertical components of forces F2, F3 and F4 respectively. The varioushorizontal components are :

F1 cos θ1 → (+)F2 cos θ2 ← (–)F3 cos θ3 ← (–)F4 cos θ4 → (+)

∴ Summation or algebraic sum of horizontal components :ΣH = F1 cos θ1 – F2 cos θ2 – F3 cos θ3 + F4 cos θ4

Similarly, various vertical components of all forces are :F1 sin θ1 ↑ (+)F2 sin θ2 ↑ (+)F3 sin θ3 ↓ (–)F4 sin θ4 ↓ (–)

∴ Summation or algebraic sum of vertical components :ΣV = F1 sin θ1 + F2 sin θ2 – F3 sin θ3 – F4 sin θ4

Then the resultant will be given by R = ( ) ( )Σ ΣH V2 2+ ...(1.20)And the angle (θ) made by resultant with x-axis is given by

tan θ = ( )( )ΣΣ

VH

...(1.21)

(b) Graphical method. The resultant of several forces acting at a point is foundgraphically with the help of the polygon law of forces, which may be stated as

“If a number of coplanar forces are acting at a point such that they can berepresented in magnitude and direction by the sides of a polygon taken in the sameorder, then their resultant is represented in magnitude and direction by the closing sideof the polygon taken in the opposite order.

Let the four forces F1, F2, F3 and F4 act at a point O as shown in Fig. 1.36. Theresultant is obtained graphically by drawing polygon of forces as explained below andshown in Fig. 1.37.


NOTES

Basics and Staticsof ParticlesF1

F2

O

F3F4

F2

F3

F4

F1

c

b

d

e aR

Resultant

Fig. 1.36 Fig. 1.37

(i) Choose a suitable scale to represent the given forces.(ii) Take any point a. From a, draw vector ab parallel to force OF1. Cut ab = force

F1 to the scale.(iii) From point b, draw bc parallel to OF2. Cut bc = force F2.(iv) From point C, draw cd parallel to OF3. Cut cd = force F3.(v) From point d, draw de parallel to OF4. Cut de = force F4.

(vi) Join point a to e. This is the closing side of the polygon. Hence ae represents theresultant in magnitude and direction.

Magnitude of resultant R = Length ae × scale.The resultant is acting from a to e.Problem 10. Two forces of magnitude 240 N and 200 N are acting at a point O

as shown in Fig. 1.38. If the angle between the forces is 60°, determine the magnitudeof the resultant force. Also determine the angle β and γ as shown in the figure.

ααγ

ββ

Q =

200

N

O P = 240 N

R β

αγ (180°–)α

O P

QR

Fig. 1.38 Fig. 1.39

Sol. Given :Force P = 240 N, Q = 200 NAngle between the forces, α = 60°The magnitude of resultant R is given by,

R = P Q PQ2 2 2+ + cos α

= 240 200 2 240 200 602 2+ + × × × °cos

= 57600 40000 48000+ + = 381.57 N. Ans.

Now refer to Fig. 1.39. Using sine formula, we getP Q R

sin sin sin ( )β γ α= =

° −180 ...(i)

orP R

sin sin ( )β α=

° −180

∴ sin β = P

Rsin ( ) sin ( )180 240 180 60

381.57° − = −α

(� P = 240 N, α = 60°, R = 381.57 N)


NOTES


240 120381.57× °sin

= 0.5447

∴ β = sin–1 0.5447 = 33°. Ans.

From equation (i), also we have Q R

sin sin ( )γ α=

−180

∴ sin γ = Q

Rsin ( )180 − α

= 200 180 60

381.57200 120

381.57sin ( ) sin− = °

= 0.4539

γ = sin–1 0.4539 = 26.966°. Ans.

��

If a single force is acting on a particle, the particle cannot be in equilibrium. Butif two forces or three forces or more than three forces are acting on the particle, theparticle can be in equilibrium under certain conditions.

When two forces are acting. If two forces are acting on a particle, the particlewill be in equilibrium when the two forces are equal, opposite and collinear.

When three forces are acting. If three forces are acting on a particle, theparticle will be in equilibrium when the three forces are concurrent and

(i) Resultant force in x and y direction is zero i.e.,ΣFx = 0 and ΣFy = 0

(ii) The three forces obey Lame’s Theorem which states that “If a particle (or abody) is in equilibrium under the action of three forces, each force is proportional to thesine of the angle between other two.”

Or(iii) The three forces form a closed triangle.When more than three forces are acting. If more than three forces are acting

on a particle, the particle will be in equilibrium when the resultant force in x and ydirections is zero i.e.,

ΣFx = 0 and ΣFy = 0.

��

When some external forces (which may be concurrent or parallel) are acting on astationary body, the body may start moving or may start rotating about any point. Butif the body does not start moving and also does not start rotating about any point, thenthe body is said to be in equilibrium.

Principle of Equilibrium. The principle of equilibrium states that, a station-ary body which is subjected to coplanar forces (concurrent or parallel) will be inequilibrium if the algebraic sum of all the external forces is zero and also the algebraicsum of moments of all the external forces about any point in their plane is zero.Mathematically, it is expressed by the equations :

ΣF = 0 ...(1.22)ΣM = 0 ...(1.23)

The sign Σ is known as sigma which is a Greek letter. This sign represents thealgebraic sum of forces or moments.


NOTES


The equation (1.22) is also known as force law of equilibrium whereas the equation(1.23) is known as moment law of equilibrium.

The forces are generally resolved into horizontal and vertical components. Henceequation (1.22) is written as

ΣFx = 0 ...(1.24)and ΣFy = 0 ...(1.25)where ΣFx = Algebraic sum of all horizontal componentsand ΣFy = Algebraic sum of all vertical components.

Equations of Equilibrium for Non-concurrent Force Systems. A non-con-current force systems will be in equilibrium if the resultant of all forces and moment iszero.

Hence the equations of equilibrium areΣFx = 0, ΣFy = 0 and ΣM = 0.

Equations of Equilibrium for Concurrent Force System. For the concur-rent forces, the lines of action of all forces meet at a point, and hence the moment ofthose force about that very point will be zero or ΣM = 0 automatically.

Thus for concurrent force system, the condition ΣM = 0 becomes redundant andonly two conditions, i.e., ΣFx = 0 and ΣFy = 0 are required.

Force Law of Equilibrium. Force law ofequilibrium is given by equation (1.22) or by equations(1.24) and (1.25). Let us apply this law to the followingimportant force system :

(i) Two force system (ii) Three force system(iii) Four force system.Two Force System. When a body is subjected to two forces, then the body will be

in equilibrium if the two forces are collinear, equal and opposite as shown in Fig. 1.40.If the two forces acting on a body are equal and

opposite but are parallel, as shown in Fig. 1.41, then thebody will not be in equilibrium. This is due to the factthat the three conditions of equilibrium will not besatisfied. This is proved as given below :

(i) Here ΣFx = 0 as there is no horizontal forceacting on the body. Hence first condition of equilibriumis satisfied.

(ii) Also here ΣFy = 0 as F1 = F2.Hence second condition of equilibrium is also

satisfied.(iii) ΣM about any point should be zero. The resultant moment about point A is

given byMA = – F2 × AB (– ve sign is due to clockwise moment)

But MA is not equal to zero. Hence the third condition is not satisfied.Hence a body will not be in equilibrium under the action of two equal and opposite

parallel forces.Two equal and opposite parallel forces produce a couple and moment of the

couple is – F1 × AB (Fig. 1.41).

F1 F2F = F1 2

Fig. 1.40

F1

AB

x

F = F2 1

Fig. 1.41


NOTES

Engineering Mechanics F2

R

O

F3

F1

Fig. 1.42

F1F3

B

A C

F2

Fig. 1.43

Three Force System. The three forces actingon a body which is in equilibrium may be eitherconcurrent or parallel. Let us first consider that thebody is in equilibrium when three forces, acting on thebody, are concurrent. This is shown in Fig. 1.42.

(a) When three forces are concurrent. The threeconcurrent forces F1, F2 and F3 are acting on a bodyat point O and the body is in equilibrium. The resultantof F1 and F2 is given by R. If the force F3 is collinear,equal and opposite to the resultant R, then the body willbe in equilibrium. The force F3 which is equal andopposite to the resultant R is known as equilibrant.Hence for three concurrent forces acting on a body whenthe body is in equilibrium, the resultant of the two forcesshould be equal and opposite to the third force.

(b) When three forces are parallel. Fig. 1.43shows a body on which three parallel forces F1, F2 andF3 are acting and the body is in equilibrium. If threeforces F1, F2 and F3 are acting in the same direction,then there will be a resultant R = F1 + F2 + F3 and body will not be in equilibrium. Thethree forces are acting in opposite direction and their magnitude is so adjusted that thereis no resultant force and body is in equilibrium. Let us suppose that F2 is acting inopposite direction as shown in Fig. 1.43.

Now let us apply the three conditions of equilibrium :(i) ΣFx = 0 as there is no horizontal force acting on the body

(ii) ΣFy = 0 i.e., F1 + F3 = F2

(iii) ΣM = 0 about any point.Taking the moments of F1, F2 and F3 about point A,

ΣMA = – F2 × AB + F3 × AC

(Moment of F3 is anti-clockwise whereas moment of F2 is clockwise)For equilibrium, ΣMA should be zero

i.e., – F2 × AB + F3 × AC = 0If the distances AB and AC are such that the above equation is satisfied, then

the body will be in equilibrium under the action of three parallel forces.Four Force System. The body will be in equilibrium if the resultant force in

horizontal direction is zero (i.e., ΣFx = 0), resultant force in vertical direction is zero(i.e., ΣFy = 0) and moment of all forces about any point in the plane of forces is zero(i.e., ΣM = 0).

Problem 11. Two forces F1 and F2 are acting on abody and the body is in equilibrium. If the magnitude ofthe force F1 is 100 N and its acting at O long x-axis asshown in Fig. 1.44, then determine the magnitude anddirection of force F2.

Sol. Given :Force, F1 = 100 NThe body is in equilibrium under the action of two

forces F1 and F2.

F = 100 N1O

Fig. 1.44


NOTES


When two forces are acting on a body and the body is in equilibrium, then thetwo forces should be collinear, equal and opposite.

∴ F2 = F1 = 100 NThe force F2 should pass through O, and would be acting in the opposite direction

of F1.

��

When a force is represented in three dimensions,then the force is known as ‘Force in space’. Forces inspace are generally represented in vector form. InFig. 1.45, the force F is represented by vector OA.Through point A, draw planes parallel to co-ordinateplanes (i.e., Co-ordinate planes are xy, yz and zx asshown in Fig. 1.45).

These planes along with co-ordinate planes makea rectangular box. The force F is then represented bythe diagonal of the box and its three components Fx ,Fy and Fz by its edges.

Let θx = Angle made by force F with x-axisθy = Angle made by force F with y-axisθz = Angle made by force F with z-axis

Now Fx = F cos θx

Fy = F cos θy ...(1.26)Fz = F cos θz

Also F = F F Fx y z2 2 2+ + ...(1.27)

The cosines of θx, θy and θz are known as the direction cosines of the force F anddenoted by

l = cos θx, m = cos θy and n = cos θz ...(1.27A)The three angles are related by

cos2 θx + cos2 θy + cos2 θz = 1 ...(1.28)or l2 + m2 + n2 = 1

Let i = vector of unit length in the positive x-direction,j = vector of unit length in the positive y-direction,

and k = vector of unit length in the positive z-direction then force vector F isrepresented by

F = Fx i + Fy j + Fz k ...(1.29)Magnitude of the vector F is given by

F = | F | = F F Fx y z2 2 2+ +

But Fx = F cos θx, Fy = F cos θy and Fz = F cos θz. Substituting these values inequation (1.29), we get

F = (F cos θx) i + (F cos θy) j + (F cos θz) k ...(1.30)

θyθx

θz

Y

Z

OX

FA

Fig. 1.45

��


NOTES

Engineering Mechanics Unit Vector. The vector having a unit length (or unit magnitude) is known asunit vector. Hence unit vector corresponding to force vector F is equal to F ÷ Magnitudeof vector F.

∴ Unit vector = Force vector

Magnitude of vectorF F

F F=

= F

F F Fx y z2 2 2+ +

...(1.31)

=+ +

+ +

F F F

F F F

x y z

x y z

i j k2 2 2 (� F = Fxi + Fy j + Fzk)

Note. Force vector is represented by THICK, BOLD FACE and magnitude is repre-sented by capital letters (without thickness) only.

Components of a Force when Two Points on its Line of Action are Given.Fig. 1.46 shows a force F, which is defined by two points A(x1, y1, z1) and B(x2, y2, z2).

O

Z

X

Y

d

dx

dy

dz

F

A (x , y , z )1 1 1

A (x , y , z )2 2 2

dx = (x – x )2 1

dz = (z – z )2 1

dy = (y – y )2 1

Fig. 1.46

Let d = Distance between A and Bθx = Angle made by force F with x-axisθy = Angle made by force F with y-axisθz = Angle made by force F with z-axis

dx = Distance along x-axis between A and Bdy = Distance along y-axis between A and Bdz = Distance along z-axis between A and B.

Then dx = (x2 – x1), dy = (y2 – y1)and dz = (z2 – z1) ...(1.32)

Also dx = d cos θx, dy = d cos θy

and dz = d cos θz

and d = dx dy dz2 2 2+ + ...(1.33)

Also we know thatFx = F cos θx, Fy = F cos θy

and Fz = F cos θz


NOTES

Basics and Staticsof ParticlesNow

Fdx

Fd

Fd

x x

x= =

coscos

θθ

Similarly F

dyy

= Fd

andFdz

z = Fd

∴ Fdx

F

dyFdz

Fd

x y z= = = ...(1.34)

or Fx x

F

y yF

z zFd

x y z

( ) ( ) ( )2 1 2 1 2 1−=

−=

−= .

∴ Fx = (x2 – x1) Fd

, Fy = (y2 – y1) Fd

and F2 = (z2 – z1) Fd

...(1.34A)

From equation (1.34A) the components of the force can be obtained, provided weknow ‘F ’ and ‘d’.

Position Vector of a Given Point. The position vector of a point can be givenwith respect to origin or with respect to another point.

(i) The position vector of a point A with respect to origin O is the vector OA. It isrepresented by r. It is given by (see Fig. 1.47)

r = xi + yj + zk

r

x

zX

Z

Y

y

z

O

A(x, y, z)

B(x , y , z )1 1 1

A(x , y , z )2 2 2

X

Z

Y

Fig. 1.47 Fig. 1.48

(ii) The position vector of a point A with co-ordinates (x2, y2, z2) with respect topoint B(x1, y1, z1) is given by (see Fig. 1.48)

r = (x2 – x1) i + (y2 – y1) j + (z2 – z1) k.Problem 12. A force vector is represented by a line AB. The co-ordinates of

point A are (2, 4, 3) and of point B(1, – 5, 2) respectively. If the magnitude of force =10 N, then determine :

(i) the components of the force along x, y and z-axis,(ii) angles with the x, y and z-axis, and

(iii) specify the force vector.Sol. Given :Co-ordinates of point A are (2, 4, 3)∴ x1 = 2, y1 = 4 and z1 = 3Co-ordinates of point B are (1, – 5, 2)∴ x2 = 1, y2 = – 5 and z2 = 2.Magnitude of force, F = 10 N


NOTES

Engineering Mechanics Here the magnitude of the force and co-ordinates of two points through whichforce passes, are given. The force components can be obtained if we know ‘d’ (i.e., distance

between two points). But d = dx dy dz2 2 2+ + .

From equation (1.32), dx = x2 – x1 = 1 – 2 = – 1 dy = y2 – y1 = – 5 – 4 = – 9 dz = z2 – z1 = 2 – 3 = – 1

and from equation (1.33),

d = dx dy dz2 2 2+ + = − + − + −( ) ( ) ( )1 9 12 2 2

= 1 81 1 83 9 11+ + = = . .

(i) Components of the force along the axesLet Fx = Component of the force F along x-axis,

Fy = Component of the force F along y-axis, and Fz = Component of the force F along z-axis.

Now using equation (1.34), we get

Fdx

F

dyFdz

Fd

x y z= = =

Substituting the values of dx, dy, dz, F and d, we get

F F Fx y z

−=

−=

−=

1 9 110

9 11.

∴ Fx = (– 1) × − = − = −109 11

109 11. .

1.1 N. Ans.

Fy = (– 9) × 10

9 1190

9 11. .–= − = 9.88 N. Ans.

Fz = (– 1) × 10

9 1110

9 11. .= − = − 1.1 N. Ans.

(ii) Angles with x, y and z-axesLet θx = The angle that the force F makes with x-axis,

θy = The angle made by F with y-axis, andθz = The angle made by F with z-axis.

Using equation (1.26),Fx = F cos θx

∴ cos θx = FF

x =−

= −1.1

100 11.

∴ θx = cos–1 (– 0.11) = 96.3°. Ans.Similarly, Fy = F cos θy

∴ cos θy = F

Fy = − =9 88

100 988

.– .

∴ θy = cos–1 (– 0.988) = 171.09°. Ans.Also Fz = F cos θz

∴ cos θz = FF

z =−

= −1.09710

0 11. or θz = cos–1 (– 0.11) = 96.3°. Ans.


NOTES


(iii) Specify the force vectorThe force vector F is represented by equation (1.29) as

F = Fxi + Fy j + Fzk = – 1.1 i + (– 9.88) j + (– 1.1) k(� Fx = – 1.1, Fy = – 9.88 and Fz = – 1.1)

= – 1.1 i – 9.88 j – 1.1 k. Ans.IInd MethodThe force vector F can also be obtained as given below :The vector joining the points (2, 4, 3) and (1, – 5, 2) is given by

= dx i + dx j + dz k = (x2 – x1) i + (y2 – y1) j + (z2 – z1) k= (1 – 2) i + (– 5 – 4) j + (2 – 3) k = – 1 i – 9 j – 1 k.

Now find the unit vector in the direction of above vector i.e., in the direction ofvector (– 1 i – 9 j – 1 k).

This unit vector is obtained from equation (1.31) and is equal to the given vectordivided by the square root of the sum of squares of its components.

∴ Unit vector in the direction of the vector (– 1 i – 9 j – 1 k) is( )

( ) ( ) ( )

( )− − −

− + − + −= − − −

+ +1 9

1 9 1

1 9 1

1 81 12 2 2

i j k i j k1 = − − −( )

.1 9 1

9 11i j k

The force vector F = Magnitude of force times unit vector in the direction of force

= 10 1 9 1

9 11( )

.− − −i j k = − × − × − ×10 1

9 1110 99 11

10 19 11. . .

i j k

= – 1.1 i – 9.88 j – 1.1 k. Ans.Problem 13. A bar AB is acted upon by six forces as shown in Fig. 1.49. The

forces are :

F1 = 80i, F2 = 130k,F3 = – 50j, F4 = 210k,

F5 = – 100i + 190j – 50k. F6 = 315j.

A(0, 16, 0)

16 cm

F1

Y

F2

8 cm

2cm

F3

F4

F6

B(2, 0, 8)

(1, 8, 4)C

Z

F5

OX

Fig. 1.49


NOTES

Engineering Mechanics All forces are in Newtons. Determine the equivalent force and moment actingat A.

Sol. Given :F1 = 80i, F2 = 130k F3 = – 50j,

F4 = – 210k F5 = – 100i + 190j – 50k, F6 = 315j.Position vector of point A = 16jPosition vector of point B = (2i + 8k)Position vector of point C = (i + 8j + 4k).(i) To find the equivalent force

Equivalent force is the resultant force.The resultant force R is given by

R = F1 + F2 + F3 + F4 + F5 + F6

= 80i + 130k + ( – 50j) + (– 210k) + (– 100i + 190j – 50k) + 315j

= (80 – 100)i + (– 50 + 190 + 315)j + (130 – 210 – 50)k= – 20i + 455j – 130k. Ans.

(ii) To find the moment of all forces about point ALet r1 = Position vector of force F1 from point A

= 0 (as force F1 is passing through A)r2 = Position vector of force F2 from point A = 0r3 = Position vector of force F3 from point A

= Position vector of point C – position vector of point A= (i + 8j + 4k) – (16j) = (i – 8j + 4k)

r4 = Position vector of force F4 with respect to point A= Position vector of point B – position vector of point A= (2i + 8k) – (16j) = (2i – 16j + 8k)

r5 = Position vector of force F5 from point A= 2i – 16j + 8k

r6 = Position vector of force F6 from point A= 2i – 16j + 8k.

(Force F5 and force F6 are acting at point B, hence their position vector will besame as of force F4.)

∴ Moment of all forces about point A is given by,MA = r1 × F1 + r2 × F2 + r3 × F3 + r4 × F4 + r5 × F5 + r6 × F6

= 0 + 0 + (i – 8j + 4k) × (– 50j) + (2i – 16j + 8k) × (– 210k)+ (2i – 16j + 8k) × (– 100i + 190j – 50k) + (2i – 16j + 8k) × (315j)

= (– 50i × j + 400j × j – 200k × j) + (– 420i × k + 3360j × k– 1680k × k) + (– 200i × i + 380i × j – 100i × k + 1600j × i

– 16 × 190 × j × j + 800j × k – 800k × i + 1520k × j– 400k × k) + (630i × j – 16 × 315 × j × j + 2520 k × j)

But i × j = k, j × j = 0, i × i = 0, k × k = 0, k × j = – i, i × k = – j,j × k = i, j × i = – k, k × i = j, k × j = – i.

∴ MA = [– 50k + 0 – 200 (– i)] + [– 420 (– j) + 3360i – 0]+ [0 + 380k – 100(– j) + 1600 (– k) – 0 + 800i – 800j

+ 1520 (– i) – 0] + [630k – 0 + 2520 (– i)]


NOTES


= – 50k + 200i + 420j + 3360i + 380k + 100j – 1600k+ 800i – 800j – 1520i + 630k – 2520i

= (200 + 3360 + 800 – 1520 – 2520)i + (420 + 100 – 800)j+ (– 50 + 380 – 1600 + 630)k

= 320i + 290j – 640k. Ans.

��

When a particle (or a rigid body) is in space, the forces acting on the rigid body(or on the particle) may be concurrent or non-concurrent. If the forces acting areconcurrent, then the equations of equilibrium are

ΣFx = 0, ΣFy = 0 and ΣFz = 0i.e., the resultant force in x, y and z directions are zero.

But if the forces acting are non-concurrent then the resultant force in x, y and zdirections should be zero also the resultant moment about x, y and z axis should bezero. i.e.,

ΣFx = 0, ΣFy = 0 and ΣFz = 0Also ΣMxy = 0, ΣMxz = 0 and ΣMyz = 0Thus there will be six equations of equilibrium, three equations of equilibrium

will be for force and three equations of equilibrium will be for moment. In vector form,these equations can be written as

ΣF = 0 and ΣM = 0.

��

We have learnt that a force can be replaced by a force and a couple. When anumber of forces are acting on a body in space then each force is transferred to anarbitrary reference point along with a couple for each force transferred. Thus, we get

(i) a system of concurrent forces and(ii) a system of couples.The system of concurrent forces can be added to get the resultant single force

(R). Also the system of couples can be added to get the resultant moment (M).Hence the equations become as

R = F1 + F2 + F3 + ......M = M1 + M2 + M3 + ......

An equivalent system for a given system of coplanar forces, is a combination of aforce passing through a given point and a moment about that point. The force is theresultant of all forces acting on the body. And the moment is the sum of all the momentsabout that point.

Hence equivalent system consists of :(i) a single force R passing through the given point P and

(ii) a single moment MR

where R = the resultant of all force acting on the body.MR = sum of all moments of all the forces about point P.


NOTES

Engineering Mechanics Problem 14. Three external forces are acting on a L-shaped body as shown inFig. 1.50. Determine the equivalent system through point O.

200 mm

30°

2000 N

X

A

C

1000 N

200 mm

100 mm1500 N

B90°

O

Y

Fig. 1.50

Sol. Given :Force at A = 2000 N, Angle = 30°Force at B = 1500 NForce at C = 1000 NDistance OA = 200 mm, OB = 100 mm

and BC = 200 mmAngle COA = 90°Determine the equivalent system through O. This means find(i) single resultant force, R

(ii) single moment through O.Taking x-axis along OA and y-axis along OC.The force at A is resolved into two components.Component along x-axis = 2000 × cos 30° = 1732 NComponent along y-axis = 2000 × sin 30° = 1000 NResolving all forces along X-axis i.e.,

ΣFx = 2000 cos 30° – 1500 – 1000 = – 768 NSimilarly ΣFy = – 2000 × sin 30° = – 1000 (– ve sign is due to downward)

∴ Resultant, R = Σ ΣF Fx y2 2 2 2768 1000+ = − + −( ) ( )

= 589824 1000000+ = 1260.88 NTaking moments of all forces about point O,

M0 = (– 2000 sin 30) × 200 + 1500 × 100 + 1000 × 300= – 200000 + 150000 + 300000= 250000 Nmm = 250 Nm

∴ Equivalent system through point O isR = 1260.88 NM = 250 Nm.


NOTES

Basics and Staticsof Particles��

The principle of transmissibility of forces has been already defined in Art. 1.3.5(see page 16). It states that “If a force, acting at a point on a rigid body, is shifted toany other point which is on the line of action of the force, the external effect of the forceon the body remains unchanged”. For more details and explanation, please refer to Art.1.3.5.

��

When a number of forces are acting on a rigid body, then these forces can bereplaced by a single force which has the same effect on the rigid body as that of all theforces acting together, then this single, force is known as ‘Single Equivalent Force’.This single equivalent force is also known as resultant of several forces. Hence a singleforce which can replace a number of forces acting on a rigid body, without causing anychange in the external effects on the body, is known as single equivalent force (or resultantforce).


NOTES

Engineering Mechanics��

1. Engineering mechanics is divided into statics and dynamics. The study of a bodyat rest is known as statics whereas the study of a body in motion is known asdynamics.

2. A quantity which is completely specified by magnitude and direction is known asvector quantity.

3. A particle is a body of infinitely small volume and is considered to be concentratedat a point.

4. Law of parallelogram of forces states that “If two forces, acting at a point be repre-sented in magnitude and direction by the two adjacent sides of a parallelogram,then their resultant is represented in magnitude and direction by the diagonal ofthe parallelogram passing through that point.

5. According to Lame’s theorem, “If three forces acting at a point are equilibrium,each force will be proportional to the sine of the angle between the other two forces.”

6. The relation between newton and dyne is given by One newton = 105 dyne.7. Moment of a force about a point = Force × perpendicular distance of the line of

action of the force from that point.8. Coplanar forces means the forces are acting in one plane.9. Polygon law of forces states that if a number of coplanar forces are acting at a point

such that they can be represented in magnitude and direction by the sides of apolygon taken in the same order, then their resultant is represented in magnitudeand direction by the closing side of the polygon taken in the opposite order.

10. The moment of a force about any point is the product of force and perpendiculardistance between the point and line of action of force.

11. The resultant of two like parallel forces is the sum of the two forces and acts at apoint between the line in such a way that the resultant divides the distance in theratio inversely proportional to the magnitudes of the forces.

12. If the resultant of a number of parallel forces is not zero, the system can be reducedto a single force, whose magnitude is equal to the algebraic sum of all forces. Thepoint of application of this single force is obtained by equating the moment of thissingle force about any point to the algebraic sum of moments of all forces acting onthe system about the same point.

13. If the resultant of a number of parallel forces is zero, then the system may have aresultant couple or may be in equilibrium. If the algebraic sum of moments of allforces about any point is not zero, then system will have a resultant couple. But ifthe algebraic sum of moments of all forces about any point is zero, the system willbe in equilibrium.

��

• Vector quantity. Quantity specified by both magnitude and direction.• Scalar quantity. Quantity specified only by magnitude.• Particle. It is a body of infinitely small volume.


NOTES


• Dot product. It is also known as scalar product. It is the product of two vector, Aand B and represented as A . B.

• Cross product. It is a product of two vectors A and B and denoted by A × B.• Single equivalent force. A single force which can replace a number of forces

acting on a rigid body, without causing any change in external effects on the body.

��

1. What do you mean by scalar and vector quantities ?2. Define the law of parallelogram of forces. What is the use of this law ?3. Write the S.I. units of : Force, moment and velocity.4. A number of coplanar forces are acting at a point making different angles with x-

axis. Find an expression for the resultant force. Find also the angle made by theresultant force with x-axis.

5. Define and explain the following terms :(i) Coplanar and non-coplanar forces (ii) Collinear and concurrent forces

(iii) Parallel and non-parallel forces.6. Explain the procedure of resolving a given force into two components at right angles

to each other.7. Three collinear forces F1, F2 and F3 are acting on a body. What will be the resultant

of these forces, if(a) all are acting in the same direction(b) force F3 is acting in opposite direction.

8. State the law of parallelogram of forces and show that the resultant R = P Q2 2+

when the two forces P and Q are acting at right angles to each other. Find thevalue of R if the angle between the forces is zero.

9. Define the terms : Coplanar parallel forces, like parallel forces and unlike parallelforces.

10. Define and explain the moment of a force. Differentiate between clockwise momentand anti-clockwise moment.

11. Indicate whether the following statements are True or False :(i) Force is an agency which tends to cause motion.

(ii) The value of g reduces slightly as we move from poles towards the equator.(iii) Coplanar forces are those which have the same magnitude and direction.(iv) A couple consists of two unequal and parallel forces acting on a body, having

the same line of action.(v) A vector diagram of a force represents its magnitude, direction, sense and point

of application.(vi) The force of gravitation on a body is called its weight.

(vii) The centre of gravity of a body is the point, through which the resultant ofparallel forces passes in whatever position may the body be placed.

[Ans. (i) True (ii) True (iii) False (iv) False (v) False (vi) True (vii) True]


NOTES

Engineering Mechanics 12. Determine the magnitude of the resultant of the two forces of magnitude 12 N and9 N acting at a point when the angle between the two forces is 30°. [Ans. 20.3 N]

13. Find the magnitude of two equal forces acting at a point with an angle of 60°

between them, if the resultant is equal to 30 × 3 N. [Ans. 30 N]

14. A beam AB of span 6 m carries a point load of 100 N at a distance 2 m from A.Determine the beam reaction. [Ans. RA = 66.67 N ; RB = 33.33 N]

15. Four forces of magnitudes 20 N, 30 N, 40 N and 50 N are acting respectively alongthe four sides of a square taken in order. Determine the magnitude, direction and

position of the resultant force. Ans. 20 2 2257

2 2× °

×

�

�

�

��N, ,

a

16. The four coplanar forces are acting at a point asshown in Fig. 1.51. One of the forces is unknownand its magnitude is shown by P. The resultant ishaving a magnitude 500 N and is acting along x-axis. Determine the unknown force P and its incli-nation with x-axis.

[Ans. P = 286.5 N and θ = 53° 15′]

17. Four forces of magnitudes 20 N, 40 N, 60 Nand 80 N are acting respectively along the foursides of a square ABCD as shown in Fig. 1.52.Determine the resultant moment about pointA.

Each side of square is 2 m.

[Ans. 200 Nm anti-clockwise]

18. Four parallel forces of magnitudes 100 N, 200N, 50 N and 400 N are shown in Fig. 1.53.Determine the magnitude of the resultant andalso the distance of the resultant from pointA. [Ans. R = 350 N, 3.07 m]

19. A system of parallel forces are acting on a rigid bar as shown in Fig. 1.54. Reducethis system to :(i) a single force, [Ans. (i) R = 120 N at 2.83 m from A

(ii) a single force and a couple at A (ii) R = 120 N and MA = – 340 Nm

(iii) a single force and a couple at B. (iii) R = 120 N and MB = 120 Nm]

2 m

2 m20 N

40 N

C60 N D

BA

80 N

Fig. 1.52

45°θ20° Resultant

= 500 N

500 N

P

200 N

200 N

Fig. 1.51

A B D

100 N 200 N 50 N 400 N

C

1 m 1 m1.5 m

Fig. 1.53


NOTES


20 N 40 N 80 N100 N

1 m 1 m 2 m

A C D B

Fig. 1.54

��

• Rajasekaran, S, Sankarasubramanian, G., ‘‘Fundamentals of EngineeringMechanics’’, Vikas Publishing House Pvt. Ltd., (2000).

• Hibbeller, R.C., ‘‘Engineering Mechanics’’, Vol. 1 Statics, Vol. 2 Dynamics, PearsonEducation Asia Pvt. Ltd., (2000).

• Ashok Gupta, ‘‘Interactive Engineering Mechanics – Statics – A Virtual Tutor(CDROM)’’, Pearson Education Asia Pvt., Ltd., (2002).


NOTES

Engineering Mechanics U N I T

2EQUILIBRIUM OF

RIGID BODIES

STRUCTURE

2.1 Introduction

2.2 Free Body Diagram

2.3 Types of Supports and their Reactions

2.4 Requirements of Stable Equilibrium

2.5 Moments and Couples

2.6 Moment of a Force about a Point and about an Axis

2.7 Vectorial Representation of Moments and Couples

2.8 Varignon’s Theorem (Or Principle of Moments)

2.9 Equilibrium of Rigid Bodies in Two Dimensions and in Three Dimensions

2.10 Methods for Finding Out the Reactions of a Beam

2.11 Problems for Equilibrium of Rigid Bodies in Two-Dimension andThree Dimensions

• Summary

• Glossary

• Review Questions

• Further Readings

��

After going through this unit, you should be able to :• illustrate the concepts of free body diagram, different types of support reactions

and their determinations.• elaborate the conditions of equilibrium for concurrent forces, parallel forces,

moment of force about a point and about an axis.• define vertical representation of moments and couples, scalar components of

a moment and varignon’s theorems.


NOTES

Equilibrium ofRigid Bodies��

When some external forces (which may be concurrent or parallel) are actingon a stationary body, the body may start moving or may start rotating about anypoint. But if the body does not start moving and also does not start rotating aboutany point, then the body is said to be in equilibrium. In this chapter, the conditionsof equilibrium for concurrent forces (i.e., forces meeting at a point), for parallel forces,moment of a force about a point and about an axis, vectorial representation ofmoments and couples and Varignon’s theorem will be described. Also the concept offree body diagram, different types of support reactions and determination of reactionswill be explained.

Equations of Equilibrium. Equilibrium of rigid bodies has been describedearlier. A stationary body which is subjected to coplanar forces (concurrent or parallel)will be in equilibrium if the algebraic sum of all the external forces is zero and alsothe algebraic sum of moments of all the external forces about any point in their planeis zero. Mathematically, it is expressed by the equations :

ΣF = 0 ...(2.1)ΣM = 0 ...(2.2)

The sign Σ is known as sigma which is a Greek letter. This sign representsthe algebraic sum of forces or moments.

The equation (2.1) is also known as force law of equilibrium whereas theequation (2.2) is known as moment law of equilibrium.

The forces are generally resolved into horizontal and vertical components.Hence equation (2.1) is written as

ΣFx = 0and ΣFy = 0where ΣFx = Algebraic sum of all horizontal componentsand ΣFy = Algebraic sum of all vertical components.

Equations of Equilibrium for Non-concurrent Force Systems. A non-con-current force systems will be in equilibrium if the resultant of all forces and momentis zero.

Hence the equations of equilibrium areΣFx = 0, ΣFy = 0 and ΣM = 0.

Equations of Equilibrium for Concurrent Force System. For the concur-rent forces, the lines of action of all forces meet at a point, and hence the moment ofthose force about that very point will be zero or ΣM = 0 automatically.

Thus for concurrent force system, the condition ΣM = 0 becomes redundant andonly two conditions, i.e., ΣFx = 0 and ΣFy = 0 are required.

Action and Reaction. From the Newton’s third law of motion, we know thatto every action there is equal and opposite reaction. Hence reaction is always equaland opposite to the action.

Fig. 2.1(a) shows a ball placed on a horizontal surface (or horizontal plane)such that it is free to move along the plane but cannot move vertically downward.Hence the ball will exert a force vertically downwards at the support as shown inFig. 2.1(b). This force is known as action. The support will exert an equal forcevertically upwards on the ball at the point of contact as shown in Fig. 2.1(c).


NOTES

Engineering Mechanics W

A

W

A

Action

Support

W

A

RA

(a) (b) (c)

Fig. 2.1

The force, exerted by the support on the ball, is known as reaction. Hence ‘anyforce on a support causes an equal and opposite force from the support so that actionand reaction are two equal and opposite forces’.

��

The equilibrium of the bodies which are placed on the supports can beconsidered if we remove the supports and replace them by the reactions whichthey exert on the body. In Fig. 2.1 (a), if we remove the supporting surface and replaceit by the reaction RA that the surface exerts on the balls as shown in Fig. 2.1 (c), weshall get free-body diagram.

The point of application of the reaction RA will be the point of contact A, andfrom the law of equilibrium of two forces, we conclude that the reaction RA must bevertical and equal to the weight W.

Hence Fig. 2.1 (c), in which the ball is completely isolated from its support andin which all forces acting on the ball are shown by vectors, is known a free-bodydiagram. Hence to draw the free-body diagram of a body we remove all the supports(like wall, floor, hinge or any other body) and replace them by the reactions whichthese support exert on the body. Also the body should be completely isolated.

Problem 1. Draw the free body diagram of ball of weight W supported by astring AB and resting against a smooth vertical wall at C as shown in Fig. 2.2(a).

Sol. Given :Weight of ball = W

W

A String

BC

W

F

BRC

(a) (b)

Fig. 2.2


NOTES

Equilibrium ofRigid Bodies

The ball is supported by a string AB and is resting against a vertical wall at C.To draw the free-body diagram of the ball, isolate the ball completely (i.e.,

isolate the ball from the support and string). Then besides the weight W acting at B,we have two reactive forces to apply one replacing the string AB and another replacingthe vertical wall AC. Since the string is attached to the ball at B and since a stringcan pull only along its length, we have the reactive force F applied at B and parallelto BA. The magnitude of F is unknown.

The reaction RC will be acting at the point of contact of the ball with verticalwall i.e., at point C. As the surface of the wall is perfectly smooth, the reaction RCwill be normal to the vertical wall (i.e., reaction RC will be horizontal in this case)and will pass through the point B. The magnitude of RC is also unknown. The completefree-body diagram is shown in Fig. 2.2(b).

Problem 2. A roller of radius 40 cm, weighing 3000 N is to be pulled over arectangular block of height 20 cm as shown in Fig. 2.3, by a horizontal force appliedat the end of a string wound round the circumference of the roller. Find the magnitudeof the horizontal force which will just turn the roller over the corner of the rectangularblock. Also determine the magnitude and direction of reactions at A and B. All surfacesmay be taken as smooth.

Sol. Given :Radius of roller = 40 cmWeight, W = 3000 NHeight of block = 20 cmFind horizontal force P, reaction RA

and reaction RB when the roller just turnsover the block.

When the roller is about to turn overthe corner of the rectangular block, the rollerlifts at the point A and then there will be nocontact between the roller and the point A.Hence reaction RA at point A will become zero.

Now the roller will be in equilibrium under the action of the following threeforces :

(i) its weight W acting vertically downward(ii) horizontal force P

(iii) reaction RB at point B. The direction of RB is unknown.For the equilibrium, these three forces should pass through a common point.

As the force P and weight W is passing through point C, hence the reaction RB mustalso pass through the point C. Therefore, the line BC gives the direction of the reactionRB.

In ∆ BOD, BO = Radius = 40 cm,OD = OA – AD = 40 – 20 = 20 cm

∴ BD = BO OD2 2 2 240 20 1200− = − = = 34.64

Now in ∆ BCD, tan θ = BDCD CO OD

=+

=+

34 64 34 6440 20

. .( )

= 0.5773

∴ θ = tan–1 0.5773 = 29.999° ~ 30°

Fig. 2.3

C

P

O

40cm

20cm

20cm

D

A RB

θ

θ

Block

W B


NOTES

Engineering Mechanics Resolving forces horizontally, we get P – RB sin θ = 0or P = RB sin θ = RB × sin 30° = 0.5 RB ...(i)

Resolving forces vertically, we get W – RB cos θ = 0 or 3000 – RB × cos 30° = 0

or RB = 3000

30cos ° = 3464.2 N. Ans.

Substituting this value of RB in equation (i), we getP = 0.5 × 3464.2 = 1732.1 N. Ans.

Alternate MethodThis problem can also be solved by taking moments of all the three forces about

the point B (i.e., corner of the rectangular block) as shown below :P × CD = W × BD

or P × 60 = 3000 × 34.64 (� BD = 34.64)

∴ P = 3000 34 64

60× .

= 1732.0 N. Ans.

Problem 3. A body weighing 2000 N is suspended with a chain AB 2 m long.It is pulled by a horizontal force of 320 N as shown in Fig. 2.4. Find the force in thechain and the lateral displacement (i.e., x) of the body.

Sol. Given :

Weight suspended at B = 2000 N

Length AB = 2 m

Horizontal force at B = 320 NFind : Force in AB and value of xLet F = Force in chain AB

θ = Angle made by AB with horizontal.The free-body diagram of the point B is shown in Fig. 2.4(b).

320 N

2000 N

θBxC

A

2 m

320 N

2000 N

θB

F

(a) (b)

Fig. 2.4

The point B is in equilibrium under the action of three forces. Hence usingLami’s theorem, we get

Fsin sin ( ) sin ( )90

2000180

32090°

=−

=+θ θ


NOTES

Equilibrium ofRigid Bodiesor

F1

2000 320= =sin cosθ θ

[� sin (180 – θ) = sin θ, sin (90 + θ) = cos θ ]or F sin θ = 2000 ...(i)and F cos θ = 320 ...(ii)

Dividing equation (i) by equation (ii), we get

tan θ = 2000320

= 6.25

∴ θ = tan–1 6.25 = 80.9°Substituting this value of θ in equation (i), we get

F sin 80.9° = 2000 or F = 2000

80 9sin . = 2025.5 N. Ans.

Now from Fig. 2.4 (a), cos θ = x2

or x = 2 × cos θ = 2 × cos 80.9°

= 0.3163 m. Ans.

��

Though there are many types of supports, yet the following are important fromthe subject point of view :

1. Simple supports or knife edge supports2. Roller support3. Pin-joint (or hinged) support4. Smooth surface support5. Fixed or built-in support.Simple Support or Knife Edge Support. A beam supported on the knife edges

A and B is shown in Fig. 2.5(a). The reactions at A and B in case of knife edge sup-port will be normal to the surface of the beam. The reactions RA and RB with free-body diagram of the beam is shown in Fig. 2.5(b).

(a)BEAM

A B

(b)

RA RB

(a)A B

(b)

RA RB

Fig. 2.5 Fig. 2.6

Roller Support. A beam supported on the rollers at points A and B is shownin Fig. 2.6(a). The reactions in case of roller supports will be normal to the surfaceon which rollers are placed as shown in Fig. 2.6(b).

Pin Joint (or Hinged) Support. A beam,which is hinged (or pin-joint) at point A, is shown inFig. 2.7. The reaction at the hinged end may be eithervertical or inclined depending upon the type of loading.If the load is vertical, then the reaction will also bevertical. But if the load is inclined, then the reactionat the hinged end will also be inclined.

A

Fig. 2.7


NOTES

Engineering Mechanics Smooth Surface Support. Fig. 2.8 shows a body in contact with a smoothsurface. The reaction will always act normal to the support as shown in Fig. 2.8 (a) and2.8(b).

ASmoothsurface

Body

RA

Surface

RBB

(a) (b)

Fig. 2.8

Fig. 2.9 shows a rod AB resting inside asphere, whose surface are smooth. Here the rodbecomes body and sphere becomes surface. Thereactions on the ends of the rod (i.e., at point A andB) will be normal to the sphere surface at A and B.The normal at any point on the surface of the spherewill always pass through the centre of the sphere.Hence reactions RA and RB will have directions AOand BO respectively as shown in Fig. 2.9.

Fixed or Built-in Support. Fig. 2.10 showsthe end A of a beam, which is fixed. Hence the supportat A is known as a fixed support. In case of fixedsupport, the reaction will be inclined. Also the fixedsupport will provide a couple.

Types of Loading. The following are theimportant types of loading :

(a) Concentrated or point load,(b) Uniformly distributed load, and(c) Uniformly varying load.(a) Concentrated or point load. Fig. 2.11

shows a beam AB, which is simply supported atthe ends A and B. A load W is acting at the point C. This load is known as point load(or concentrated load). Hence any load acting at a point on a beam, is known as pointload.

In actual practice, it is not possible to apply a load at a point (i.e., at amathematical point) as it must have some contact area. But this area in comparisonto the length of the beam is very very small (or area is negligible).

Fig. 2.9

W

A

B

OBody

W

A BC

RA RB

A

Fig. 2.10

Fig. 2.11


NOTES


(b) Uniformly distributed load. If a beamis loaded in such a way, that each unit length ofthe beam carries same intensity of the load, thenthat type of load is known as uniformly distributedload (which is written as U.D.L.). Fig. 2.12 showsa beam AB, which carries a uniformly distributedload.

For finding the reactions the total uniformly distributed load is assumed toact at the C.G. of the load.

(c) Uniformly varying load. Fig. 2.13shows a beam AB, which carries load in such away that the rate of loading on each unit lengthof the beam varies uniformly. This type of load isknown as uniformly varying load. The total loadon the beam is equal to the area of the loaddiagram. The total load acts at the C.G. of the loaddiagram.

��

For stable equilibrium of a body, the algebraic sum of all the external forcesshould be zero and also the algebraic sum of moments of all the external forces aboutany point in their plane is zero. Mathematically, it is expressed by the equations :

ΣF = 0ΣM = 0

The sign Σ is known as sigma which is a Greek letter. This sign representsthe algebraic sum of forces or moments.

For two-dimensional bodies, the forces are generally resolved into horizontaland vertical components. Hence we have

ΣFx = 0and ΣFy = 0where ΣFx = Algebraic sum of all horizontal componentsand ΣFy = Algebraic sum of all vertical components.

Similarly for moments, we haveΣMx = 0 and ΣMy = 0

��

Moment. The moment is the product of the forceand the perpendicular distance between the line of actionof the force and the point about which moment is to betaken.

Couple. When two equal and opposite parallelforces act on a body at some distance apart, the two forcesform a couple as shown in Fig. 2.14. This couple has thetendency to rotate the body. The perpendicular distance

A B

RBRA

A B

RBRA

Fig. 2.12

Fig. 2.13

F

A B

a

F

Body

Fig. 2.14


NOTES

Engineering Mechanics between the parallel forces is known as arm of the couple. The moment of the coupleis the porduct of either one of the forces and perpendicular distance betweenthe forces.

∴ Moment of the couple = F × a.

��

��

The product of a force and the perpendiculardistance of the line of action of the force from apoint is known as moment of the force about thatpoint.

Let F = A force acting on a body as shownin Fig. 2.15.

r = Perpendicular distance from the point O on the line of action of force F.

Then moment (M) of the force F about O is given by, M = F × rThe tendency of this moment is to rotate

the body in the clockwise direction about O.Hence this moment is called clockwise moment.If the tendency of a moment is to rotate the bodyin anti-clockwise direction, then that moment isknown as anti-clockwise moment. If clockwisemoment is taken – ve then anti-clockwise momentwill be + ve.

In S.I. system, moment is expressed in N m (Newton metre).Fig. 2.16 shows a body on which three forces F1, F2 and F3 are acting. Suppose

it is required to find the resultant moments if these forces about point O.Let r1 = Perpendicular distance from O on the line of action of force F1.r2 and r3 = Perpendicular distances from O on the lines of action of force F2

and F3 respectively.Moment of F1 about O = F1 × r1 (clockwise) (–)Moment of F2 about O = F2 × r2 (clockwise) (–)Moment of F3 about O = F3 × r3 (anti-clockwise) (+)The resultant moment will be algebraic sum of all the moments.∴ The resultant moment of F1, F2 and F3 about

O = – F1 × r1 – F2 × r2 + F3 × r3.Problem 4. Four forces of magnitude 10 N, 20 N, 30 N and 40 N are acting

respectively along the four sides of a square ABCD as shown in Fig. 2.17. Determinethe resultant moment about the point A. Each side of the square is given 2 m.

Sol. Given :Length AB = BC = CD = DA = 2 m

Fig. 2.15

Fig. 2.16

r

Line of action of force

O

Perpendiculardistance

F

O

F1

r1

r2r3F2

F3


NOTES


Force at B = 10 N,Force at C = 20 N,Force at D = 30 N,Force at A = 40 N,The resultant moment about point A is to be

determined.The forces at A and B passes through point A.

Hence perpendicular distance from A on the lines ofaction of these forces will be zero.

Hence their moments about A will be zero. Themoment of the force at C about point A

= Force at C × ⊥ distance from A on the line of action of force at C= (20 N) × (Length AB)= 20 × 2 N m = 40 N m (anti-clockwise)

The moment of force at D about point A= Force at D × ⊥ distance from A on the line of action of force at D= (30 N) × (Length AD)= 30 × 2 N m = 60 N m (anti-clockwise)

∴ Resultant moment of all forces about A= 40 + 60 = 100 N m (anti-clockwise). Ans.

��

��

Vectorial Representation of Moments. Themoment of a force is a vector which is the product ofdistance and force. Hence in case of moment* of aforce the cross-product of distance and force would betaken. Consider the Fig. 2.18.

Let F = Force vector (Fxi + Fy j + Fzk)

r = Distance (or position) vector withrespect to O

= xi + yj + zk

M = Moment of force about point O

Then M = r × F

or M = r × F = i j kx y z

F F Fx y z

∴ M = (yFz – zFy) i + (zFx – xFz) j + (xFy – yFx) k

Fig. 2.17

20 N

CD

30 N

2 m

A B

2 m

40 N

10 N

Fz

Fx

FFyY

My

r

Z

MzMx

OX

Fig. 2.18

*A quantity which is the product of two vectors and the quantity is also a vector, thencross product of the two vectors will be taken. But if the quantity is scalar, then dot productis taken.


NOTES

Engineering Mechanics The moment of the given force about x, y and z-axis are equal toMx = yFz – zFy , My = zFx – xFz , Mz = xFy – yFx

where Mx = Moment of F about x-axis My = Moment of F about y-axis, andMz = Moment of F about z-axis.

Also Mx, My, and Mz are known as scalar components of moment.Problem 5. A force F = 2i + 4j – 3k is applied at a point P(1, 1, – 2). Find the

moment of the force F about the point (2, – 1, 2).Sol. Given :Force F = 2i + 4j – 3kThe position vector r of the point P w.r.t. O.

= Position vector of point P– Position vector of point O.

= (i + j – 2k) – (2i – j + 2k)∴ r = (1 – 2)i + [1 + (1)] j + [– 2 – 2]k

= – i + 2j – 4kThe moment M is given by

M = r × F = i j k

− −−

1 2 42 4 3

= [(2)(– 3) – (– 4)(4)] i + [(– 4)(2) – (– 1) (– 3)] j + [(– 1)(4) – (2)(2)]k= (– 6 + 16) i + (– 8 – 3) j + (– 4 – 4) k = 10i – 11j – 8k. Ans.

Vectorial Representation of Couples. The moment produced by two equal,opposite parallel forces is known as couple.

Fig. 2.20 shows two equal opposite and parallel forces acting at points A andB. Let rA and rB are the position vectors of A and B with respect to O. The vectorwhich joints B to A is represented by r.

The moment of two forces about point O is givenby

Mo = rA × F – rB × F= (rA – rB) × F= r × F (� rA – rB = r)

The above equation shows that moment vectoris independent of moment centre O.

∴ M = r × FThis moment is known as couple.The effect of couple is to produce pure rotation about an axis normal of the

plane of force which constitute couple.Scalar Components of a Moment. The moment of a force about any point

O, is given by,M = r × F

=

i j kx y z

F F Fx y z

r

O (2, –1, 2)

F = 2 + 4 – 3ki j

P(1, 1, –2)

Fig. 2.19

O

A F

B F

rB

rA

r

Fig. 2.20


NOTES


= (yFz – zFy)i + (zFx – xFz)j + (xFy – yFx)k= Mxi + Myj + Mzk

where Mx = Moment of F about x-axis = (yFz – zFy),My = Moment of F about y-axis = (zFx – xFz) andMz = Moment of F about z-axis = (xFy – yFx).

Also Mx, My and Mz are known as scalar components of Moment M.

��! ��"�� #��

�� $

Varignon’s Theorem states that the moment of a force about any point is equalto the algebraic sum of the moments of its components about that point.

Principle of moments states that the moment of the resultant of a number offorces about any point is equal to the algebraic sum of the moments of all the forcesof the system about the same point.

Proof of Varignon’s Theorem

C

BF2

F1

R

rr2

r1

O O′

A90°

C

H

G

DE

A

F1

r1

r

r2θ2

O′FO θ1θ

F2

B R

x

(a) (b)

Fig. 2.21

Fig. 2.21 (a) shows two forces F1 and F2 acting at point O. These forces arerepresented in magnitude and direction by OA and OB. Their resultant R isrepresented in magnitude and direction by OC which is the diagonal of parallelogramOACB. Let O′ is the point in the plane about which moments of F1, F2 and R are tobe determined. From point O′, draw perpendiculars on OA, OC and OB.

Let r1 = Perpendicular distance between F1 and O′. r = Perpendicular distance between R and O′.r2 = Perpendicular distance between F2 and O′.

Then according to Varignon’s principle ;Moment of R about O′ must be equal to algebraic sum of moments of F1 and

F2 about O′.or R × r = F1 × r1 + F2 × r2

Now refer to Fig. 2.21 (b). Join OO′ and produce it to D. From points C, A andB draw perpendiculars on OD meeting at D, E and F respectively. From A and Balso draw perpendiculars on CD meeting the line CD at G and H respectively.


NOTES

Engineering Mechanics Let θ1 = Angle made by F1 with OD, θ = Angle made by R with OD, andθ2 = Angle made by F2 with OD.

In Fig. 2.21 (b), OA = BC and also OA parallel to BC, hence the projection ofOA and BC on the same vertical line CD will be equal i.e., GD = CH as GD is theprojection of OA on CD and CH is the projection of BC on CD.

Then from Fig. 2.34 (b), we haveP1 sin θ1 = AE = GD = CH

F1 cos θ1 = OEF2 sin θ1 = BF = HD

F2 cos θ2 = OF = ED � (OB = AC and also OB || AC. Henceprojections of OB and AC on the same

horizontal line OD will be equali.e., OF = ED)

R sin θ = CD

R cos θ = ODLet the length OO′ = x.Then x sin θ1 = r1, x sin θ = r and x sin θ2 = r2

Now moment of R about O′= R × (⊥ distance between O′ and R) = R × r= R × x sin θ (� r = x sin θ)= (R sin θ) × x= CD × x (� R sin θ = CD)= (CH + HD) × x= (F1 sin θ1 + F2 sin θ2) × x

(� CH = F1 sin θ1 and HD = F2 sin θ2)= F1 × x sin θ1 + F2 × x sin θ2

= F1 × r1 + F2 × r2 (� x sin θ1 = r1 and x sin θ2 = r2)= Moment of F1 about O′ + Moment of F2 about O′.

Hence moment of R about any point in the algebraic sum of moments of itscomponents (i.e., F1 and F2) about the same point. Hence Varignon’s principle isproved.

The principle of moments (or Varignon’s principle) is not restricted to only twoconcurrent forces but is also applicable to any coplanar force system, i.e., concurrentor non-concurrent or parallel force system.

Problem 6. A force of 100 N is acting at a point A as shown in Fig. 2.22.Determine the moments of this force about O.

Sol. Given :Force at A = 100 NDraw a perpendicular from O on the line of action of force 100 N. Hence OB is

the perpendicular on the line of action of 100 N as shown in Fig. 2.22.


NOTES


1st MethodTriangle OBC is a right-angled triangle. And angle

OCB = 60°.

∴ sin 60° = OBOC

∴ OB = OC sin 60°= 3 × 0.866 = 2.598 m

Moment of the force 100 N about O= 100 × OB = 100 × 2.598= 259.8 N m (clockwise). Ans.

2nd MethodThe moment of force 100 N about O, can also be

determined by using Varignon’s principle. The force 100N is replaced by its two rectangular components at anyconvenient point. Here the convenient point is chosenas C. The horizontal and vertical components of force100 N acting at C are shown in Fig. 2.23.

(i) The horizontal component= 100 × cos 60° = 50 N

But this force is passing through O and hence hasno moment about O.

The vertical component= 100 × sin 60° = 100 × 0.866 = 86.6 N

This force is acting vertically downwards at C. Moment of this force about O= 86.6 × OC = 86.6 × 3 (� OC = 3 m)= 259.8 N (clockwise). Ans.

��% �� &��

��

For the equilibrium of rigid bodies in two-dimensions or in three dimensionsthe algebraic sum of all the external forces should be zero and also the algebraic sumof moments of all the external forces about any point in their plane should be zero.Mathematically it is expressed by the equations :

ΣF = 0 and ΣM = 0The sign Σ is known as sigma which represents the algebraic sum of forces or moments

(i) For two-dimensional bodies, the forces are written as ΣFx = 0 and ΣFy = 0

(ii) For three-dimensional bodies, the forces are expressed asΣFx = 0, ΣFy = 0 and ΣFz = 0

where ΣFx = Algebraic sum of all components of forces in x-direction,ΣFy = Algebraic sum of all components of forces in y-direction, andΣFz = Algebraic sum of all components of forces in z-direction.

Fig. 2.22

Fig. 2.23

100 N

AY

O C60°

3 m

BX

100 NY

O C X60°

100 cos 60°

100 sin 60°

100 N


NOTES

Engineering Mechanics��' ��

��

The following are the methods of finding out the reactions at the two supportsof a beam :

1. Analytical method, and2. Graphical method.Analytical Method. Fig. 2.24 shows a beam AB of length L and is simply sup-

ported at the ends A and B. The beam carries two point loads W1 and W2 at a distanceL1 and L2 from the end A.

Let RA = Reaction at Aand RB = Reaction at B

As the beam is in equilibrium, theequations of the equilibrium i.e., ΣFx = 0, ΣFy = 0and ΣM = 0 should be satisfied. In this case thereis no horizontal force, hence the equations ofequilibrium are ΣFy = 0 and ΣM = 0.

For Σy = 0, we haveRA + RB = W1 + W2 ...(i)

For ΣM = 0, the moments about any point of all the forces should be zero.Taking the moments about point A, we get

W1 × L1 + W2 × L2 – RB × L = 0

or W1 × L1 + W2 × L2 = RB × L or RB = W L W L

L1 1 2 2× + ×

As W1, W2, L1, L2 and L are given, hence value of RB can be calculated.Now from equation (i), we have RA = (W1 + W2) – RB.Graphical Method for Finding Out the Reactions of a Beam. The graphi-

cal method consists of the following steps :(a) Construction of space diagram ;(b) Use of Bow’s notations ; and(c) Construction of vector diagram.The given beam is drawn to a suitable scale along with the loads and the

reactions RA and RB. This step is known as construction of space diagram.The different loads and forces (i.e., reactions RA and RB) are named by two

capital letters, placed on their either side of the space diagram as shown in Fig. 2.25.This step is known as Bow’s notation. The load W1 is named by PQ, W2 by QR, reactionRB by SR and reaction RA by SP.

Now the vector diagram is drawn according to the following steps :(i) Choose a suitable scale to represent the various loads. Now take any point p

and draw pq parallel and equal to the load PQ (i.e., W1) vertically downward to thesame scale.

(ii) Now through q, draw qr parallel and equal to QR vertically downward to thesame scale.

(iii) Select any suitable point O. Now join the point O to points p, q and r asshown in Fig. 2.25 (b).

L1

L2

L

RA RB

W1 W2

A B

Fig. 2.24


NOTES


(iv) Now in Fig. 2.25 (a), extend the lines of action of the loads and the tworeactions. Take any point 1, on the line of action of the reaction RA. Through 1, drawthe line 1-2 parallel to pO, intersecting the line of action of load W1 at 2.

(v) Now from point 2, draw line 2-3 parallel to qO, intersecting the line of ac-tion of load W2 at 3. Similarly, from point 3, draw the line 3-4 parallel to rO,intersecting the line of action of reaction RB at point 4.

L1

L2

LRA RB

W1 W2

A B

Closing line 4

1

2

3

S

P Q R

O

p

sq

r

(a) Space diagram (b) Vector diagram

Fig. 2.25

(vi) Now join the point 1 to point 4. The line 1–4 is known as closing line. Nowfrom point O (i.e., from vector diagram) draw line Os parallel to line 1–4.

(vii) Now in the vector diagram the length sp represents the magnitude of reac-tion RA to the same scale. Similarly, the length rs represents the magnitude of reactionRB to the same scale.

��

&(��

Problems on simple supported beamProblem 7. A simply supported beam AB of span 6 m carries point loads of 3

kN and 6 kN at a distance of 2 m and 4 m from the left end A as shown in Fig. 2.26.Find the reactions at A and B analytically and graphically.

Sol. Given :Span of beam = 6 mLet RA = Reaction at A

RB = Reaction at B(a) Analytical method. As the beam is

in equilibrium, the moments of all the forcesabout any point should be zero.

Now taking the moment of all forcesabout A, and equating the resultant moment tozero, we get

RB × 6 – 3 × 2 – 6 × 4 = 0

2 m

RA RB

3 kN 6 kN

A B

4 m

6 m

Fig. 2.26


NOTES

Engineering Mechanics or 6RB = 6 + 24 = 30

or RB = 306

= 5 kN. Ans.

Also for equilibrium, ΣFy = 0∴ RA + RB = 3 + 6 = 9∴ RA = 9 – RB = 9 – 5 = 4 kN. Ans.(b) Graphical method. First of all draw the space diagram of the beam to a

suitable scale. Let 1 cm length in space diagram represents 1 m length of beam.Hence take AB = 6 cm, distance of load 3 kN from A = 2 cm and distance of 6 kNfrom A = 4 cm as shown in Fig. 2.27 (a).

Now name all the loads and reactions according to Bow’s notation i.e., load 3kN is named by PQ, load 6 kN by QR, reaction RB by SR and reaction RA by SP.

Now the vector diagram is drawn according to the following steps : [Refer to Fig.2.27 (b)].

1. Choose a suitable scale to represent various loads. Let 1 cm represents 1 kNload. Hence load PQ (i.e., 3 kN) will be equal to 3 cm and load QR (i.e., 6 kN) 6 cm.

2. Now take any point p and draw line pq parallel to load PQ (i.e., 3 kN). Takepq = 3 cm to represent the load of 3 kN.

3. Through q, draw line qr parallel to load QR (i.e., 6 kN). Cut qr equal to 6 cmto represent the load of 6 kN.

6 cmRA RB

3 kN 6 kN

A B

Closing line

41

2

3

S

P Q R

4 cm2 cm

O

p

q

s

r

(a) Space diagram (b) Vector diagramFig. 2.27

4. Now take any point O. Join the point O to the points p, q and r as shown inFig. 2.27 (b).

5. Now in Fig. 2.27 (a), extend the lines of action of the loads (3 kN and 6 kN),and the two reactions. Take any point 1, on the line of action of the reaction RA.Through 1, draw the line 1-2 parallel to pO, intersecting the line of action of load 3kN at point 2.

6. From point 2, draw line 2-3 parallel to qO, intersecting the line of action ofload 6 kN at 3. Similarly, from point 3, draw a line 3-4 parallel to rO, intersectingthe line of action of reaction RB at point 4.

7. Join 1 to 4. The line 1-4 is known as closing line. From the vector diagram,from point O, draw line Os parallel to line 1–4.


NOTES


8. Measure the length sp and rs. The length sp represents the reaction RA andlength rs represents the reaction RB.

By measurement, sp = 4 cm and rs = 5 cm∴ RA = Length sp × scale = 4 × 1 kN = 4 kN. Ans.

RB = Length rs × scale = 5 × 1 kN = 5 kN. Ans.Problems on Roller and Hinged Supported Beams. In case of roller sup-

ported beams, the reaction on the roller end is always normal to the support. All thesteel trusses of the bridges is generally having one of their ends supported on rollers.The main advantage of such a support is that beam, due to change in temperature,can move easily towards left or right, on account of expansion or contraction.

In case of a hinged supported beam, the reaction on the hinged end may beeither vertical or inclined, depending upon the type of loading. The main advantageof a hinged end is that the beam remains stable. Hence all the steel trusses of thebridges, have one of their end on rollers and the other end as hinged.

Problem 8. A beam AB 1.7 m long is loaded as shown in Fig. 2.28. Determinethe reactions at A and B.

Sol. Given :Length of beam = 1.7 mLet RA = Reaction at A

and RB = Reaction at B.

RAY RB

1.7 m

70 cm40 cm 40 cm20 cm

C D E

RAXA

50 N 20 N 30 N 15 N

80°

B45°60°

Fig. 2.28

Since the beam is supported on rollers at B, therefore the reaction RB will bevertical.

The beam is hinged at A, and is carrying inclined load, therefore the reactionRA will be inclined. This means reaction RA will have two components, i.e., verticalcomponent and horizontal component.

Let RAX = Horizontal component of reaction RA

RAY = Vertical component of reaction RA.First resolve all the inclined loads into their vertical and horizontal components.(i) Vertical component of load at D

= 20 sin 60° = 20 × 0.866 = 17.32 Nand its horizontal component

= 20 cos 60° = 10 N ←(ii) Vertical component of load at E

= 30 sin 45° = 21.21 N


NOTES

Engineering Mechanics and its horizontal component= 30 cos 45° = 21.21 N →

(iii) Vertical component of load at B= 15 sin 80° = 14.77 N

and its horizontal component= 15 cos 80° = 2.6 N ←

From condition of equilibrium, ΣFx = 0or RAX – 10 + 21.21 – 2.6 = 0or RAX = 10 – 21.21 + 2.6 = – 8.61 N

– ve sign shows that the assumed direction of RAX (i.e., horizontal componentof RA) is wrong. Correct direction will be opposite to the assumed direction. Assumeddirection of RAX is towards right. Hence correct direction of RAX will be towards leftat A.

∴ RAX = 8.61 N ←To find RB, take moments of all forces about A.For equilibrium, ΣMA = 0∴ 50 × 20 + (20 sin 60°) × (20 + 40) + (30 × sin 45°)

× (20 + 40 + 70) + (15 sin 80°) × (170) – 170 RB = 0or 1000 + 1039.2 + 2757.7 + 2511 – 170 RB = 0or 7307.9 – 170 RB = 0

∴ RB = 7307 9

170.

= 42.98 N. Ans.

To find RAY, apply condition of equilibrium, ΣFy = 0or RAY + RB = 50 + 20 sin 60° + 30 sin 45° + 15 sin 80°or RAY + 42.98 = 50 + 17.32 + 21.21 + 14.77

= 103.3∴ RAY = 103.3 – 42.98 = 60.32 N ↑

∴ Reaction at A, RA = R RAX AY2 2+

= 8 61 60 322 2. .+ = 60.92 N

The angle made by RA with x-direction is given by

tan θ = RR

AY

AX=

60 328 61

..

= 7.006

∴ θ = tan–1 7.006 = 81.87°. Ans.Problems when Beams are Subjected to Couples. In this section, the re-

actions of the beam will be calculated when beams are subjected to clockwise or anti-clockwise couple along with the other loads. While taking the moments about anypoint, the magnitude and sense of the couple is taken into consideration. But whenthe total load on the beam is calculated the magnitude and sense of the couple is notconsidered.

Problem 8. A simply supported beam AB of 7 m span is subjected to : (i) 4 kNm clockwise couple at 2 m from A, (ii) 8 kN m anti-clockwise couple at 5 m from Aand (iii) a triangular load with zero intensity at 2 m from A increasing to 4 kN per mat a point 5 m from A. Determine reactions at A and B.

Fig. 2.29

R = 8.61 NAXAθ

RA

R = 60.32AY


NOTES


Sol. Given :Span of beam = 7 mCouple at C (i.e., at 2 m from A) = 4 kN m (clockwise)Couple at D (i.e., at 5 m from A) = 8 kN m (anti-clockwise)Triangular load from C to D with :Vertical load at C = 0Vertical load at D = 4 kN/m

∴ Total load on beam = Area of triangle CDE = CD DE× = ×

23 4

2 = 6 kN

2 m

E

4 kN/m4 kN/mO

C D

A B

3 m5 m

7 m

8 kN/m

Fig. 2.30

This load will be acting at the C.G. of the ∆CDE i.e., at a distance of 23

× CD =

23

× 3 = 2 m from C or 2 + 2 + 4 m from end A.

Let RA = Reaction at A.RB = Reaction at B.

Taking the moments of all forces about point A and equating the resultantmoment to zero (i.e., ΣMA = 0 and considering clockwise moment positive and anti-clockwise moment negative), we get

– RB × 7 + 4 – 8 + (Total load on beam) × (Distance of total load from A) = 0or – 7 RB + 4 – 8 + 6 × 4 = 0

– 7 RB + 4 – 8 + 24 = 0or 20 = 7 RB

or RB = 207

kN. Ans.

Also for the equilibrium of the beam ΣFy = 0or RA + RB = Total load on the beam = 6 kN

∴ RA = 6 – RB = 6 – 207

= 227

kN. Ans.


NOTES


1. The principle of equilibrium states that a stationary body will be in equilibriumif the algebraic sum of all the forces is zero and also the algebraic sum of momentsof all the external forces is zero.

2. The conditions of equilibrium are written mathematically as ΣFx = 0, ΣFy = 0,and ΣM = 0. The sign Σ is known as sigma and this sign represents the algebraicsum.

3. Free body diagram of a body is a diagram in which the body is completely isolatedfrom its support and the supports are replaced by the reactions which thesesupports exert on the body.

4. A load, acting at a point on a beam, is known as point load or concentrated load.5. If each unit length of the beam carries same intensity of load, then that type of

load is known as uniformly distributed load which is written as U.D.L.6. The reactions of a beam can be determined by analytical method and graphical

method.7. The reactions by analytical method are obtained by using equations of equilib-

rium, i.e., ΣFx = 0, ΣFy = 0 and ΣM = 0.8. If a beam is loaded with inclined loads, then the inclined loads are resolved

normal to the beam and along the beam. Now the equations of equilibrium areused for finding reactions.

9. The reaction on a roller support is at right angles to the roller base.10. The forces in the members of a frame are determined by :

(i) Method of joints (ii) Method of sections, and(iii) Graphical method.

11. The force in a member will be compressive if the member pushes the joint towhich it is connected whereas the force in the member will be tensile if themember pulls the joint to which it is connected.

12. If three forces act at a joint and two of them are along the same straight linethen third force would be zero.

��

• Stable equilibrium. It is a state of a body when the algebraic sum of all externalforce should be zero and also the algebraic sum of moment of all the externalforces about any point in their plane is zero.

• Moment. It is the product of the force and the perpendicular distances betweenthe line of action of the force and the point about which moment is to be taken.

• Couple. When two equal and opposite forces act on a body at some distanceapart, the two forces form a couple.

• Varignon’s theorem. The moment of force about any point is equal to thealgebraic sum of the moments of its components about that point.


NOTES

Equilibrium ofRigid Bodies��

1. Define and explain the terms (i) Principle of equilibrium, (ii) Force law ofequilibrium (iii) Moment law of equilibrium.

2. A number of forces are acting on a body. What are conditions of equilibrium, sothat the body is in equilibrium ?

3. Two forces are acting on a body and the body is in equilibrium. What conditionsshould be fulfilled by these two forces ?

4. How will you prove that a body will not be in equilibrium when the body issubjected to two forces which are equal and opposite but are parallel ?

5. Explain the statement “Two equal and opposite parallel forces produces a couple”.6. (a) What is a frame ? State the difference between a perfect frame and an im-

perfect frame.(b) What are the assumptions made in finding out the forces in a frame ?

7. What are the different methods of analysing (or finding out the forces) a perfectframe ? Which one is used where and why ?

8. How will you find the forces in the members of a truss by method of joints when(i) the truss is supported on rollers at one end and hinged at other end and

carries vertical loads.(ii) the truss is acting as a cantilever and carries vertical loads.

(iii) the truss is supported on rollers at one end and hinged at other end andcarries horizontal and vertical loads.

(iv) the truss is supported on rollers at one end and hinged at other end andcarries inclined loads.

9. Three parallel forces F1, F2 and F3 areacting on a body as shown in Fig. 2.31and the body is in equilibrium. If forceF1 = 300 N and F3 = 1000 N and thedistance between F1 and F2 = 2.0 m, thendetermine the magnitude of force F2 anddistance of F3 from force F2.

[Ans. 1300 N, 0.6 m]

10. Three forces of magnitude 40 kN, 15 kN and 20 kN are acting at a point O. Theangles made by 40 kN, 15 kN and 20 kN forces with x-axis are 60°, 120° and240° respectively. Determine the magnitude and direction of the resultant force.

[Ans. 30.41 kN and 85.28° with x-axis]11. A lamp weighing 10 N is suspended from the ceiling by a chain.

It is pulled aside by a horizontal cord until the chain makesan angle of 60° with the ceiling. Find the tensions in the chainand the cord by applying Lami’s theorem and also by graphicalmethod. [Ans. 11.54 N and 5.77 N]

12. A circular roller of weight 1000 N and radius 20 cm hangs bya tie rod AB = 40 cm and rests against a smooth vertical wallat C as shown in Fig. 2.84. Determine the tension in the tierod and reaction RC at point C. [Ans. 1154.7 N and 577.3 N]

2 m x

F = 300 N1 F = 1000 N3

F2

Fig. 2.31

A

BC

40 cm

Fig. 2.32


NOTES

Engineering Mechanics 13. In problem 6 if radius of ball = 5 cm, length of string AB = 10 cm, weight of ballW = 40 N and the horizontal force F = 30 N, then find the tension the string andvertical reaction RC at point C. [Ans. 34.64 N and 57.32 N]

14. A smooth circular cylinder of weight 1000 Nand radius 10 cm rests in a right-angled groovewhose sides are inclined at an angle of 30° and60° to the horizontal as shown in Fig. 2.32.Determine the reaction RA and RC at the pointsof contact. [Ans. RA = 500 N, RC = 866.6 N]

15. If in the above problem, the sides of the groove makes an angle of 45° with thehorizontal, then find the reactions RA and RC. [Ans. RA = RC = 707 N]

16. A beam 6 m long is simply supported at the ends and carries a uniformlydistributed load of 1.5 kN/m and three concentrated loads 1 kN, 2 kN and 3kN acting respectively at a distance of 1.5 m, 3 m and 4.5 m from the left end.Calculate the reactions at both ends. [Ans. 7 kN, 8 kN]

17. A simply supported beam of span 10 m carries a uniformly varying load fromzero at the left end to 1200 N/m at the right end. Calculate the reactions at bothends of the beam. [Ans. 2000 N and 4000 N]

��

• Palanichamy, M.S., Nagam, S., ‘‘Engineering Mechanics – Statics & Dynamics’’,Tata McGraw-Hill, (2001).

• Irving H. Shames, ‘‘Engineering Mechanics – Statics and Dynamics’’, IV Edition– Pearson Education, Asia Pvt. Ltd., (2003).

• Dr. I.S. Gujral, ‘‘Engineering Mechanics’’, Laxmi Publications (P) Ltd.

Fig. 2.33

1000 N

B

CA

60°30°


NOTES

Properties of Surfacesand Solids

U N I T

3PROPERTIES OF SURFACES

AND SOLIDS

STRUCTURE

3.1 Introduction3.2 First Moment of Area and Centroid of Sections—Rectangle, Circle,

Triangle from Integration3.3 Centroid of T-section, I-section, Angle-section, Hollow-section etc.3.4 Centroid of Volume3.5 Second Moment of Area (Or Area Moment of Inertia)3.6 Theorem of the Perpendicular Axis3.7 Theorem of Parallel Axis3.8 Determination of Second Moment of Area (or Area Moment of Inertia)

of Plane Area Like Rectangle, Triangle, Circle etc. from Integration3.9 Moment of Inertia of T-section, I-section, Angle-section, Hollow Section

etc. by using Standard Formula3.10 Polar Moment of Inertia3.11 Product of Inertia3.12 Principal Axes3.13 Principal Moments of Inertia3.14 Mass Moment of Inertia3.15 Derivation of Mass Moment of Inertia for Rectangle Section, Prism,

Sphere etc. from First Principal• Summary• Glossary• Review Questions• Further Readings

��

After going through this unit, you should be able to :• determine areas of two-dimensional bodies and volume of three dimensional

bodies.


NOTES

Engineering Mechanics • describe the first as well as second moment of areas such as rectangle, triangle,circle, T-section and I-section.

• illustrate the concepts of mass moment of inertia, principal moment of inertiaand principal axis of inertia.

��

Surfaces are a two-dimensional whereas the solids are three dimensional bodies.For two-dimensional bodies, area is to be determined but for three dimensional bodiesvolume is to be determined. This chapter deals with determination of areas of two-dimensional bodies and volume of three dimensional bodies. Also the first momentof area and second moment of areas of plane sections such as rectangle, triangle, circle,T-section and I-section are discussed. The concept of mass moment of inertia, principalmoment of inertia and principal axes of inertia is also given.

Centre of Gravity. Centre of gravity of a body is the point through whichthe whole weight of the body acts. A body is having only one centre of gravity for allpositions of the body. It is represented by C.G. or simply G.

Centroid. The point at which the total area of a plane figure (like rectangle,square, triangle, quadrilateral, circle etc.) is assumed to be concentrated, is knownas the centroid of that area. The centroid is also represented by C.G. or simply G.The centroid and centre of gravity are at the same point.

Centroid or Centre of Gravity of Simple Plane Figures. (i) The centre ofgravity (C.G.) of a uniform rod lies at its middle point.

(ii) The centre of gravity of a triangle lies at the pointwhere the three medians of the triangle meet.

(iii) The centre of gravity of a rectangle or of aparallelogram is at the point, where its diagonal meet eachother. It is also the point of intersection of the lines joiningthe middle points of the opposite sides.

(iv) The centre of gravity of a circle is at its centre.Centroid (or Centre of Gravity) of Areas of

Plane Figures by the Method of Moments. Fig. 3.1shows a plane figure of total area A whose centre of gravityis to be determined. Let this area A is composed of anumber of small areas a1, a2, a3, a4, ...... etc.

∴ A = a1 + a2 + a3 + a4 + ...Let x1 = The distance of the C.G. of the area a1 from axis OY

x2 = The distance of the C.G. of the area a2 from axis OYx3 = The distance of the C.G. of the area a3 from axis OY

x4 = The distance of the C.G. of the area a4 from axis OY and so on.The moment of small area a1 about the axis OY

= area a1 × distance of C.G. of area a1 from axis OY= a1 × x1

The above equation is known as first moment of area about the axis OY.This first moment of area is used to determine the centroid of the area.

Fig. 3.1

x1

x2

x3

x4

G

Area a3Area a2Area a1

Area a4

xO X

Y


NOTES


The moments of all small areas about the axis OY= a1x1 + a2x2 + a3x3 + a4x4 + ... ...(i)

Let G is the centre of gravity of the total area A whose distance from the axisOY is x .

Then moment of total area about OY = Ax ...(ii)The moments of all small areas about the axis OY must be equal to the moment

of total area about the same axis. Hence equating equations (i) and (ii), we get a1x1 + a2x2 + a3x3 + a4x4 + ... = Ax

or x = a x a x a x a x

A1 1 2 2 3 3 4 4+ + + + ...

...(3.1)

where A = a1 + a2 + a3 + a4 ...If we take the moments of the small areas about the axis OX and also the

moment of total area about the axis OX, we will get

y = a y a y a y a y

A1 1 2 2 3 3 4 4+ + + + ...

...(3.2)

where y = The distance of G from axis OXy1 = The distance of C.G. of the area a1 from axis OX

y2, y3, y4 = The distance of C.G. of area a2, a3, a4 from axis OX respectively.

Centre of Gravity of Areas of Plane Figures by Integration Method. Theequations (3.1) and (3.2) can be written as

x = Σ

Σa xai i

iand y =

ΣΣa yai i

i

where i = 1, 2, 3, 4, ..... xi = Distance of C.G. of area ai from axis OY and yi = Distance of C.G. of area ai from axis OX.The value of i depends upon the number of small areas. If the small areas are

large in number (mathematically speaking infinite in number), then the summationsin the above equations can be replaced by integration. Let the small areas arerepresented by dA instead of ‘a’, then the above equations are written as :

x = ��x dA

dA*

...(3.3)

and y = ��y dA

dA*

...(3.4)

where ∫ x* dA = Σxiai

∫ dA = Σai

∫ y*dA = Σyiai

Also x* = Distance of C.G. of area dA from axis OY,y* = Distance of C.G. of area dA from axis OX.

x*dA = First moment of area dA about the axis OY, andy*dA = First moment of area dA about the axis OX.

Centroid (or Centre of Gravity) of a Line. The centre of gravity of a linewhich may be straight or curve, is obtained by dividing the given line, into a largenumber of small lengths as shown in Fig. 3.2.


NOTES

Engineering Mechanics The centre of gravity is obtained byreplacing dA by dL in equations (3.3) and(3.4).

Then these equations become

x = ��x dL

dL*

...(3.5)

and y = ��y dL

dL*

...(3.6)

where x* = Distance of C.G. of length dL fromy-axis, and

y* = Distance of C.G. of length dL fromx-axis.

If the lines are straight, then the above equations are written as :

x = L x L x L x

L L L1 1 2 2 3 3

1 2 3

+ + ++ + +

.............

...(3.6A)

and y = L y L y L y

L L L1 1 2 2 3 3

1 2 3

+ + ++ + +

............

...(3.6B)

Important Points. (i) The axis, about which moments of areas are taken, isknown as axis of reference. In the above article, axis OX and OY are called axis ofreference.

(ii) The axis of reference, of plane figures, is generally taken as the lowest line ofthe figure for determining y , and left line of the figure for calculating x .

(iii) If the given section is symmetrical about X-X axis or Y-Y axis, then the C.G.of the section will lie on the axis is symmetry.

��

��

��

The first moment of area as defined in Art. 3.1will be used to determine the centroid of the followingsections by the method of integration :

1. Rectangular section, 2. Circular section, and3. Triangular section.Centroid of a Rectangular Section by

Integration. Fig 3.3 shows a rectangular sectionABCD having width = b and depth = d. Consider arectangular elementary strip of thickness ‘dy’ at adistance y from the axis OX.

Let dA = Area of strip = b . dy

Moment of the area dA about axis OX= dA × y = (b . dy) × y (� dA = b . dy)= by × dy

Fig. 3.2

A

dL

BL

x∗

y∗

Y

O X

Fig. 3.3

b

Y

A B

d

O D C X

dy

y


NOTES


The moment of the whole area about axis OX (or first moment of the wholearea about axis OX) will be obtained by integrating the above equation between thelimits 0 to d.

∴ Moment of the whole rectangular section about axis OX

= 0

dby dy� × = b

0

dy dy�

(� b is constant and can be taken outside the integral sign)

= b y

d2

02

�

��

�� =

bd2

2...(i)

Let A = Total area of rectangular section

= 0 0

d ddA b dy� �= . = b y

d�

��

��

0[� dA = b . dy]

= b × d

y = Distance of the centroid of the rectangular section from axis OX.∴ Moment of total area of rectangular section about axis OX

= A × y = (b × d) × y ...(ii)Equating the equations (i) and (ii), we get

(b × d) × y = bd2

2

∴ y = bd2

2 ×

1bd

= d2

Similarly, the distance of the centroid of the rectangular section from the axisOY is given by

x = b2

Refer to Fig. 3.4 (a)Area of strip, dA = d . dxMoment of this area dA about axis OY

= dA × x∴ Moment of the whole area about axis OY

= 0 0

b bdA x d dx x� �× = ( . )

[� dA = d . dx]

= d 0

2

0

2

2 2

bb

x dx dx d b

� =�

��

�� = ×

...(iii)

Let x = Distance of centroid of whole area from axis OY. A = Total area = b × d

∴ Moment of total area A about axis OY= A × x ...(iv)

Equating equations (iii) and (iv), we get

A × x = d b× 2

2

or x = d b× 2

2 ×

12

12

Ad b

b d= × ×

× =

b2

.

Fig. 3.4

b

Y

d

O Xx

dx


NOTES

Engineering Mechanics Centroid of Circular Section by Integration.Fig. 3.5 shows a circular section of radius R with O ascentre. The equation of the circle is

x2 + y2 = R2

Consider a rectangular elementary strip ofthickness ‘dy’ at a distance of y from the axis OX.

The area of strip, dA = 2x . dy.Moment of this area dA about x-axis

= dA . y= (2x . dy) . y (� dA = 2 x . dy)= 2xy . dy ...(i)

But the equation of the circle isx2 + y2 = R2

or x = R y2 2−Substituting the above value of x in equation (i), we getMoment of area dA about x-axis

= 2 R y2 2− . y . dy

Moment of total area A about x-axis will be obtained by integrating the aboveequation form – R to R.

∴ Moment of area A about x-axis

= −

+

� −R

RR y2 2 2 . y . dy (� y varies from – R to + R)

= – −

+

−� − − = − −�

��

�

��

R

R

R

R

R y y dyR y2 2

2 2 3 22

3 2( ) .

( )/

/

= – 23

[(R2 – R2)3/2 – {R2 – (– R)2}3/2]

= − 23

[0 – 0] = 0 ...(ii)

Also the moment of total area A about x-axis

= A × y ...(iii)

where y = Distance of centroid of total area A from x-axis.

Equating the two values given by equations (ii) and(iii), we get

A × y = 0 or y = 0

This means that the centroid of the circle is at the centre of the circle.Similarly, taking a strip of thickness dx parallel to y-axis at a distance x from

y-axis as shown in Fig. 3.6, it can be proved that

x = 0Centroid of a Triangular Section by Integration. Fig. 3.7 shows a trian-

gular section AOB of base width = b and height = h. Consider a small strip of thicknessdy at a distance y from the axis OX.

Area of strip, dA = Length DE × dy ...(i)

Fig. 3.5

Fig. 3.6

x x

Y

y

O

dy

x + y = R2 2 2

X

y

y

Y

O

X

xdx


NOTES


The distance DE in terms of y, b and h isobtained from two similar triangles ADE and AOBas

DEOB

ADAO

=

where OB = b, AO = h and AD = (h – y)

∴DEb

= ( )h y

h−

or DE = b h y

h( )−

Substituting this value of DE in equation (i), we get

Area of strip, dA = b h y

h( )−

. dy

Moment of this area dA about axis OX

= dA . y = b h y

h( )−

. dy . y

= bh

(h – y) . y . dy

The moment of the total area A of the triangular section is obtained byintegrating the above equation between the limits 0 to h.

∴ Moment of total area A about axis OX

= 0

h bh

h y y dy� −( ) . . = bh

h y y dyh

0� −( ) . .

[� b and h are constants and can be takenoutside the integral sign]

= bh

hy y dyh

0

2� −( ) = bh

hy yh2 3

02 3

−�

��

�

��

= bh

h h h. 2 3

2 30−

�

�

� −

�

��

�

�� = b

hh h3 3

2 3−

�

�

� =

bh

h×3

6...(ii)

Let y = Distance of centroid of total area A from axis OX

A = Total area = 0 0

h hdA

bh

h y dy� �= −( ) . � dAbh

h y dy= −��

��

( ) .

= bh

hyy b

hh

h bh

h bhh

−�

��

�

�� = −

�

��

�

�� = =

2

0

22 2

2 2 2 2.

Then moment of total area A about axis OX= A × y

= b h×�

�� 2 × y ...(iii)

Equating the two values given by equations (ii) and (iii), we get

b h×�

�� 2 × y =

bh

h.

3

6

Fig. 3.7

hD E

AY

dy

BO

b X

y


NOTES


or y = bh

hb h

h.

3

61

23

�

�

� ×

×�

��

=

Similarly, taking a strip of thickness dxparallel to y-axis at a distance of x from the y-axis(Refer to Fig. 3.8), it can be proved that

x = b3

.

Problems of Finding Centre of Gravity of Areas by Integration MethodProblem 1. Determine the co-ordinates of the C.G. of the area OAB shown in

Fig. 3.9, if the curve OB represents the equation of a parabola, given byy = kx2

in which OA = 6 unitsand AB = 4 units.

Sol. The equation of parabola is y = kx2 ...(i)First determine the value of constant k. The point B is lying on the curve

and having co-ordinatesx = 6 and y = 4

Substituting these values of equation (i), we get4 = k × 62 = 36 k

∴ k = 436

19

=

Substituting the value of k in equation (i), weget

y = 19

x2 ...(ii)

or x2 = 9y

or x = 3 y ...(iii)Consider a strip of height y and width dx as shown in Fig. 3.9. The area dA of

the strip is given by dA = y × dx

The co-ordinates of the C.G. of this area dA are x and y2

∴ Distance of C.G. of area dA from y-axis = x

and distance of C.G. of area dA from x-axis = y2

∴ x* = x and y* = y2

Let x = Distance of C.G. of total area OAB from axis OYy = Distance of C.G. of total area OAB from axis OX.

Using equation (3.3), we get

x = ��

��

=×x dA

dA

x ydx

ydx

*0

6

0

6 (� dA = ydx, x* = x)

Fig. 3.8

Fig. 3.9

h

Y

Ob X

x dx

Y B

O A Xdx

6

4.0

y

y = kx2

y/2


NOTES

Properties of Surfacesand SolidsBut y =

x2

9 from equation (ii).

∴ x = 0

6 2

0

6 29

9

��

× ×xx

dx

xdx

=

1919

0

63

0

62

��

x dx

x dx

= 0

63

0

62

��

x dx

x dx =

x

x

4

0

6

3

0

6

4

3

4

3

14

6

13

6

�

��

��

�

��

��

=×

×

= 14

31

× × 6 = 4.5. Ans.


y = ��y dA

dA

*

where y* = distance of C.G. of area dA from x-axis

= y2

(here)

dA = ydx

∴ � y*dA = y2� × dA =

0

6

2�y

× ydx = 0

6 2

2�y

dx

= 12

0

6

� y2dx = 12

0

6

� x2 2

9

�

�

� dx � y

x=�

�

� 2

9

= 12

0

6

� x4

81 dx =

12

× 1

81

0

6

� x4 dx = 12

× 1

81

x5

0

6

5

�

��

��

= 12

× 1

81 ×

65

6810

5 5

=

Also dA� = 0

6

� ydx = 0

6

� x2

9 dx =

19 3

3

0

6x�

��

�� =

19

63

627

3 3

× =

∴ y = ��y dA

dA

* =

6810627

27810

66

5

3

5

3= ×

= 1

30 × 62 = 36

3065

= . Ans.

Problems of Finding Centroid or Centre of Gravity of Line-segment byIntegration Method

Problem 2. Determine the centre of gravity of a quadrant AB of the arc of acircle of radius R as shown in Fig. 3.10.


NOTES

Engineering Mechanics Sol. The centre of gravity of the line AB,which is an arc of a circle radius R, is obtainedby dividing the curved line AB into a large numberof elements of length dL as shown in Fig. 3.10.

The equation of curve AB is the equation ofcircle of radius R.

∴ The equation of curve AB is given byx2 + y2 = R2

Differentiating the above equation,2x dx + 2y dy = 0 [� R is constant]

or 2y dy = – 2x dx

or dy = − = −22x dxy

x dxy

...(i)

Consider an element of length dL as shown in Fig. 3.10. The C.G. of the lengthdL is at a distance x* from y-axis and y* from x-axis.

Now using equation (3.6) for y , we get

y = ��y dL

dL*

...(ii)

Let us express dL in terms of dx and dy.

But dL = dx dy2 2+

= dxx dxy

22

+ −�

��

� From ( ),i dyx dxy

=−�

��

= dxxy

dx22

22+ = dx 1

2

2+ xy

= dx y x

y

2 2

2+

= dx Ry

2

2 (� x2 + y2 = R2)

= Ry

. dx.

Substituting the value of dL in equation (ii),

y = y

Ry

dx

dL

* .×��

= y

Ry

dx

dL

×��

.(� y* = y)

= R dx

dL

��

= R dx

dL

R

0��

=

R x

R

R��

�

��

0

24π

(� ∫ dL is total length of arc of one quadrant of a circle)

= R RR

R×=

24

2π π

. Ans.

Fig. 3.10

dx

dydL

Y

R

y∗

B

Ox∗

R

A X


NOTES


Similarly, the value of x can be calculated.

Due to symmetry this value will also be equal to 2Rπ

.

∴ x = y = 2Rπ

. Ans.

Alternate MethodHere dL = R dθ

y* = R sin θ x* = R cos θ

Now y = y dL

dL

*��

= 0

2

0

2

π

π

θ θ

θ

/

/

( sin ) ( )��

×R R d

R d

= 0

22

0

2

π

π

θ θ

θ

/

/

sin��

R d

R d =

R d

R d

2

0

2

0

2

π

π

θ θ

θ

/

/

sin��

=

R

R

2

0

2

0

2

−�

��

�

��

��

cos/

/

θ

θ

π

π =

− ��

−�

��

�

��

−��

��

R cos cosπ

π2

0

20

= − −R [0 ]1

2π

= 2Rπ

. Ans.

Similarly,

x = x dL

dL

*��

= 0

2

0

2

π

π

θ θ

θ

/

/

( sin ) ( )��

×R R d

R d =

R d

R d

2

0

2

0

2

π

π

θ θ

θ

/

/

cos��

= R

R

sin/

/

θ

θ

π

π

��

��

��

0

2

0

2 =

R [sin sin ]90 0

20

° − °

−�

��

π

= Rπ2��

= 2Rπ

. Ans.

��

��

The centroid of structural sections like T-section, I-section, L-section etc. areobtained by splitting them into rectangular components. Then equations (3.1) and(3.2) are used.

Fig. 3.11

dL

Y

R

B

O

R

A X

y∗

x∗

x∗

R

θ

dθ


NOTES

Engineering Mechanics Problem 3. Find the centre of gravity of the T-sectionshown in Fig. 3.12.

Sol. The given T-section is split up into two rectanglesABCD and EFGH as shown in Fig. 3.13. The given T-sectionis symmetrical about Y-Y axis. Hence the C.G. of the sectionwill lie on this axis. The lowest line of the figure is line GF.Hence the moments of the areas are taken about this line GF,which is the axis of reference in this case.

Let y = The distance of the C.G. of the T-section fromthe bottom line GF (which is axis of refer-ence)

a1 = Area of rectangle ABCD = 12 × 3 = 36 cm2

y1 = Distance of C.G. of area a1 from bottom lineGF

= 10 + 32

= 11.5 cm

a2 = Area of rectangle EFGH = 10 × 3 = 30 cm2

y2 = Distance of C.G. of area a2 from bottom lineGF

=102

= 5 cm.

Using equation (3.2), we have

y = a y a y

Aa y a y

a a1 1 2 2 1 1 2 2

1 2

+=

++

(� A = a1 + a2)

= 36 11.5 30 5

36 30414 150

66× + ×

+= +

= 8.545 cm. Ans.Problem 4. Find the centre of gravity of the I-section shown in Fig. 3.14 (a).Sol. The I-section is split up into three rectangles ABCD, EFGH and JKLM

as shown in Fig. 3.14 (b). The given I-section is symmetrical about Y-Y axis. Hencethe C.G. of the section will lie on this axis. The lowest line of the figure line is ML.Hence the moment of areas are taken about this line, which is the axis of reference.

Let y = Distance of the C.G. of the I-section from the bottom line ML


2 cm

15 cm

2 cm

2 cm

10 cm

20 cm

2 cm

15 cm

2 cm

2 cm

10 cmA B

D CEH

Y1

3

2

J G F K

LM20 cm

(a) (b)

Fig. 3.14

12 cm

3 cm

3 cm

10 cm

Fig. 3.12

Fig. 3.13

12 cm

3 cm

10 cm

BA

D H E C

1

2

Y

G F3 cm

Y


NOTES


y1 = Distance of C.G. of rectangle ABCD from bottom line ML

= 2 + 15 + 22

= 18 cm


y2 = Distance of C.G. of rectangle EFGH from bottom line ML = 2 + 152

= 2 + 7.5 = 9.5 cma3 = Area of rectangle JKLM = 20 × 2 = 40 cm2

y3 = Distance of C.G. of rectangle JKLM from bottom line ML

= 22

= 1.0 cm

Now using equation (3.2), we have

y = a y a y a y

A1 1 2 2 3 3+ +

= a y a y a y

a a a1 1 2 2 3 3

1 2 3

+ ++ +

(� A = a1 + a2 + a3)

= 20 18 30 9 5 40 1

20 30 40× + × + ×

+ +.

= 360 285 40

9068590

+ + = = 7.611 cm. Ans.

Problem 5. Find the centre of gravity of the L-section shown in Fig. 3.15.

Sol. The given L-section is not symmetrical aboutany section. Hence in this case, there will be two axisof references. The lowest line of the figure (i.e., line GF)will be taken as axis of reference for calculating y . Andthe left line of the L-section (i.e., line AG) will be takenas axis of reference for calculating x .

The given L-section is split up into two rectanglesABCD and DEFG, as shown in Fig. 3.15.

To Find y

Let y = Distance of the C.G. of the L-section from bottom line GF


y1 = Distance of C.G. of rectangle ABCD from bottom line GF

= 2 + 102

= 2 + 5 = 7 cm

a2 = Area of rectangle DEFG = 8 × 2 = 16 cm2

y2 = Distance of C.G. of rectangle DEFG from bottom line GF

=22

= 1.0 cm.

Using equation (3.2), we have

y = a y a y

A1 1 2 2+

where A = a1 + a2

= a y a y

a a1 1 2 2

1 2

++

= 20 7 16 1

20 16× + ×

+ =

140 1636

+

= 15636

= 133

= 4.33 cm.

2 cm

10 cm

12 cm

A B2 cm

D C

8 cmG F

E

2

1

Fig. 3.15


NOTES

Engineering Mechanics To find x

Let x = Distance of the C.G. of the L-section from left line AG

x1 = Distance of the rectangle ABCD from left line AG = 22

= 1.0 cm

x2 = Distance of the rectangle DEFG from left line AG = 82

= 4.0 cm.


x = a x a x

A1 1 2 2+

where A = a1 + a2

= a x x y

a a1 1 2 2

1 2

++

= 20 1 16 4

20 16× + ×

+ (� a1 = 20 and a2 = 16)

= 20 64

36+

= 8436

= 73

= 2.33 cm.

Hence the C.G. of the L-section is at a distance of 4.33 cm from the bottomline GF and 2.33 cm from the left line AG. Ans.

Problem 6. Using the analytical method, determine the centre of gravity of theplane uniform lamina shown in Fig. 3.16.

Sol. Let y be the distance between C.G. of thelamina and the bottom line AB.

Area 1a1 = 10 × 5 = 50 cm2

y1 = 52

= 2.5 cm

Area 2

a2 = π2

× r2 = π2

× 2.52 = 9.82 cm2

y2 = 52

= 2.5 cm

Area 3

a2 = 5 5

2×

= 12.5 cm2

y2 = 5 + 53

= 6.67 cm.

Using the relation,

y = a y a y a y

a a a1 1 2 2 3 3

1 2 3

+ ++ +

= 50 2 5 9 82 2 5 12 5 6 67

50 9 82 12 5× + + + ×

+ +. . . . .

. . cm =

232 972 32

..

= 3.22 cm.

Similarly, let x be the distance between C.G. of the lamina and the left line CD.Area 1

a1 = 50 cm2

x1 = 2.5 + 102

= 7.5 cm

Fig. 3.16

5 cm

5 cm

2.5cm

2.5cm

5 cm

2.5cm

3

12

C

A B10 cm

12.5 cmD


NOTES


Area 2a2 = 9.82 cm2

x2 = 2.5 – 43

rπ

= 2.5 – 4 253.π

cm = 1.44 cm

Area 3a3 = 12.5 cm2

x3 = 2.5 + 5 + 2.5 = 10 cm.Now using the relation,

x = a x a x a x

a a a1 1 2 2 3 3

1 2 3

+ ++ +

= 50 7 5 9 82 144 12 5 10

50 9 82 12 5× + × + ×

+ +. . . .

. . cm

= 514 1472 32

..

= 7.11 cm.

Hence the C.G. of the uniform lamina is at a distance of 3.22 cm from thebottom line AB and 7.11 cm from the left line CD. Ans.

Problem 7. From a rectangular lamina ABCD 10 cm × 12 cm a rectangularhole of 3 cm × 4 cm is cut as shown in Fig. 3.17.

Find the C.G. of the remainder lamina.Sol. The section shown in Fig. 3.17, is having

a cut hole. The centre of gravity of a section with acut hole is determined by considering the mainsection first as a complete one, and then subtractingthe area of the cut-out hole, i.e., by taking the areaof the cut-out hole as negative.

Let y is the distance between the C.G. of thesection with a cut hole from the bottom line DC.

a1 = Area of rectangle ABCD

= 10 × 12 = 120 cm2

y1 = Distance of C.G. of the rectangleABCD from bottom line DC

=122

= 6 cm

a2 = Area of cut-out hole, i.e., rectangle EFGH,= 4 × 3 = 12 cm2

y2 = Distance of C.G. of cut-out hole from bottom line DC

= 2 + 42

= 2 + 2 = 4 cm.

Now using equation (3.2) and taking the area (a2) of the cut-out hole asnegative, we get

y = a y a y

A1 1 2 2−�

��

*

where A = a1 – a2

= a y a y

a a1 1 2 2

1 2

−−

(– ve sign is taken due to cut-out hole)

= 120 6 12 4

120 12720 48

108× − ×

−= −

= 6.22 cm.

Fig. 3.17

3cm

1cm

1cm

E F

H G

10 cm

12cm

A B

D C

4 cm

2 cm


NOTES

Engineering Mechanics To Find x

Let x = Distance between the C.G. of the section with a cut hole from the leftline AD

x1 = Distance of the C.G. of the rectangle ABCD from the left line AD

= 102

= 5 cm

x2 = Distance of the C.G. of the cut-out hole from the left line AD

= 5 + 1 + 32

= 7.5 cm.

Using equation (3.1) and taking area (a2) of the cut hole as negative, we get

x = a x a x

a a1 1 2 2

1 2

−−

(� A = a1 – a2)

= 120 5 12 7 5

120 12× − ×

−.

= 600 90

108510108

− = = 4.72 cm.

Hence the C.G. of the section with a cut hole will be at a distance of 6.22 cmfrom bottom line DC and 4.72 cm from the line AD. Ans.

��

Centroid of volume is the point at which the total volume of a body is assumedto be concentrated. The volume is having three dimensions i.e., length, width andthickness. Hence volume is measured in [length]3. The centroid [i.e., or centre ofgravity] of a volume is obtained by dividing the given volume into a large number ofsmall volumes as shown in Fig. 3.18. Similar method was used for finding the centroidof an area in which case the given area was divided into large number of small areas.The centroid of the volume is hence obtained by replacing dA by dv in equations (3.3)and (3.4).

Then these equations becomes as

x = x dv

dv

*��

...(3.7)

and y = y dv

dv

*��

...(3.8)

As volume is having three dimen-sions, hence third equation is written as

z = z dv

dv

*��

...(3.9)

where x* = Distance of C.G. of small volume dv from y-z plane (i.e., from axis OY)y* = Distance of C.G. of small volume dv from x-z plane (i.e., from axis OX)z* = Distance of C.G. of small volume dv from x-y plane

and x , y , z = Location of centroid of total volume.

Fig. 3.18

C.G.

dv V

O

Z

Y

Z*

y*x*

X

z

y

x


NOTES


Note. If a body has a plane of symmetry, the centre of gravity lies in that plane. If ithas two planes of symmetry, the line of intersection of the two planes gives the position ofcentre of gravity. If it has three planes of symmetry, the point of intersection of the threeplanes gives the position of centre of gravity.

Problem 8. A right circular cone of radius R at the base and of height h isplaced as shown in Fig. 3.19. Find the location of the centroid of the volume of thecone.

Sol. Given :

Radius or cone = R

Height of cone = h

In the Fig. 3.19, the axis ofthe cone is along x-axis. The centroidwill be at the x-axis. Hence y = 0and z = 0.

To find x , consider a smallvolume dv. For this, take a thincircular plate at a distance x from O.Let the thickness of the plate is dxas shown in figure and radius of the plate is r. The centroid of the plate is at a distance‘x’ from O. Hence x* = x.

Now volume of the thin plate,

dv = πr2 × dx ...(i)

Let us find the value of r in terms of x.

From similar triangles, we get

Rr

= hx

or r = R x

h×

Substituting the value of r in equation (i), we get

dv = π R x

h×�

��

2

dx ...(ii)

Now x is given by equation (3.7) as

x = x dv

dv

*��

= x dv

dv

��

[� Here x* = x]

= x

R xh

dx

R xh

dx

. π

π

�

�

×�

��

×�

��

2

2

� dvR x

hdx ii=

×�

��

�

��

�

��

π2

from equation ( )

Fig. 3.19

Y

X

Z

dxx

h

R

Or


NOTES


=

π

π

×

×

��

Rh

x dx

Rh

x dx

h

h

2

23

02

2 0

2

[� Limits of integration are w.r.t. x and x varies from 0 to h]

=

x

x

h4

3

0

4

3

�

�

� �

�

��

�

�

��

= 3h4

. Ans.

��

��

Consider a thick lamina of area A as shown inFig. 3.20.

Let x = Distance of the C.G. of area A fromthe axis OY.

y = Distance of the C.G. of area A fromthe axis OX.

Then moment of area about the axis OY= Area × perpendicular distance of C.G.

of area from axis OY= Ax ...(3.10)

Equation (3.10) is known as first moment of area about the axis OY. This firstmoment of area is used to determine the centroid of the area.

If the moment of area given by equation (3.10) is again multiplied by theperpendicular distance between the C.G. of the area and axis OY (i.e., distance x),then the quantity (Ax). x = Ax2 is known as moment of the moment of area or secondmoment of area or area moment of inertia about the axis OY. This second momentof area is used in the study of mechanics of fluids and mechanics of solids.

Similarly, the moment of area (or first moment of area) about the axis OX =Ay.

And second moment of area (or area moment of inertia) about the axis OX= (Ay) . y = Ay2.

If, instead of area, the mass (m) of the body is taken into consideration thenthe second moment is known as second moment of mass. This second moment of massis also known as mass moment of inertia.

Hence moment of inertia when mass is taken into consideration aboutthe axis OY = mx2 and about the axis OX = my2.

Hence the product of the area (or mass) and the square of the distance of thecentre of gravity of the area (or mass) from an axis is known as moment of inertia ofthe area (or mass) about that axis. Moment of inertia is represented by I. Hencemoment of intertia about the axis OX is represented by IXX whereas about the axisOY by IYY.

Fig. 3.20

xy

C.G.

O X

Lamina ofarea A

Y


NOTES


The product of the area (or mass) and the square of the distance of the centreof gravity of the area (or mass) from an axis perpendicular to the plane of the area isknown as polar moment of inertia and is represented by J.

Consider a plane area which is split up into small areas a1, a2, a3, ... etc. Letthe C.G. of the small areas from a given axis be at a distance of r1, r2, r3, ... etc. asshown in Fig. 3.21.

Then the moment of inertia of the plane area about the given axis is given byI = a1r1

2 + a2r22 + a3r3

2 + ... ...(3.11)or I = Σar2. ...(3.12)

Radius of Gyration. Radius of gyration of a body (ora given lamina) about an axis is a distance such that its squaremultiplied by the area gives moment of inertia of the areaabout the given axis.

For the Fig. 3.21, the moment of inertia about the givenaxis is given by equation (3.11) as

I = a1r12 + a2r2

2 + a3r32 + ... ...(i)

Let the whole mass (or area) of the body is concentratedat a distance k from the axis of reference, then the moment ofinertia of the whole area about the given axis will be equal toAk2.

If Ak2 = I, then k is known as radius of gyration about the given axis.

∴ k = IA

. ...(3.13)

��! �� "��"��#��

Theorem of the perpendicular axis states thatif IXX and IYY be the moment of inertia of a planesection about two mutually perpendicular axis X-Xand Y-Y in the plane of the section, then the momentof inertia of the section IZZ about the axis Z-Z,perpendicular to the plane and passing through theintersection of X-X and Y-Y is given by

IZZ = IXX + IYY.The moment of inertia IZZ is also known as

polar moment of inertia.Proof. A plane section of area A and lying in plane x-y is shown in Fig. 3.22.

Let OX and OY be the two mutually perpendicular axes, and OZ be the perpendicularaxis. Consider a small area dA.

Let x = Distance of dA from the axis OY

y = Distance of dA from axis OXr = Distance of dA from axis OZ

Then r2 = x2 + y2.Now moment of inertia of dA about x-axis

= dA × (Distance of dA from x-axis)2

= dA × y2.

r1

r2

r3

Area a3Area a2

Area a1

Givenaxis

Fig. 3.21

Fig. 3.22

Z

X

xOr

y

YdA Plane

sectionof area A


NOTES

Engineering Mechanics ∴ Moment of inertia of total area A about x-axis, IXX = ΣdAy2.Similarly, moment of inertia of total area A about y-axis, IYY = ΣdAx2

and moment of inertia of total area A about z-axis, IZZ = ΣdAr2

= ΣdA [x2 + y2] (� r2 = x2 + y2)= ΣdA x2 + ΣdA y2

= IYY + IXX

or IZZ = IXX + IYY. ...(3.14)The above equation shows that the moment of inertia of an area about an axis

at origin normal to x, y plane is the sum of moments of inertia about the correspondingx and y-axis.

��$ ��"��#��

It states that if the moment of inertia of a plane areaabout an axis in the plane of area through the C.G. of theplane area be represented by IG, then the moment of theinertia of the given plane area about a parallel axis AB inthe plane of area at a distance h from the C.G. of the area isgiven by

IAB = IG + Ah2.where IAB = Moment of inertia of the given area about AB

IG = Moment of inertia of the given area about C.G. A = Area of the section h = Distance between the C.G. of the section and the axis AB.Proof. A lamina of plane area A is shown in Fig. 3.23.Let X-X = The axis in the plane of area A and passing through the C.G. of

the area.AB = The axis in the plane of area A and parallel to axis X-X. h = Distance between AB and X-X.

Consider a strip parallel to X-X axis at a distance y from the X-X axis.Let the area of the strip = dAMoment of inertia of area dA about X-X axis = dAy2.∴ Moment of inertia of the total area about X-X axis,

IXX or IG = ΣdAy2 ...(i)Moment of inertia of the area dA about AB

= dA(h + y)2

= dA[h2 + y2 + 2hy].∴ Moment of inertia of the total area A about AB,

IAB = ΣdA[h2 + y2 + 2hy]= ΣdAh2 + ΣdAy2 + ΣdA2hy.

As h or h2 is constant and hence they can be taken outside the summation sign.Hence the above equation becomes

IAB = h2ΣdA + ΣdAy2 + 2hΣdAy.

G

X X

Planearea A

h

y

A B

Fig. 3.23


NOTES


But ΣdA = A. Also from equation (i), ΣdAy2 = IG. Substituting these values inthe above equation, we get

IAB = h2. A + IG + 2hΣdAy. ...(ii)But dA . y represents the moment of area of strip about X-X axis. And ΣdAy

represents the moments of the total area about X-X axis. But the moments of thetotal area about X-X axis is equal to the product of total area (A) and the distance ofthe C.G. of the total area from X-X axis. As the distance of the C.G. of the total areafrom X-X axis is zero, hence ΣdAy will be equal to zero.

Substituting this value in equation (ii), we get IAB = h2. A + IG + 0

or IAB = IG + Ah2 ...(3.15)Thus if the moment of inertia of an area with respect to an axis in the plane

of area (and passing through the C.G. of the area) is known, the moment of inertiawith respect to any parallel axis in the plane may be determined by using the aboveequation.

��% ��

�� "��&�

��

��

The area moment of inertia of the following sections will be determinedby the method of integration :

1. Moment of inertia of a rectangular section,2. Moment of inertia of a circular section,3. Moment of inertia of a triangular section,4. Moment of inertia of a uniform thin rod.Moment of Inertia of a Rectangular Section1st Case. Moment of inertia of the rectangular

section about the X-X axis passing through the C.G.of the section.

Fig. 3.24 shows a rectangular section ABCD havingwidth = b and depth = d. Let X-X is the horizontal axispassing through the C.G. of the rectangular section. Wewant to determine the moment of inertia of the rectangularsection about X-X axis. The moment of inertia of the givensection about X-X axis is represented by IXX.

Consider a rectangular elementary strip of thicknessdy at a distance y from the X-X axis as shown in Fig. 3.24.

Area of the strip = b . dy.Moment of inertia of the area of the strip about X-X axis

= Area of strip × y2

= (b . dy) × y2 = by2dy.Moment of inertia of the whole section will be obtained by integrating the above

equation between the limits – d2

to d2

.

Fig. 3.24

A Bb

XXd

dy

y

d2

d2

D C


NOTES


∴ IXX = −� d

d

2

2 by2dy = b −� d

d

2

2 y2dy

(� b is constant and can be taken outside the integral sign)

= b y

d

d3

2

2

3

�

��

��

−

= b3

d b2 2

3 2��

− −�

��

�

��

�

��

= b3

d d8 8

3 3

− −�

�

� �

��

�

�� =

b3

d d8 8

3 3

+�

��

�

��

= b3

. 28 12

3 3d bd= . ...(3.16)

Similarly, the moment of inertia of the rectangularsection about Y-Y axis passing through the C.G. of thesection is given by

IYY = db3

12...(3.17)

Refer to Fig. 3.25Area of strip, dA = d × dx

M.O.I. of strip above Y-Y axis = dA × x2

= (d × dx) × x2 (� dA = d . dx)= d × x2 × dx

∴ IYY = −� b

b

2

2 d × x2 × dx = d x

b

b3

2

2

3

�

��

��

−

= d3

b b2 2

3 3��

− −�

��

�

��

�

��

= d3

b b8 8

3 3

+�

��

�

�� =

d b db3 4 12

3 3

. =

2nd Case. Moment of inertia of the rectangular section about a linepassing through the base.

Fig. 3.26 shows a rectangular section ABCD havingwidth = b and depth = d. We want to find the moment of inertiaof the rectangular section about the line CD, which is the baseof the rectangular section.

Consider a rectangular elementary strip of thickness dyat a distance y from the line CD as shown in Fig. 3.26.

Area of strip = b . dy.Moment of inertia of the area of strip about the line CD

= Area of strip . y2

= b . dy . y2 = by2 dy.Moment of inertia of the whole section about the line CD is obtained by

integrating the above equation between the limits o to d.

Fig. 3.25

b

d

A B

dy

D C

y

Fig. 3.26

b2

A Bb

d

DC

x dxb2

Y

Y


NOTES


∴ Moment of inertia of the whole section about the line CD

= 0

d

� by2dy = b 0

d

� y2dy

= b y

d3

03

�

��

�� =

bd3

3...(3.18)

3rd Case. Moment of inertia of a hollow rectangular section.Fig. 3.27 shows a hollow rectangular section in which ABCD is the main section

and EFGH is the cut-out section.The moment of inertia of the main section ABCD

about X-X axis is given by equation (3.14),

= bd12

where b = Width of main section d = Depth.The moment of inertia of the cut-out section

EFGH about X-X axis

= b d1 1

3

12where b1 = Width of the cut-out section and

d1 = Depth of the cut-out section.Then moment of inertia of hollow rectangular section about X-X axis,IXX = Moment of inertia of rectangle ABCD about X-X axis – moment of inertia

of rectangle EFGH about X-X axis

= bd b d3

1 13

12 12− .

Moment of Inertia of a Circular Section. Fig. 3.28 shows a circular sectionof radius R with O as centre. Consider an elementary circular ring of radius ‘r’ andthickness ‘dr’. Area of circular ring

= 2πr. dr.

In this case first find the moment of inertia of thecircular section about an axis passing through O andperpendicular to the plane of the paper. This moment ofinertia is also known as polar moment of inertia. Let this axisbe Z-Z. (Axis Z-Z is not shown in Fig. 3.28). Then from thetheorem of perpendicular axis, the moment of inertia aboutX-X axis or Y-Y axis is obtained.

Moment of inertia of the circular ring about an axispassing through O and perpendi-cular to the plane of thepaper

= (Area of ring) × (radius of ring from O)2

= (2πr . dr) . r2

= 2πr3dr ...(i)Moment of inertia of the whole circular section is obtained by integrating

equation (i) between the limit O to R.

Fig. 3.27

Fig. 3.28

A Bb

d

D C

X X

b1

E F

H G

d1

drR

X X

Y

Y

O

r


NOTES

Engineering Mechanics ∴ Moment of inertia of the whole section about an axis passing through Oand perpendicular to the plane of paper is given as

IZZ = 0

R

� 2πr3 dr = 2π 0

R

� r3dr

= 2π r

R4

04

�

��

�� = 2π

R R4 4

4 2=

π.

But R = D2

where D = Diameter of the circular section

∴ IZZ = π2

× D2

4�� =

πD4

32...(3.19)

But from the theorem of perpendicular axis given by equation (3.14), we haveIZZ = IXX + IYY.

But due to symmetry, IXX = IYY

∴ IXX = IYY = IZZ/2.0

= πD4

32 ×

12

= πD4

64...(3.20)

Moment of inertia of a hollow circular sectionFig. 3.29 shows a hollow circular section.Let D = Diameter of outer circle, and

d = Diameter of cut-out circle.Then from equation (3.20), the moment of inertia

of the outer circle about X-X axis = π

64 D4.

And moment of inertia of the cut-out circle aboutX-X axis

= π

64 d4.

∴ Moment of inertia of the hollow circular section, about X-X axis,IXX = Moment of inertia of outer circle – moment of inertia of cut-out circle

= π

64 D4 –

π64

d4 = π

64 [D4 – d4]

Similarly, IYY = π

64 [D4 – d4].

Moment of Inertia of a Triangular Section1st Case. Moment of inertia of a triangular

section about its base.Fig. 3.30 shows a triangular section ABC of

base width = b and height = h. Consider a small stripof thickness dy at a distance y from the vertex A.

Area of the strip = DE . dy ...(i)The distance DE in terms of y, b and h is

obtained from two similar triangles ADE and ABC as

DEBC

= yh

Fig. 3.29

X X

Y

Y

O

dD

Fig. 3.30

Ddy

y

hE

B Cb

A


NOTES

Properties of Surfacesand Solids∴ DE = BC .

yh

= b y

h.

(� BC = b)

Substituting this value of DE in equation (i), we get

Area of strip = byh

. dy.

Distance of the strip from the base = (h – y)∴ Moment of inertia of the strip about the base

= Area of strip × (Distance of strip from base)2

= byh

. dy . (h – y)2 = byh

(h – y)2 . dy.

The moment of inertia of the whole triangular section about the base (IBC) isobtained by integrating the above equation between the limits O to h.

∴ IBC = 0

h

� byh

[h – y]2 dy

= bh

0

h

� y(h – y)2 dy

(� b and h are constants and can be taken outside the integral sign)

= bh

0

h

� y (h2 + y2 – 2hy) dy

= bh

0

h

� (yh2 + y3 – 2hy2) dy

= bh

y h y hy

h2 2 4 3

02 4

22

+ −�

��

�

��

= bh

h h h h h2 2 4 3

2 42

3. .+ −

�

��

�

�� =

bh

h y h4 4 4

2 42

3+ −

�

��

�

��

= bh

. h4 6 3 8

12+ −�

�� = bh3.

112

= bh3

12...(3.21)

2nd Case. Moment of inertia of the triangularsection about an axis passing through the C.G. andparallel to the base.

Consider a triangular section of base = b andheight = h as shown in Fig. 3.31. Let X-X is the axispassing through the C.G. of the triangular section andparallel to the base.

The distance between the C.G. of the triangular

section and base AB = h3

.

Now from the theorem of parallel axis, given by equation (3.15), we haveMoment of inertia about

BC = Moment of inertia about C.G. + Area× (Distance between X-X and BC)2

Fig. 3.31

h

B Cb

A

h3

X X

C.G


NOTES


or IBC = IG + A × h3

2��

∴ IG = IBC – A × h3

2��

= bh3

12 –

b h×�

�� 2 .

h3

2��

� Ibh b h

BC = = ×�

�

� 3

12 2and Area

= bh3

12 –

bh3

18 =

bh3 3 236( )−

= bh3

36. ...(3.22)

Moment of Inertia of a Uniform Thin Rod

Consider a uniform thin rod AB of length L as shownin Fig. 3.32.

Let m = Mass per unit length of rod, andM = Total mass of the rod

= m × L ...(i)Suppose it is required to find the moment of inertia

of the rod about the axis Y-Y. Consider a strip of lengthdx at a distance x from the axis Y-Y.

Mass of the strip= Length of strip × Mass per unit length= dx . m or m . dx.

Moment of inertia of the strip about Y-Y axis= Mass of strip × x2

= (m . dx) . x2 = mx2dx.Moment of inertia of the whole rod (IYY) will be obtained by integrating the

above equation between the limits O to L.

∴ IYY = 0

L

� mx2dx = m 0

L

� x2 dx (� m is constant)

= m x

L3

03

�

��

�� =

mL3

3

= mL L. 2

3 =

mL2

3 [� m . L = M from equation (i)]

Moment of Inertia of Area under a Curve of Given Equation

Fig. 3.33 shows an area under a curve whose equation is parabolic and is givenby

x = ky2

in which y = b when x = aSuppose it is required to find the moment of inertia of this area about y-axis.

Consider a strip of thickness dx at a distance x from y-axis.The area of strip, dA = y dx ...(i)

dxx

L

A B

Y

Fig. 3.32


NOTES


Let us substitute the value of y in terms of x in the above equation. The equationof curve is x = ky2 ...(ii)

y

a

x

b

xO

y

dx

x = ky2

Fig. 3.33

First find the value of k.When y = b, x = a. Hence above equation becomes

a = kb2

or k = a

b2

Substituting the value of ‘k’ in equation (ii), we get

x = a

b2 . y2 or y2 = b xa

2

or y = b xa

2 1/2�

�

� =

b

a x ...(iii)

Substituting this value of y in equation (i), we get

dA = b

a . x . dx

The moment of inertia of elemental area (dA) about y-axis

= x2. dA = x2 . b

a . x dx

∴ Moment of inertia of the total area about y-axis is obtained by integratingthe above equation between the limits O to a.

(� x varies from O to a)

∴ IYY = 0

a

� x2 . b

a . x . dx = b

a .

0

a

� x5/2. dx

= b

a

xa7 2

07 2

/

/

�

��

�

�� =

27

. b

a . a7/2 =

27

ba2. Ans.

To find the moment of inertia of the given area about x-axis, the elementshown in Fig. 3.33 can be considered to be a rectangle of thickness dx. The momentof inertia of this element about x-axis is equal to the moment of inertia of the rectangleabout its base.


NOTES

Engineering Mechanics ∴ Moment of inertia of the element about x-axis

= dx y. 3

3 � it is where and

bdb dx d y

3

3= =

�

��

�

��

The moment of inertia of the given area about x-axis is obtained by integratingthe above equation between the limits O to a.

∴ IXX = 0

a

� dx y. 3

3 =

0

a

� y3

3 . dx

= 0

a

�

ba

x dx.�

��

�

��

3

3 � y

ba

x iiifrom equation ( )�

��

�

��

= b

a

3

3 23 / 0

a

� . x3/2 dx = b

a

3

3 23 / x

a5 2

05 2

/

/

�

��

�

��

= ba

3

3 23 . / . 25

. a5/2 = 2

15 b3 . a

= 2

15 ab3. Ans.

��

��

��

First of all, the location of the centroid of the given section is determined. Thenthe given section is splitted into rectangles or triangles. The moment of inertia ofthe rectangles is determined about its centroid. Then this moment of inertia istransferred about the axis passing through the centroid of the given section, usingtheorem of parallel axis.

Problem 9. Fig. 3.34 shows a T-section ofdimensions 10 × 10 × 2 cm. Determine the moment ofinertia of the section about the horizontal and verticalaxes, passing through the centre of gravity of the section.

Sol. First of all, find the location of centre ofgravity of the given T-section. The given section issymmetrical about the axis Y-Y and hence the C.G. ofthe section will lie on Y-Y axis. The given section is splitup into two rectangles ABCD and EFGH for calculatingthe C.G. of the section.

Let y = Distance of the C.G. of the section fromthe bottom line GF


y2 = Distance of C.G. of the area a1 from the bottom line GF

= 8 + 1 = 9 cma2 = Area of rectangle EFGH = 8 × 2 = 16 cm2

y2 = Distance of C.G. of rectangle EFGH from the bottom line GF

= 82

= 4 cm

Fig. 3.34

10 cm

2 cm

10 cm

BA

D H E C

1

2

G F2 cm


NOTES

Properties of Surfacesand SolidsUsing the relation, y =

a y a ya a

1 1 2 2

1 2

++

= 20 9 16 4

20 16180 64

36× + ×

+= +

= 24436

= 6.777 cm.

Hence the C.G. of the given section lies at a distance of 6.777 cm from GF.Now find the moment of inertia of the T-section.

Now, Let IG1= Moment of inertia of rectangle 1 about the horizontal axis and

passing through its C.G.

IG2= Moment of inertia of rectangle 2 about the horizontal axis and

passing through the C.G. of the rectangle 2

h1 = The distance between the C.G. of the given section and theC.G. of the rectangle 1

= y1 – y = 9.0 – 6.777 = 2.223 cm

h2 = The distance between the C.G. of the given section and the C.G.of the rectangle 2

= y – y2 = 6.777 – 4.0 = 2.777 cm.

Now IG1 =

10 22

3× = 6.667 cm4

IG2 =

2 812

3× = 85.333 cm4.

From the theorem of parallel axes, the moment of inertia of the rectangle 1

about the horizontal axis passing through the C.G. of the given section

= IG1 + a1h1

2 = 6.667 + 20 × (2.223)2

= 6.667 + 98.834 = 105.501 cm4.

Similarly, the moment of inertia of the rectangle 2 about the horizontal axispassing through the C.G. of the given section

= IG2 + a2 h2

2 = 85.333 + 16 × (2.777)2

= 85.333 + 123.387 = 208.72 cm4.∴ The moment of inertia of the given section about the horizontal axis passing

through the C.G. of the given section= 105.501 + 208.72 = 314.221 cm4. Ans.

The moment of inertia of the given section about the vertical axis passingthrough the C.G. of the given section

= 2 10

12

3× +

8 212

3×

= 166.67 + 5.33 = 172 cm4. Ans.Problem 10. Find the moment of inertia of the section shown in Fig. 3.35 about

the centroidal axis X-X perpendicular to the web.Sol. First of all find the location of centre of gravity of the given figure. The

given section is symmetrical about the axis Y-Y and hence the C.G. of the section


NOTES

Engineering Mechanics will lie on Y-Y axis. The given section is split up intothree rectangles ABCD, EFGH and JKLM. The centreof gravity of the section is obtained by using

y = a y a y a y

a a a1 1 2 2 3 3

1 2 3

+ ++ +

...(i)

where y = Distance of the C.G. of the section from

the bottom line MLa1 = Area of rectangle ABCD = 10 × 2 = 20 cm2

y1 = Distance of the C.G. of the rectangle ABCDfrom the bottom line ML

= 2 + 10 + 22

= 12 + 1 = 13 cm


y2 = Distance of the C.G. of rectangle EFGH from the bottom line ML

= 2 + 102

= 2 + 5 = 7 cm

a3 = Area of rectangle JKLM = 20 × 2 = 40 cm2

y3 = Distance of the C.G. of rectangle JKLM from the bottom line ML

= 22

= 1.0 cm.

Substituting the above values in equation (i), we get

y = 20 13 20 7 40 1

20 20 40× + × + ×

+ +

= 260 140 40

80+ +

= 44080

= 5.50 cm.

The C.G. of the given section lies at a distance of 5.50 cm from the bottom lineML. We want to find the moment of inertia of the given section about a horizontalaxis passing through the C.G. of the given section.

Let IG1= Moment of inertia of rectangle 1 about the horizontal axis passing

through its C.G.

IG2= Moment of inertia of rectangle 2 about the horizontal axis passing

through the C.G. of rectangle 2

IG3= Moment of inertia of rectangle 3 about the horizontal axis

passing through the C.G. of rectangle 3

h1 = The distance between the C.G. of the rectangle 1 and the C.G. ofthe given section

= y1 – y = 13.0 – 5.50 = 7.50 cm

h2 = The distance between the C.G. of rectangle 2 and the C.G. of thegiven section

= y2 – y = 7.0 – 5.50 = 1.50 cm

h3 = The distance between the C.G. of the rectangle 3 and the C.G. ofthe given section

= y – y3 = 5.50 – 1.0 = 4.5 cm

Fig. 3.35

2 cm

10 cm

2 cm

2 cm

10 cmA B

D CFE1

3

2

J H G K

LM20 cm


NOTES

Properties of Surfacesand SolidsNow, IG1

= 10 2

12

3× = 6.667 cm4

IG2 =

2 1012

3× = 166.667 cm4

IG3 =

20 212

3× = 13.333 cm4.

From the theorem of parallel axes, the moment of inertia of the rectangle 1

about the horizontal axis passing through the C.G. of the given section

= IG1 + a1h1

2 = 6.667 + 20 × (7.5)2

= 6.667 + 1125 = 1131.667 cm4.

Similarly, the moment of inertia of the rectangle 2 about the horizontal axispassing through the C.G. of the given section

= IG2 + a2h2

2 = 166.667 + 20 × 1.52

= 166.667 + 45 = 211.667 cm4.

And moment of inertia of the rectangle 3 about the horizontal axis, passingthrough the C.G. of the given section

= IG3 + a3h3

2 = 13.333 + 40 × 4.52

= 13.333 + 810 = 823.333 cm4

Now moment of inertia of the given section about the horizontal axis, passingthrough the C.G. of the given section

= Sum of the moment of inertia of the rectangles 1 , 2 and

3 about the horizontal axis, passing through the C.G. of

the given section= 1131.667 + 211.667 + 823.333 = 2166.667 cm4. Ans.

Problem 11. Determine the polar moment of inertia of I-section shown inFig. 3.36. (All dimensions are in mm).

Sol. Let us first find the location of C.G. of thegiven section. It is symmetrical about the vertical axis,hence C.G. lies on this section.

Now, A1 = Area of first rectangle = 80 × 12 = 960mm2

A2 = Area of second rectangle [(150 – 12 –10) × 12]

= 128 × 12 = 1536 mm2

A3 = Area of third rectangle = 120 × 10 =1200 mm2

y1 = Distance of C.G. of area A1 frombottom line

= 150 – 122

= 144 mm

y2 = Distance of C.G. of area A2 from bottom line

= 10 + 1282

= 74 mm

150150128128

8080

120120

12

12

10

2

1

3

Fig. 3.36


NOTES

Engineering Mechanicsy3 = Distance of C.G. of area A3 from bottom line =

102

= 5 cm.

y = Distance of C.G. of the given section from bottom line.

The C.G. of the section is obtained by using,

y = A y A y A y

A A A1 1 2 2 3 3

1 2 3

+ ++ +

= 960 144 1536 74 1200 5

960 1536 1200× + × + ×

+ +

= 138240 113664 6000

3696+ +

= 2579043696

= 69.779 ~ 69.78 cm

Location of centroidal axis is shown in Fig. 3.37.(i) Moment of inertia of the given section

about X-XM.O.I. of the rectangle 1 about centroid axis

X-X is given by,

IXX1 = (IG1)X + A1 × h1

2 where h1 = (y1 – y )

= 80 12

12

3× + 960(144 – 69.78)2

= 5.3 × 106 mm4

M.O.I. of rectangle 2 about centroid axis X-X isgiven by,

IXX2 = ( IG2)X + A2 × h2

2 where h2 = (y2 – y )

= 12 128

12

3× + 1536 × (74 – 69.78) = 2.12 × 106 mm4

and IXX3 = ( IG3)X + A3 × h3

2 where h3 = (y3 – y )

= 120 10

12

3× + 1200 × (5 – 69.78)2 = 5.04 × 106 mm4

∴ IXX = IXX1 + IXX2 + IXX3

= 5.3 × 106 + 2.12 × 106 + 5.04 × 106 mm4

= 12.46 × 106 mm4

(ii) Moment of inertia of the given section about Y-Y

IYY1 = (IG1)Y =

12 8012

3× = 521 × 103 mm4 = 0.521 × 106 mm4

IYY2 = ( IG2)Y =

128 1212

3× = 18.432 × 103 mm4 = 0.018432 × 106 mm4

IYY3 = ( IG3)Y =

10 12012

3× = 1.44 × 106 mm4

∴ IY-Y = IYY1 + IYY2 + IYY3

= 0.521 × 106 + 0.018432 × 106 + 1.44 × 106 mm4 = 1.979 × 106 mm4

∴ Polar moment of inertia (IZZ) is given by, IZZ = IXX + IYY

= 12.46 × 106 + 1.979 × 106 mm4

= 14.439 × 106 mm4. Ans.

8080

XX

69.78 mm69.78 mm

Y

Y

Fig. 3.37


NOTES


Problem 12. Find the moment of inertia of the areashown shaded in Fig 3.38, about edge AB.

Sol. Given :Radius of semi-circle, R = 10 cmWidth of rectangle, b = 20 cmDepth of rectangle, d = 25 cmMoment of inertia of the shaded portion about AB

= M.O.I. of rectangle ABCD about AB– M.O.I. of semi-circle on DC about AB

M.O.I. of rectangle ABCD about AB

= bd3

3[see equation (3.18)]

= 20 25

12

3× = 104.167 cm4

M.O.I. of semi-circle about DC

= 12

× [M.O.I. of a circle of radius 10 cm about a diameter]

= 12

× π

644d�

�� =

12

× π

64 × 204 = 3.925 cm4

Distance of C.G. of semi-circle from DC

= 43

4 103

rπ π

= × = 4.24 cm

Area of semi-circle, A = π πr2 2

2102

= × = 157.1 cm2

M.O.I. of semi-circle about a line through its C.G. parallel to CD

= M.O.I. of semi-circle about CD – Area× [Distance of C.G. of semi-circle from DC]2

= 3925 – 157.1 × 4.242

= 3925 – 2824.28 = 1100.72 cm4

Distance of C.G. of semi-circle from AB= 25 – 4.24 = 20.76 cm

M.O.I. of semi-circle about AB = 1100.72 + 157.1 × 20.762

= 1100.72 + 67706.58 = 68807.30 cm4

∴ M.O.I. of shaded portion about AB= 104.167 – 68807.30 = 35359.7 cm4. Ans.

Problem 13. Find the moments of inertia about the centroidal XX and YY axesof the section shown in Fig. 3.39.

Sol. First find the location of the C.G. of the given figure.Let a1 = Area of complete rectangle = B × D

a2 = Area of removed rectangle portion

= B2

× D2

= BD4

Fig. 3.38

A B

25 cm

20 cm

DR =

10 cmSemicircle

C


NOTES

Engineering Mechanicsx1 =

B2

, y1 = D2

and x2 = B2

+ 12

B2�� =

34B

,

y2 = D2

+ 12

D2�� =

34D

where (x1, y1) and (x2, y2) are the co-ordinates of theC.G. of the complete rectangle and cut out rectanglerespectively. Area a2 is negative.

Now x = a x a y

a a1 1 2 2

1 2

−−

= BD

B B D B

BD

× − × ×2 4

34

34

=

B DB D

BD

22

23

1634

× − =

516

34

2BD

BD =

512

× D

Similarly y = a y a y

a a1 1 2 2

1 2

−−

= BD

D BD D

BD

× − ×2 4

34

34

=

BDBD

BD

22

23

1634

− =

516

34

512

2BD

BD= × D

Now draw the centroidal axes XX and YY as shown in Fig. 3.40.

1

YB/2B/2 B/2B/2

D/12D/12D/2D/2

D/2D/2

X

X

Y

5B/125B/12

xx

5D/125D/12y =y =

X

Y

Fig. 3.40

B/2B/2

DD

BB

D/2D/2

2

1

Y

X

Fig. 3.39


NOTES


Let IXX1 = M.O.I. of complete rectangle 1 about centroidal axis X-X

= M.O.I. of complete rectangle 1 about horizontal axis passing through

its C.G. + Area of complete rectangle 1

× Distance between X-X axis and horizontal axis passing through the C.G. of rectangle 1 (By theorem of parallel axis)

[� IXX1 = IG xx1 + A1h1

2]

= BD3

12 + (B × D) [y1 – y ]2

= BD3

12 + BD

D D2

512

2

−��

��

� yD

yD

1 252

= =��

��

,

= BD3

12 + BD

D12

2��

= BD3

12 +

BD3

144 =

13144

BD3

Similarly IXX2 = (IG2X) + A2 × h22

=

B DBD2 2

12 4

3

× ��

+ × [y2 – y ]2 � ABD

h y y2 2 24= = −�

��

, ( )

= BD3

192 +

BD4

34

512

2D D−��

��

� yD

yD

232

512

= =��

��

,

= BD3

192 +

BD4

× 412

2D�

�� =

BD3

192 +

164 144

3BD×

= BD192

3

+ BD3

36 =

3 16576

3 3BD BD+ =

19576

3BD

Now IXX = M.O.I. of given section about centroidal axis X-X

= IXX1 – IXX2 = 13

144

3BD –

19576

3BD

= 52 19

576

3 3BD BD− =

33576

3BD = 0.0573 BD3. Ans.

Similarly the M.O.I. of the given section about centroidal axis Y-Y is given by IYY = IYY1 – IYY2

where IYY1 = M.O.I. of rectangle 1 about centroidal axis Y-Y= IG1y + A1 × [x1 – x ]2

= DB3

12 + BD ×

B B2

512

2

−��

��

= DB2

12 +

BD B× 2

144 =

13144

DB3

and IYY2 = IG2y + A2[x2 – x ]2

=

D B2 2

12

3

× ��

+ BD4

34

512

2B B−��

��

= DB DB DB3 3 3

192 3619

576+ =

∴ IYY = 13

144 DB3 –

19576

3DB =

33576

DB3 = 0.0573 DB3. Ans.


NOTES


Refer to Fig. 3.41.Let A = Area of a surface in x-y plane

dA = Small areax = Distance of centroid of small area

dA from axis OYy = Distance of centroid of small area

dA from axis OXr = Distance of centroid of small area

dA from axis passing through Oand perpendicular to the plane ofarea A i.e., from axis OZ.

Ixx = Second moment of total area A (i.e.,moment of inertia) about x-axis.

Iyy = Second moment of total area A (i.e., moment of inertia) about y-axis.Izz = Second moment of total area A about z-axis.

Now second moment of small area dA about x-axis = dA . y2 or y2 . dASecond moment of small area dA about y-axis. = dA . x2 or x2 . dASecond moment of small area dA about z-axis = dA . r2 or r2 . dA

But Ixx = y dA2 .�Iyy = x dA2 .�Izz = r dA2 .�

The second moment of area about z-axis is known as polar moment of inertia.∴ Polar moment of Inertia

= Izz = r dA2 .� or Izz = � r dA2

= � +( )x y dA2 2 (� r2 = x2 + y2)

=� �+x dA y dA2 2. .

= Iyy + Ixx� x dA I y dA Iyy xx

2 2= =�

�� and

= Ixx + Iyy ...(3.23)Hence the polar moment of inertia for an area with respect to an axis

perpendicular to the plane is equal to the sum of the moments of inertia about twomutually perpendicular axes in the plane of the area.

��

The Fig. 3.42 shows a body of area A. Consider a small area dA. The momentof this area about x-axis is y . dA. Now the moment of y . dA about y-axis is xydA.

Fig. 3.41

dA

O X

Y

Area, A

y

r

x


NOTES


Then xydA is known as the product of inertia ofarea dA with respect to x-axis and y-axis. The

integral xydA� is known as the product of inertia

of area A with respect to x and y axes. This productof inertia is represented by Ixy.

∴ Ixy = ∫ xydA ...(3.23)Hence the product of inertia of the plane

area is obtained if an elemental area is multipliedby the product of its co-ordinates and is integratedfor entire area.

The product of inertia (Ixy) can also bewritten mathematically as

Ixy = Σxi yi Ai = x1y1A1 + x2 y2A2 + ...... ...(3.24)

where xi yi = co-ordinates of the C.G. of area Ai.Note. (i) The product of inertia may be

positive, negative or zero depending upon distancex and y which could be positive, negative or zero.

(ii) If area is symmetrical with respect to oneor both of the axes, the product of inertia will bezero as shown in Fig. 3.43. The total area A issymmetrical about y-axis. The small area dA whichis symmetrical about y-axis has co-ordinates (x, y)and (– x, y). The corresponding product of inertiafor small area are xydA and – xydA respectively.Hence product of inertia for total area becomes zero.

(iii) The product of inertia with respect tocentroidal axis will also be zero.

Problem 14. Fig. 3.44 shows a plane area. Determine the product moment ofinertia of the given area. All dimensions are in mm.

Y

X40 3070

90C.G1

C.G2(20, 45)

(50, 30)

2

1

O

(b)

70

40

90

(a)

Fig. 3.44

Sol. Divide the given area into two parts. The first part is a rectangle andsecond part is a right angled triangle. Take x-axis and y-axis as shown in the Fig.3.44. The areas and location of their C.G. are given below :

Area of rectangle, A1 = 90 × 40 = 3600 mm2.

Fig. 3.42

dA

x

O X

Y

y

dA dAx x

Y

X

Fig. 3.43


NOTES

Engineering Mechanics The co-ordinates of C.G. of rectangle 1 are : x1 = 20 mm, y1 = 45 mm.

Area of triangle,

A2 = 90 30

2×

= 1350 mm2.

The co-ordinates of C.G. of triangle 2 are :

x2 = 40 + 13

× 30 = 40 + 10 = 50 mm ; y2 = 13

× 90 = 30 mm.

The product of inertia of given area is given by equation (3.24), as Ixy = x1y1A1 + x2y2A2

= A1x1y1 + A2 x2 y2

= 3600 × 20 × 45 + 1350 × 50 × 30= 3240000 + 2025000 = 5265000 mm4. Ans.

��

The principal axes are the axes about which the product of inertia is zero.The product of inertia (Ixy) of plane area A with respect to x and y axes is given

by equation (3.23), as

Ixy = xy dA�But the moment of inertia of plane area A about x-axis [Ixx] or about y-axis

[Iyy] is given by

Ixx = y dA2� and Iyy = x dA2�The moment of inertia is always positive but product of inertia may be positive

(if both x and y are positive), may be negative (if one co-ordinate is positive and otheris negative) or may be zero (if any co-ordinate is zero).

Fig. 3.45 (a) shows a body of area A. Consider a small area dA. The product ofinertia of the total area A with respect to x and y-axes is given as

Ixy = xydA� ...(i)

y x′

x

y′

y′

y x1

y1x

dA dA

Total area A

OO

[Here x is + ve, but y is –ve]′ ′

(a) (b)

Fig. 3.45


NOTES


Let now the axes are rotated anticlockwise by 90° as shown in Fig. 3.45 (b)keeping the total area A in the same position. Let x1 and y1 are the new axes. Theco-ordinates of the same small area dA with respect to new axes are x′ and y′.

Hence the product of inertia of the total area A with respect to new axes x1and y1 becomes as

Ix y1 1 = x y dA′ ′� ...(ii)

Now let us find the relation between old and new co-ordinates. From Figs. 3.45(a) and 3.45 (b), we get

x = – y′ and y = x′or y′ = – x and x′ = y

Substituting the values of x′ and y′ in equation (ii), we get

Ix y1 1 = ( )( )y x dA xydA− = − �� = Ixy

� xydA Ixy� =�

��

The above result shows that by rotating the axes through 90°, the product ofinertia has become negative. This means that the product of inertia which was positivepreviously has now become negative by rotating the axes through 90°. Hence productof inertia has changed its sign. It is also possible that by rotating the axes throughcertain angle, the product of inertia will become zero. The new axes about whichproduct of inertia is zero, are known as principal axes.

Note. (i) The product of inertia is zero about principal axes.

(ii) As the product of inertia is zero about symmetrical axis, hence symmetrical axis isthe principal axis of inertia for the area.

(iii) The product of inertia depends upon the orientation of the axes.

��

Fig. 3.46 (a) shows a body of area A with respect to old axes (x, y) and newaxes (x1, y1). The new axes x1 and y1 have been rotated through an angle θ inanticlockwise direction. Consider a small area dA. The co-ordinates of the small areawith respect to old axes is (x, y) whereas with respect to new axes, the co-ordinatesare x′ and y′. The new co-ordinates (x′, y′) are expressed in terms of old co-ordinates(x, y) and angle θ as [Refer to Figs. 3.46 (b) and 3.46 (c)]

x′ = y sin θ + x cos θ ...(i)and y′ = y cos θ – x sin θ ...(ii)

The moment of inertia and product of inertia of area A with respect to old axesare

Ixx = y dA2 ,� Iyy = x dA2� and Ixy = xydA� ...(3.25)

Also the moment of inertia and product of inertia of area A with respect tonew axes will be

I y dAx x1 1

2= ′� ( ) , I x dAy y1 1

2= ′� ( ) and I x y dAx x1 1= ′ ′�


NOTES

Engineering Mechanicsy1

y

xO

θ

θ

θ

x′

x

y′ x1

Total area A

Small area dA

(a)

y

y

xO

θ

θx

y′ X1

y sinθ

(x sin)θ

y cosθ

x′

x cos θ

y sinθ

dA(x, y)

x cos θ(x sin )θ

(c)

y cosθ

θ

Y1

X1

y

y sin θ

y

xO

(b)

Fig. 3.46

Let us substitute the values of x′, y′ from equation (i) and (ii) in the aboveequations, we get

Ix x1 1 = ( )y dA′� 2

= ( cos sin )y x dAθ θ−� 2 [� y′ = y cos θ – x sin θ]

= ( cos sin cos sin )y x xy dA2 2 2 2 2θ θ θ θ+ −�= y dA x dA xy dA2 2 2 2 2cos sin cos sinθ θ θ θ+ −� ��= cos sin cos sin2 2 2 2 2θ θ θ θy dA x dA xy dA+ −� ��

(� After rotation, the angle θ is constant and hencecos2 θ, sin2 θ and 2 cos θ sin θ are constant)

= (cos2 θ)Ixx + (sin2 θ)Iyy – (2 cos θ sin θ)Ixy ...(3.26)

� y dA I x dA I xydA Ixx yy xy2 2= = =�

�� , and

Similarly, Iy y1 1 = ( )x dA′� 2

= ( sin cos )y x dAθ θ+� 2 [� x′ = y sin θ + x cos θ]

= ( sin cos sin cos )y x xy dA2 2 2 2 2θ θ θ θ+ +�


NOTES

Properties of Surfacesand Solids= y dA x dA xy dA2 2 2 2 2sin cos sin cosθ θ θ θ� � �+ +

= sin cos sin cos2 2 2 2 2θ θ θ θy dA x dA xydA� � �+ +

(� θ is constant and hence sin θ and cos θ are constants)

= sin2 θ . Ixx + cos2 θ Iyy + 2 sin θ cos θ Ixy ...(3.27)

� y dA I x dA I xydA Ixx yy xy2 2� � �= = =�

��, and

Adding equations (3.26) and (3.27), we get

I Ix x y y1 1 1 1+ = Ixx [sin2 θ + cos2 θ] + Iyy [sin2 θ + cos2 θ]

+ 2 sin θ cos θ Ixy – 2 sin θ cos θ Ixy

= Ixx + Iyy [� sin2 θ + cos2 θ = 1] ...(3.28)

The equation (3.28) shows that sum of moments of inertia about old axes (x, y)and new axes (x1, y1) are same. Hence the sum of moments of inertia of area A is

independent of orientation of axes. Now let us find the value of I Ix x y y1 1 1 1− .

Subtracting equation (3.27) from equation (3.26), we get

I Ix x y y1 1 1 1− = cos2 θ Ixx + sin2 θ Iyy – 2 cos θ sin θ Ixy

– [sin2 θ Ixx + cos2 θ Iyy + 2 cos θ sin θ Ixy]

= Ixx (cos2 θ – sin2 θ) + Iyy (sin2 θ – cos2 θ) – 4 cos θ sin θ . Ixy

= Ixx (cos2 θ – sin2 θ) – Iyy (cos2 θ – sin2 θ) – 4 cos θ sin θ Ixy

= (Ixx – Iyy) (cos2 θ – sin2 θ) – 2 × 2 cos θ sin θ × Ixy

= (Ixx – Iyy) cos 2θ – 2Ixy sin2 θ ...(3.29)

� coscos

, sincos2 21 2

21 2

2θ θ θ θ= +�

��= −

∴ cos2 θ – sin2 θ = cos 2θ and 2 sin θ cos θ = sin 2θ)

Now let us find the values of Ix x1 1 and Iy y1 1

in terms of Ixx, Iyy and θ.

Adding equations (3.28) and (3.29), we get

21 1

Ix x = [Ixx + Iyy] + [(Ixx – Iyy) cos 2θ – 2Ixy sin 2θ]

or Ix x1 1 =

( ) ( )I I I Ixx yy xx yy++

−2 2

cos 2θ – Ixy sin 2θ ...(3.30)

To find the values of Iy y1 1, subtract equation (3.29) from (3.28). Now subtracting

equation (3.29) from equation (3.28), we get

21 1

Iy y = (Ixx + Iyy) – [(Ixx – Iyy) cos 2θ – 2Ixy sin 2θ]

∴ Iy y1 1 =

( ) ( )I I I Ixx yy xx yy+−

−2 2

cos 2θ + Ixy sin 2θ ...(3.31)

Product of inertia about new axes

Let us now find the value of Ix y1 1 in terms of Ixy and angle θ.

We know that Ix y1 1 = ( )( )x y dA′ ′�


NOTES

Engineering Mechanics Substituting the values of x′ and y′, we get

Ix y1 1 = ( sin cos )( cos sin )y x y x dAθ θ θ θ+ −�

(� x′ = y sin θ + x cos θ and y′ = y cos θ – x sin θ)

or Ix y1 1 = ( sin cos sin cos cos sin )y xy xy x dA2 2 2 2θ θ θ θ θ θ− + −�= y dA xy dA xy dA x dA2 2 2 2sin cos sin cos cos sinθ θ θ θ θ θ� � ��− + − −

= sin cos sin cos cos sinθ θ θ θ θy dA xydA xydA x dA2 2 2 2� � ��− +

(� θ is constant and hence sin θ, cos θ are constants)

Ix y1 1 =

22

22

2 2 2 2sin cossin cos

cos sinθ θ θ θ θ θy dA I I x dAxy xy� �− + −

� xydA Ixy� =��

��

= sin 2

2θ

. Ixx + Ixy (cos2 θ – sin2 θ) – sin 2

2θ

Iyy

� y dA I x dA Ixx yy2 2� �= =�

��,

= Ixx

2 sin 2θ + Ixy (cos2 θ – sin2 θ) –

Iy

2 . sin θ

= ( )I Ixx yy−

2 sin 2θ + Ixy (cos2 θ – sin2 θ)

= ( )I Ixx yy−

2 sin 2θ + Ixy cos 2θ ...(3.32)

(� cos2 θ – sin2 θ = cos 2θ)

Direction of principal axes

We have already defined the principal axes. Principal axes are the axes aboutwhich the product of inertia is zero. Now the new axes (x1, y1) will become principal

axes if the product of inertia given by equation (3.32) is zero (i.e. Ix y1 1 = 0).

∴ For principal axes, Ix y1 1 = 0

or( )I Ixx yy−

2 sin 2θ + Ixy cos 2θ = 0

or I Ixx yy−

2 sin 2θ = – Ixy cos 2θ

or sincos

22

2 2θθ

=−

−=

−I

I I

I

I Ixy

xx yy

xy

yy xx

or tan 2θ = 2I

I Ixy

yy xx− ...(3.33)

The above equation will give the two values of 2θ or θ. These two values of θwill differ by 90°. By substituting the values of θ in equations (3.30) and (3.31), the

values of principal moments of inertia ( Ix x1 1 and Iy y1 1

) can be obtained. If from


NOTES


equation (3.33), the values of sin 2θ and cos 2θ in terms of Ixy, Ixx and Iyy are substi-tuted in equation (3.30), we get

Ix x1 1 =

I I I IIxx yy xx yy

xy

+±

−+

2 2

2

2

( ).

These are the values of principal moment of inertia.Problem 15. For the section shown in Fig. 4.41 (a) determine :(i) Moment of inertia about its centroid along (x, y) axis.

(ii) Moment of inertia about new axes which is turned through an angle of 30°anticlockwise to the old axis.

(iii) Principal moments of inertia about its centroid.All dimensions are in cm.Sol. Given :The Fig. 3.47 (a) shows the given section. It is symmetrical about x-axis. The

C.G. of the section lies at O (origin of the axes). To find moment of inertia of thegiven section, it is divided into three rectangles as shown in Fig. 3.47 (b). First themoment of inertia of each rectangle about its centroid is calculated. Then by usingparallel axis theorem, the moment of inertia of the given section about its centroid isobtained.

40

40 40

30

30

10

10

C.G.

y

O x

(a)

20

3030

30

30

10

10

C.G.2

C.G.1

C.G.3 10

1

2

3

(b)

Fig. 3.47

(a) Consider rectangle 1

The C.G. of rectangle 1 is at a distance of 20 cm from x-axis and at a distanceof 25 cm from y-axis.

(Ixx)1 = (IG)1x + A1(k1x)2 ...(i)

where (Ixx)1 = M.O.I. of rectangle 1 about x-axis passing through the centroid ofthe given section.

(IG)1x = M.O.I. of rectangle 1 about an axis passing through C.G. of rectangle1 and parallel to x-axis

= bd3

12

= 10 30

12

3× (Here b = 10 and d = 30)

= 2.25 × 104 cm4

A1 = Area of rectangle 1 = 10 × 30 = 300


NOTES

Engineering Mechanics (k1x) = Distance of C.G. of rectangle 1 from x-axis= 20

Substituting the above values in equation (i), we get (Ixx)1 = 2.25 × 104 + 300 × 202

= 2.25 × 104 + 12 × 104

= 14.25 × 104 cm4 ...(A)

Similarly the M.O.I. of rectangle 1 about y-axis passing through the centroidof the given figure is given by,

(Iyy)1 = (IG)1y + A1(k1y)2 ...(ii)

where (IG)1y = bd3

12 =

30 1012

3× = 0.25 × 104 cm4

(k1y) = Distance of C.G. of rectangle 1 from y-axis = 25∴ (Iyy)1 = 0.25 × 104 + 300 × 25 (� A1 = 300)

= 0.25 × 104 + 18.75 × 104

= 19 × 104 cm4 ...(B)

(b) Consider rectangle 2

The C.G. of this rectangle coincides with the C.G. of the given section. Hence

(Ixx)2 = bd3 3

1260 10

12= ×

= 0.5 × 104 cm4 ...(C)

and (Iyy)2 = 10 60

12

3× = 18 × 104 cm4 ...(D)

(c) Consider rectangle 3

The C.G. of rectangle 3 is at a distance of 20 cm from x-axis and at a distanceof 25 cm from y-axis. Hence k3x = 20 cm and k3y = 25 cm.

Now (Ixx)3 = (IG)3x + A3(k3x)2

= 10 30

12

3× + (10 × 30)(20)2

= 2.25 × 104 + 12 × 104 = 14.25 × 104 cm4

and (Iyy)3 = (IG)3y + A3(k3 y)2

= 30 10

12

3× + 300 × 252

= 0.25 × 104 + 18.75 × 104 = 19 × 104 cm4.(i) Moment of inertia of complete section about its centroid

Ixx = (Ixx)1 + (Ixx)2 + (Ixx)3

= 14.25 × 104 + 0.5 × 104 + 14.25 × 104 cm4

= 29 × 104 cm4. Ans.and Iyy = (Iyy)1 + (Iyy)2 + (Iyy)3

= 19 × 104 + 18 × 104 + 19 × 104

= 56 × 104 cm4. Ans.(ii) Moment of inertia of complete section about new axes which is turned through

an angle of 30° anticlockwise.


NOTES


Here θ = 30°.Let us first calculate the product of

inertia of whole area about old axes x, y.

(a) Consider rectangle 1 ,

A1 = 10 × 30 = 300.

The C.G. of rectangle 1 is at a distanceof 20 cm above x-axis and at a distance of 25 cmfrom y-axis. Hence co-ordinates of this C.G. are

x1 = – 25 cm and y1 = 20 cm.

(b) For rectangle 2 ,

A2 = 10 × 60 = 600 cm2. The C.G. of rectangle 2 lies on the origin (O). Hencex2 = 0 and y2 = 0.

(c) For rectangle 3 , A3 = 10 × 30 = 300 cm2

The C.G. of rectangle 3 is at a distance of 20 cm below x-axis and at a distanceof 25 cm from y-axis.

Hence co-ordinate of this C.G. are : x3 = 25 cm and y3 = (– 20 cm).The product of inertia (Ixy) of the whole figure is given by equation (3.24) as

Ixy = A1x1y1 + A2x2 y2 + A3x3 y3

= 300 × (– 25) × 20 + 600 × 0 × 0 + 300 × 25 × (– 20)= – 15 × 104 + 0 + (– 15 × 104)= – 30 × 104 cm4

Now the moment of inertia of the complete section about the new axes (x1, y1)can be obtained from equations (3.30) and (3.31) as

Ix x1 1 =

( )I Ixx yy+2

+ ( )I Ixx yy−

2 cos 2θ – Ixy sin 2θ

where Ixx = 29 × 104 cm4, Iyy = 56 × 104 cm4, Ixy = – 30 × 104 cm4 and θ = 30°

∴ Ix x1 1 =

29 10 56 102

4 4× + ×

+ 29 10 56 10

2

4 4× − × cos 60°

– (– 30 × 104) sin 60°

= 42.5 × 104 – 13.5 × 10 × 12

+30 × 104 × 0.866

= 35.75 × 104 + 26 × 104 = 61.75× 104 cm4. Ans.

and Iy y1 1 =

I I I Ixx yy xx yy+−

−2 2

cos θ + Ixy sin 2θ

Fig. 3.47 (c)

20

20C.G.2

C.G.3

y

C.G.1

C.G.21

2

3

25

25

x

y1y

θ

θ

x1

x

Fig. 3.47 (d)


NOTES


29 10 56 102

4 4× + × –

29 10 56 102

4 4× − × cos 60°

+ (– 30 × 104) sin 60°= 42.5 × 104 + 6.75 × 104 – 26 × 104 = 23.25 × 104 cm4. Ans.

(iii) Principal moments of Inertia about the centroidThe principal moments of inertia are the moments of inertia about the principal

axes.The direction of principal axes is given by

equation (3.33) as

tan 2θ = 2I

I Ixy

yy xx−

= 2 30 10

56 10 29 10

4

4 4× − ×

× − ×( )

= − ×

×60 10

27 10

4

4 = – 2.222

As 2θ is negative, hence it lies in 2nd and4th quadrant.

∴ 2θ = tan–1 (– 2.222)= – 65.77° and 114.23°

or θ = – 32.88° and 57.12°The + ve angle is taken anti-clock and – ve angle is taken clockwise to the

existing axes x and y. The principal axes are shown as x1 and y1 in Fig. 3.47 (e). Themoment of inertia along these axes is the principal moment of inertia. Hence bysubstituting θ = – 32.88° and 57.12°, in equations (4.24) and (4.25), we get principalmoment of inertia.

∴ ( )Ix x min.max.

1 1 =

I I I Ixx yy xx yy++

−2 2

cos 2θ – Ixy sin 2θ

= 29 10 56 10

2

4 4× + × +

29 10 56 102

4 4× − ×

× cos (– 2 × 32.88) – (– 30 × 104) sin (– 2 × 32.88)

[� θ = – 32.88°]

= 42.5 × 104 – 13.5 × 104 × 0.41 + 30 × 104 × (– 0.912)

= 42.5 × 104 – 5.535 × 104 – 27.36 × 104

= 9.605 × 104 cm4

and ( )Iy y min.max.

1 1 =

I Ixx yy+2

– I Ixx yy−

2 cos 2θ + Iyy sin 2θ

= 42.5 × 104 + 5.535 × 104 + 27.36 × 104 = 75.395 × 104 cm4

Hence principal moment of inertia are

Imax. = 75.395 × 104 cm4. Ans.

Imin. = 9.605 × 104 cm4. Ans.

y1

y

x1

x

57.12°

32.88°

Fig. 3.47 (e)


NOTES


Alternate Method

The principal moments of inertia can also be obtained by

Imax.min.

= I I I I

Ixx yy xx yyxy

+±

−+

2 2

22( )

= 29 10 56 10

229 10 56 10

230 10

4 4 4 4 24 2× + × ± × − × + − ×( )

( )

= 42.5 × 104 ± ( . ) ( )− × + − ×13 5 10 30 104 2 4 2

= 42.5 × 104 ± 104 × 32.89

= (42.5 + 32.89) × 104 and (42.5 – 32.89) × 104

= 75.39 × 104 and 9.61 × 104 cm4

��

Consider a body of mass M as shown in Fig. 3.48.

Centroid

Body ofMass M

Y

XO x

y

Fig. 3.48

Let x = Distance of the centre of gravity of mass M from axis OY

y = Distances of the C.G. of mass M from axis OX

Then moment of the mass about the axis OY = M . x

The above equation is known as first moment of mass about the axis OY.

If the moment of mass given by the above equation is again multiplied by theperpendicular distance between the C.G. of the mass and axis OY, then the quantity(M . x) . x = M . x2 is known as second moment of mass about the axis OY. This secondmoment of the mass (i.e., quantity M . x2) is known as mass moment of inertia aboutthe axis OY.

Similarly, the second moment of mass or mass moment of inertia about theaxis OX

= (M . y) . y = M . y2

Hence the product of the mass and the square of the distance of the centre ofgravity of the mass from an axis is known as the mass moment of inertia about thataxis. Mass moment of inertia is represented by Im. Hence mass moment of inertiaabout the axis OX is represented by (Im)xx whereas about the axis OY by (Im)yy.

Consider a body which is split up into small masses m1, m2, m3 ...... etc. Letthe C.G. of the small areas from a given axis be at a distance of r1, r2, r3 ...... etc. as


NOTES

Engineering Mechanics shown in Fig. 3.49. Then mass moment ofinertia of the body about the given axis is givenby

Im = m1r12 + m2r2

2 + m3r32 + ......

= Σmr2

If small masses are large in number thenthe summation in the above equation can bereplaced by integration. Let the small massesare replaced by dm instead of ‘m’, then theabove equation can be written as

Im = r dm2� ...(3.34)

��

��

��

The mass moment of inertia of the following bodies will be determined by themethod of integration :

1. Mass moment of inertia of a rectangular plate,

2. Mass moment of inertia of a circular plate.

3. Mass moment of inertia of a hollow circular cylinder.

4. Mass moment of inertia of a right circular cone of base radius R, height Hand mass M about its axis.

Mass Moment of Inertia of a Rectangular Plate

(a) Mass moment of inertia of a rectangular plate about X-X axis passingthrough the C.G. of the plate. Fig. 3.50 shows a rectangular plate of width b, depth‘d’ and uniform thickness ‘t’. Consider a small element of width ‘b’ at a distance ‘y’from X-X axis as shown in Fig. 3.51.

Here X-X axis is the horizontal line passing through the C.G. of the plate.

Area of the element = b × dy

∴ Mass of the element = Density × Volume of element= ρ × [Area × thickness of element]= ρ × [b × dy × t] [� ρ = Density and

t = thickness]= ρbt dy

Mass moment of inertia of the element about X-X axis= Mass of element × y2

= (ρbt dy) × y2 = ρbt y2 dyMass moment of inertia of the plate will be obtained by integrating the above

equation between the limits – d2

to d2

.

r1

r2

r3

Mass m1

Mass m2 Mass m3

Given axis

Fig. 3.49


NOTES


d/2

d/2

d

Yb

t

X

X

Y

Ady

y

C.G.

b

d

B

d/2

d/2

X

D C

X

Fig. 3.50 Fig. 3.51

∴ (Im)xx = ρ ρbt y dy bt y dyd

d

d

d2

2

2

2

22

− −� �=/

/

/

/

[� ρ, b, t are constant and can be taken outside the integral sign]

= ρbt y

d

d3

3

2

3

�

�

��

− /

/

= ρbt d d3 2 2

3 3��

− −��

��

�

�

��

= ρ ρ ρbt d d bt d d bt d3 8 8 3 8 8 3

28

3 3 3 3 3

− −�

��

� �

�

��

= +�

�

�� = ×

= ρbt3

d3 = ρ × t bd3

12...(3.35)

But bd3

12 is the moment of inertia of the area of the rectangular section

about X-X axis. This moment of inertia of the area is represented by Ixx.∴ (Im)xx = ρ × t × Ixx ...(3.36)

where (Im)xx = Mass moment of inertia of the plate about X-X axis passing through C.G. of the plate.

Ixx = Moment of inertia of the area of the plate about X-X axis.Again from equation (4.5), we have

(Im)xx = ρ bd3

12

= ρb × d × t × d2

12

= M × d2

12(� M = Mass of the plate = ρ × Volume of the plate = ρ × [b × d × t])

= 1

12 Md2 ...(3.37)

Similarly, the mass moment of inertia of the rectangular plate about Y-Y axispassing through the C.G. of the plate is given by

(Im)yy = 1

12 Mb2. ...(3.38)


NOTES

Engineering Mechanics (b) Mass moment of inertia of therectangular plate about a line passing throughthe base. Fig. 3.52 shows a rectangular plateABCD, having width = b, depth = d and uniformthickness = t. We want to find the mass moment ofinertia of the rectangular plate about the line CD,which is the base of the plate. Consider arectangular elementary strip of width b, thicknesst and depth ‘dy’ at a distance y from the line CD asshown in Fig. 4.46.

Area of strip, dA = b . dyVolume of strip = dA × t

= b . dy . t = b . t . dyMass of the strip, dm = Density × Volume of strip

= ρ(b . t . dy) = ρ . b . t . dyMass moment of inertia of the strip about the line CD

= Mass of strip . y2

= dm . y2 = y2 . dm

Mass moment of inertia of the whole rectangular plate about the line CD isobtained by integrating the above equation between the limits 0 to d.

∴ Mass moment of inertia of the rectangular plate about the line CD

= y dm y b t dyd d

2

0

2

0� �=. ( . . . )ρ [� dm = ρ . b . t . dy]

= ρ . b . t y dyd

2

0� [� ρ, b and t are constant]

= ρ . b . t . y

d3

03

�

�

�� = ρ . b . t .

d3

3 = ρ . b . t . d .

d2

3

= M d. 2

3 [� ρ . b . t . d = Mass of rectangular plate = M]

...(3.39)(c) Mass moment of inertia of a hollow

rectangular plate. Fig. 3.53 shows a hollowrectangular plate in which ABCD is the mainplate and EFGH is the cut-out section.

The mass moment of inertia of the mainplate ABCD about X-X is given by equation

= 1

12 Md2

The mass moment of inertia of the cut-outsection EFGH about X-X axis

= 1

12 md1

2

where M = Mass of main plate ABCD

= ρ . b . d . t m = Mass of the cut-out section EFGH

= ρ . b1 . d1 . t

b

d

A

D C

B

dy

y

Fig. 3.52

b

d

A

D C

B

XX

E

H G

F

b1

d1

Fig. 3.53


NOTES


Then mass moment of inertia of hollow rectangular plate about X-X axis is givenby

(Im)xx = 1

12 Md2 –

112

md12. ...(4.34)

Mass Moment of Inertia of a CircularPlate. Fig. 3.54 shows a circular plate of radiusR and thickness t with O as centre. Consider anelementary circular ring of radius ‘r’ and widthdr as shown in Fig. 3.55 (a).

Area of ring, dA = 2πr . dr

Volume of ring = Area of ring × t = dA . t= 2πr . dr . t

Mass of ring, dm = Density × Volume of ring

= ρ(2πr.dr.t)In this case first find the mass moment of

inertia about an axis passing through O andperpendicular to the plane of the paper i.e. aboutaxis Z-Z.

∴ Mass moment of inertia of the circular ring about axis Z-Z

= (Mass of ring) × (radius of ring)2

= dm × r2 = (ρ . 2πr dr . t) × r2 = ρ . t . 2πr3 dr

The mass moment of inertia of the whole circularplate will be obtained by integrating the above equationbetween the limits O to R.

∴ Mass moment of inertia of circular plateabout Z-Z axis is given by

(Im)zz = ρ π. .t r drO

R2 3�

= 2π . ρ . t r drO

R3�

= 2πρ . t r

O

R4

4

�

�

�� = 2π . ρ . t .

R4

4

= π . ρ . t . R4

2Now mass of circular plate,

M = ρ × Volume of plate= ρ × πR2 × t [Volume of plate = Area × t = πR2 × t]

Substituting this value in above equation, we get

∴ (Im)zz = ρ × πR2 × t × R2

2 =

MR2

2...(3.41)

But from the theorem of perpendicular axis given by equation (3.14), we have Izz = Ixx + Iyy

or (Im)zz = (Im)xx + (Im)yy

r

drR

OX X

Y

Y

Fig. 3.54

Y

X

O

Rt

Fig. 3.55


NOTES

Engineering Mechanics And due to symmetry, we have (Im)xx = (Im)yy

∴ (Im)xx = (Im)yy = (Im)zz /2

= MR MR2 2

22

4

�

��

� =/ ...(3.42)

Mass Moment of Inertia of a Hollow Circular CylinderLet Ro = Outer radius of the cylinder

Ri = Inner radius of the cylinderL = Length of the cylinder

M = Mass of cylinder= Density × Volume of cylinder= ρ × π[Ro

2 – Ri2] × L ...(i)

dm = Mass of a circular ring of radius ‘r’ width ‘dr’ and length L [Refer toFig. 3.55]

= Density × Volume of ring = ρ × Area of ring × L= ρ × 2πrdr × L

Now mass moment of inertia of the circular ring about Z-Z axis= Mass of ring × (radius)2

= (ρ × 2πrdr × L) × r2

The mass moment of inertia of the hollow circular cylinder will be obtained byintegrating the above equation between the limits Ri to Ro.

∴ Mass moment of inertia of the hollow circular cylinder about Z-Z axis isgiven by,

(Im)zz = ( . )ρ πR

R

i

or dr L r� × 2 2

= ρ × 2π × L R

R

i

or dr� 3 = ρ × 2π × L r

R

R

i

o4

4

�

�

��

= ρ × 2π × L × R Ro i

4 4

4−�

�

��

= ρ × 2π × L × R Ro i

2 2

4−�

�

�� [Ro

2 + Ri2]

[� Ro4 – Ri

4 = (Ro2 – Ri

2) (Ro2 + Ri

2)]

= ρ × π[Ro2 – Ri

2] × L × ( )R Ro i

2 2

2+

= M R Ro i( )2 2

2+

[� ρ × π× (Ro

2 – Ri2) = M]

Now (Im)xx = (Im)yy = ( ) ( )I M R Rm zz o i

2 4

2 2

=+

Mass Moment of Inertia of a Right Circular Cone of Base Radius R,Height H and Mass M about its Axis

Let R = Radius of the base of the cone,H = Height of the cone,


NOTES


M = Mass of the cone= Density × Volume of cone

= ρ × 13

πR2 × H

Consider an elemental plate of thickness dy andof radius x at a distance y from the vertex (as shown inFig. 3.56).

We have, tan α = xy

RH

= ∴ x = RH

× y

Mass of the elemental plate,dm = ρ × Volume

= ρ × (πx2 × dy)

= ρ × πR y

Hdy

2 2

2×

×�

�

�� x

R yH

= ×��

��

The mass moment of inertia of the circular elemental plate about the axis ofthe cone (here axis of the cone is Z-Z axis of the circular elemental plate) is given byequation (3.41) as

(Im)zz = Mass of plate radius

2

2×

= ( )dm r dm x× = ×2 2

2 2(� r = x)

= ρπ

× ×�

�

�� ×

R yH

dyx2 2

2

2

2� dm

Ry

Hdy= ×�

��

ρ π2

= ρ π× R y

H

2 2

2 × dy × R y

H

2 2

212

�

�

�� × � x

RyH

=��

��

= ρ π× ×r y

H

4 4

42 dy

Now the total mass moment of inertia of the circular cone will be obtained byintegrating the above equation between the limits O to H.

∴ (Im)zz = ρπ ρπR yH

dyR

Hy

O

H

O

H4 4

4

4

4

5

2 2 5× = ×

�

�

��

= ρπ ρπR

HH R H4

4

5 4

2 5 2 5× = ×

×

But mass of cone, M = ρπR H4

3×

∴ (Im)zz = ρπR H2

3×

× R2 3

10×

= M × 310

R2 = 310

MR2 ...(3.43)

RRAB

dy

yy

HH

O

α

x

Fig. 3.56


NOTES


1. The point, through which the whole weight of the body acts, is known as centreof gravity.

2. The point, at which the total area of a plane figure is assumed to be concen-trated, is known as centroid of that area. The centroid and centre of gravity areat the same point.

3. The centre of gravity of a uniform rod lies at its middle point.4. The C.G. of a triangle lies at a point where the three medians of a triangle meet.5. The C.G. of a parallelogram or a rectangle is at a point where its diagonal meet

each other.6. The C.G. of a circle lies at its centre.7. The C.G. of a body consisting of different areas is given by

xa x a x a x

a a a=

+ + ++ + +

1 1 2 2 3 3

1 2 3

............ and y

a y a y a ya a a

=+ + +

+ + +1 1 2 2 3 3

1 2 3

............

where x and y = Co-ordinates of the C.G. of the body from axis of reference a1, a2, a3, ...... = Different areas of the sections of the body x1, x2, x3, ...... = Distances of the C.G. of the areas a1, a2, a3, ...... from Y-axis y1, y2, y3, ...... = Distances of the C.G. of the areas a1, a2, a3, ...... from X-axis.

8. If a given section is symmetrical about X-X axis or Y-Y axis, the C.G. of thesection will lie on the axis symmetry.

9. The moment of inertia of an area (or mass) about an axis is the product of area(or mass) and square of the distance of the C.G. of the area (or mass) from thataxis. It is represented by I.

10. Radius of gyration of a body (or a given lamina) is the distance from an axis ofreference where the whole mass (or area) of the given body is assumed to beconcentrated so as not to after the moment of inertia about the given axis. It is

represented by k. Mathematically, k = IA

.

11. According to theorem of perpendicular axis IZZ = IXX + IYY where IXX and IYY =Moment of inertia of a plane section about two mutually perpendicular axes X-X and Y-Y in the plane of the section, IZZ = Moment of inertia of the sectionperpendicular to the plane and passing through the intersection of X-X and Y-Yaxes.

12. According to the theorem of parallel axis IAB = IG + Ah2 whereIG = Moment of inertia of a given area about an axis passing through C.G.

of the areaIAB = Moment of inertia of the given area about an axis AB, which is parallel

to the axis passing through Gh = Distance between the axis passing through G and axis AB

A = Area of the section.13. Moment of inertia of a rectangular section :

(i) about an horizontal axis passing through C.G. = bd3

12

(ii) about an horizontal axis passing through base = bd3

12 .


NOTES

Properties of Surfacesand Solids14. Moment of inertia of a circular section =

πD4

64 .

15. Moment of inertia of a triangular section :

(i) about the base = bh3

12

(ii) about an axis passing through C.G. and parallel to the base = bh3

36 .

where b = Base width, and h = Height of the triangle.16. The C.G. of an area by integration method is given by

xx dA

dA= ��

*and y

x dA

dA= ��

*

where x* = Distance of C.G. of area dA from y-axisy* = Distance of C.G. of area dA from x-axis.

17. The C.G. of a straight or curved line is given by

xx dL

dL= ��

*and y

y dL

dL= ��

*.

��

• Centroid. The point at which the total area of a plane figure (like rectangleetc.) is assumed to be concentrated is called centroid.

• Centroid of volume. It is the point at which the total volume of a body assumedto be concentrated. The volume is having three dimensions i.e. length, widthand thickness, hence, measured in [Length]3.

• Radius of gyration. Radius of gyration of a body about an axis is a distancesuch that its square multiplied by the area gives moment of inertia of the areaabout the given axis.

• Principal axis. These are the axes about which the product of inertia is zero.• Mass moment of inertia. It is the product of the mass and the square of the

distance of the centre of gravity of the mass from an axis.

��

1. Define centre of gravity and centroid.

2. Derive an expression for the centre of gravity of a plane area using method ofmoments.

3. What do you understand by axes of reference ?

4. Define the terms : moment of inertia and radius of gyration.

5. State the theorem of perpendicular axis. How will you prove this theorem ?

6. State and prove the theorem of parallel axis.


NOTES

Engineering Mechanics 7. Find an expression for the moment of inertia of a rectangular section :

(i) about an horizontal axis passing through the C.G. of the rectangular sec-tion, and

(ii) about an horizontal axis passing through the base of the rectangular section.8. Prove that the moment of inertia of a circular section about an horizontal axis

(in the plane of the circular section) and passing through the C.G. of the section

is given by πD4

64 .

9. Prove that moment of inertia of a triangular section about the base of the section

= bh3

12

where b = Base of triangular section, andh = Height of triangular section.

10. Derive an expression for the moment of inertia of a triangular section about anaxis passing through the C.G. of the section and parallel to the base.

11. Show that IO = IG + Ah2, where IG is the moment of inertia of a lamina about anaxis through its centroid and lying in its plane and h is the distance from thecentroid to a parallel axis in the same plane about which its moment of inertiais IO, A being the area of the lamina.

12. State and prove the parallel axes theorem on moment of inertia for a planearea.

13. Prove that the moment of area of any plane figure about a line passing throughits centroid is zero.

13. Find the centre of gravity of the T-section shown in Fig. 3.57. [Ans. 8.272 cm]

12 cm

2 cm

12 cm

2 cm

2 cm

12 cm

2 cm

2cm

8 cm

16 cm

Fig. 3.57 Fig. 3.58

14. Find the centre of gravity of the I-section shown in Fig. 3.58. [Ans. 6.44 cm]


NOTES


15. Find the centre of gravity of the L-section shown in Fig. 3.59.

[Ans. x = 1.857, y = 3.857]

2 cm

8 cm

10 cm

2 cm

6 cm

3cm

10 cm

14 cm

5 cm

2 cm

1cm

1cm

Fig. 3.59 Fig. 3.60

16. From a rectangular lamina ABCD 10 cm × 14 cma rectangular hole of 3 cm × 5 cm is cut as shownin Fig. 3.60. Find the centre of gravity of theremainder lamina.

[Ans. x = 4.7 cm, y = 6.444 cm]17. Determine the C.G. of the uniform plane lamina

shown in Fig. 3.61. All dimensions are in cm.[Hint. The figure is symmetrical about Y-Y axis.

∴ ya y a y a y a y

a a a a= + + +

+ + +1 1 2 2 3 3 4 4

1 2 3 4

where a1 = 40 × 30 = 1200 cm2, y1 = 302

= 15 cm,

a2 = 30 × 20 = 600 cm2, y2 = 30 + 302

= 45 cm

a3 = – π × 10

2

2 = – 50π, y3 =

43

4 103

403

rπ π π

= × =

a4 = – 20 10

2×

= – 100, y4 = 60 – 103

= 1703

∴ y =× + × − × − ×

+ − −

1200 15 600 45 50403

100170

31200 600 50 100

ππ

π

= 18000 27000 666 7 5666 7

1700 5038666 61542 92

+ − −−

=. . ..π

= 25.06 cm from Origin 0. Ans.]

18. From a circular plate of diameter 100 mm acircular part of diameter 50 mm is cut as shownin Fig. 3.62. Find the centroid of the remainder.[Hint. Fig. 4.57 is symmetrical about x-axis.Hence centroid lies on x-axis.∴ y = 0.6. The value of x is given by

x = a x a x

a a1 1 2 2

1 2

+−

x¢ x

y¢

y 100 mm

50 mm

Fig. 3.62

Fig. 3.61

90°4

Y

10

220

1010

1

30

103

1010 20O X


NOTES

Engineering MechanicsBut a1 =

π4

1002× = 7853.98 mm2, x1 = 1002

= 50 mm

a2 = – π4

502×��

��

= – 1963.5 mm2, x2 = 100 – 25 = 75 mm

∴ x = 7853.98 50 1963.5 75

7853.98 1963.5× − ×

− = 41.67 mm

Hence centroid is at (41.67 mm, 0). Ans.]

��

• Hibbeller, R.C., ‘‘Engineering Mechanics’’, Vol. 1 Statics, Vol. 2 Dynamics,Pearson Education Asia Pvt. Ltd., (2000).

• Palanichamy, M.S., Nagam, S., ‘‘Engineering Mechanics – Statics & Dynamics’’,Tata McGraw-Hill, (2001).

• Irving H. Shames, ‘‘Engineering Mechanics – Statics and Dynamics’’ IV Edition– Pearson Education Asia Pvt. Ltd., (2003).

• Ashok Gupta, ‘‘Interactive Engineering Mechanics – A Virtual Tutor (CDROM)’’,Pearson Education Asia Pvt., Ltd., (2002).


NOTES

Dynamics of ParticlesU N I T

4DYNAMICS OF PARTICLES

STRUCTURE

4.1 Introduction4.2 Velocity4.3 Acceleration4.4 Relationship of Velocity, Acceleration and Displacement4.5 Relative Motion4.6 Curvilinear Motion4.7 Equations of Motions along a Circular Path4.8 Newton’s Laws4.9 Work Energy Equation of Particles4.10 Impulse and Momentum4.11 Impact of Elastic Bodies• Summary• Glossary• Review Questions• Further Readings

��

After going through this unit, you should be able to :• deal with velocity, acceleration and this relationship.• illustrate the concepts of relative motion, curvilinear motion, Newton’s laws,

work energy equation of particles.• define impulse and momentum.• determine the impact of elastic bodies.

��

Dynamics is the branch of science which deals with the study of a particle (orof a body), when the particle (or body) is in motion. The dynamics is further dividedinto kinematics and kinetics. The study of body in motion, when forces which cause


NOTES

Engineering Mechanics motion are not considered, is called kinematics whereas if the forces are consideredfor the body in motion, that branch of science is called kinetics. This chapter dealswith velocity, acceleration, their relationship, relative motion, curvilinear motion,Newton’s laws, work energy equation of particles, impulse and momentum and im-pact of elastic bodies.

��

Velocity is defined as the rate of change of displacement of a body moving in astraight line. It is measured in metre per second. Velocity is a vector quantity. It isdenoted by v.

Let s = Distance travelled by a body in a straight linet = Time taken to travel the distance.

Then velocity of the body = st

.

Mathematically, it can also be written as

Velocity, v = dsdt

...(4.1)

��

Acceleration is defined as the rate of change of velocity of a body. It is measuredin metre per second square and is written as m/sec2 in M.K.S. system and m/s2 inS.I. system.

∴ Acceleration = Change of velocity

TimeAcceleration is denoted by ‘a’. Acceleration is also a vector quantity.

Mathematically, it can also be written as

Acceleration, a = dvdt

...(4.2)

= ddt

dsdt��

� vdsdt

=��

��

from equation .1(4 )

= d s

dt

2

2...(4.2A)

But dvdt

can also be written as

dvdt

dvds

dsdt

dvds

= =. . v �

dsdt

v=��

��

Butdvdt

= Acceleration = a

∴ a = dvds

. v ...(4.3)


NOTES

Dynamics of Particles��

��

The relationship of velocity, acceleration and displacement is obtained fromequations of motion in a straight line.

Let us consider a body, which is moving in a straight line.Let u = Initial velocity of the body in m/s

v = Final velocity of the body in m/st = Time in seconds, during which velocity changes from u to vs = Distance travelled by the body in time ta = Acceleration of the body.

Equation for Final VelocityChange of velocity = Final velocity – Initial velocity

= (v – u)

Rate of change of velocity = Change of velocity

Time( – )= v u

t.

But rate of change of velocity = Acceleration = a

∴ a = ( )v u

t–

...(4.4)

or at = v – u

∴ v = u + at ...(4.5)

Equation of Motion for Distance Covered (s)

We know average velocity = Initial velocity Final velocity

2+

= u v+

2.

Distance covered, s = Average velocity × Time

= ( )u v+

2 × t.

Substituting the value of v from equation (4.5),

s = ( ) ( )u u at

tu at

t u at+ + × = + × = +�

��2

22

12

× t

∴ s = ut + 12

at2 ...(4.6)

Derivation of v2 – u2 = 2as. The value of ‘t’ from equation (4.5) is given as

t = ( – )v u

a.

Substituting the value of ‘t’ in equation (4.6), we get

s = u v ua

av u

a– –�

��

+ ��

��

12

2

= (v – u) ua a

v u+��

��

12

( – ) = (v – u) ( – )2

2u v u

a+�

��

= ( – ) ( ) –v u v u

av u

a+ =

2 2

2 2

or v2 – u2 = 2as ...(4.7)


NOTES

Engineering Mechanics In equations (4.5), (4.6) and (4.7), a is the linear acceleration. If instead of ac-celeration the retardation is given, then the value of ‘a’ in these equations should betaken negative. Retardation is – ve of acceleration.

2nd Method for deriving equations of motion for a body moving in astraight line

The equations of motion of a body moving in a straight line may also be derivedby integration as given below :

1. Derivation of s = ut + 12

at2

Let a body is moving with a uniform acceleration ‘a’.Then from equation (4.2), we have

d s

dt

2

2 = a or

ddt

dsdt��

= a

or d dsdt�� = a dt.

Integrating the above equation,

ddsdt

a ��

= dt ordsdt

= at + C1 ...(i)

where C1 is the constant of integration.

Butdsdt

= Velocity at any instant

when t = 0, the velocity is known as initial velocity which is represented by u.

∴ at t = 0,dsdt

= Initial velocity = u

Substituting these values in equation (i).u = a × 0 + C1

∴ C1 = u.Substituting the value of C1 in equation (i),

dsdt

= at + u ...(ii)

Integrating the above equation, we get

s = at2

2 + ut + C2 ...(iii)

when C2 is another constant of integration.When t = 0, then s = 0.Substituting these values in equation (iii), we get

0 = a2

× 0 + u × 0 + C2 = 0 + 0 + C2

∴ C2 = 0.Substituting this value of C2 in equation (iii), we get

s = 12 at2 + ut

the above equation is the same as equation (4.6).2. Derivation of v = u + at.

From equation (ii), we have dsdt

= at + u.


NOTES

Dynamics of ParticlesBut

dsdt

represents the velocity at any time. After the time ‘t’ the velocity is

known as final velocity, which is represented by v.

∴ dsdt

after time ‘t’ = Final velocity = v.

Substituting the value ofdsdt

= v, we get

v = at + u.3. Derivation of v2 = u2 + 2as.From equation (4.3), acceleration ‘a’ is given by

a = v dvds

orv dvds

= a or v dv = ads.

Integrating, we get

v2

2 = as + C3 ...(iv)

where C3 is constant of integration.When s = 0, the velocity is known as initial velocity.∴ At s = 0, v = u.Substituting these values in equation (iv), we get

u2

2 = a × 0 + C3

∴ C3 = u2

2.

Substituting this value in equation (iv), we get

v2

2 = as +

u2

2or v2 = 2as + u2.

Problem 1. A body is moving with a velocity of 2 m/s. After 4 seconds thevelocity of the body becomes 5 m/s. Find the acceleration of the body.

Sol. Given :Initial velocity, u = 2 m/sFinal velocity, v = 5 m/sTime, t = 4 secLet a = Acceleration of the body.Using equation (4.5), v = u + at

or 5.0 = 2.0 + a × 4.0

∴ a = ( – )5 2

434

= = 0.75 m/s2. Ans.

Problem 2. On a straight road, a smuggler’s car passes a police station witha uniform velocity of 10 m/s. After 10 seconds, a police party follows in pursuit in ajeep with a uniform acceleration of 1 m/s2. Find the time necessary for the jeep tocatch up with the smuggler’s car.

Sol. Given :Uniform velocity of smuggler’s car = 10 m/s


NOTES

Engineering Mechanics Uniform acceleration of police jeep = 1 m/s2

Let S = Distance from police station to the point where police jeep catch upthe smuggler’s car.

t = Time taken by smuggler’s car to cover distance S.t′ = Time taken by police party to catch up with the smuggler’s car or to

cover up the distance S metre.= (t – 10) seconds.

(i) First consider the motion of smuggler’s car

Uniform velocity, u = 10 m/sDistance S covered in time ‘t’ is given by, S = ut = 10 × t ...(i)(ii) Now consider the motion of police jeepAcceleration, a = 1 m/s2

Initial velocity, u = 0 (jeep starts from rest)Distance = S

Time, t′ = (t – 10)

Using the relation, S = ut′ + 12

at′2 (� Here time = t′)

= 0 × (t – 10) + 12

× 1 × (t – 10)2 = 0 + 12

(t – 10)2

= 12

(t2 + 100 – 20t) ...(ii)

Equating the two values of S given by equations (i) and (ii), we get

12

(t2 + 100 – 20t) = 10t

or t2 + 100 – 20t = 20t

or t2 – 40t + 100 = 0The solution of the above quadratic equation is

t = 40 40 4 1002

40 1600 4002

40 34 642

2± × = ± = ±– – .

= 74.642

and 5.362

= 37.32 seconds and 2.68 seconds

The time 2.68 seconds is impossible, as jeep starts after (t – 10) secondsi.e., 2.68 – 10 = – 7.32 seconds. This gives absurd result.

∴ The time necessary for the jeep to catch up with smuggler’s car is givenby

t′ = (t – 10) seconds = (37.32 – 10) = 27.32 seconds. Ans.Distance Travelled in the nth SecondLet u = Initial velocity of a body

a = Acceleration of the bodySn = Distance covered in n seconds

Sn – 1 = Distance covered in (n – 1) secondsThen distance travelled in the nth seconds

= Distance travelled in n seconds– Distance travelled in (n – 1) seconds

= Sn – Sn – 1.


NOTES

Dynamics of ParticlesDistance travelled in n seconds is obtained by substituting t = n in equation

s = ut + 12 at2.

∴ Sn = u × n + 12 an2.

Similarly Sn–1 = u(n – 1) + 12

a(n – 1)2 (Put t = n – 1 in s = ut + 12

at2)

∴ Distance travelled in the nth seconds= Sn – Sn–1

= (un + 12 an2) – [u(n – 1) + 1

2 a(n – 1)2]

= un + 12 an2 – [un – u + 1

2 a(n2 + 1 – 2n)]

= un + 12 an2 – un + u – 1

2 an2 – 1

2 a + 1

2 a × 2n

= an + u – 12 a = u +

a2

(2n – 1) ...(4.8)

Problem 3. A body is moving with uniform acceleration and covers 15 m infifth second and 25 m in 10th second. Determine :

(i) the initial velocity of the body, and(ii) acceleration of the body.Sol. Given :Distance covered in 5th second = 15 m.Distance covered in 10th second = 25 m.Let u = Initial velocity, and

a = Acceleration of the body.

Using equation (4.8) for the distance covered in nth second = u + a2

(2n – 1).

∴ Distance covered in 5th second

= u + a2

[2 × 5 – 1] (� n = 5)

or 15 = u + 92a

...(i)

Distance covered in 10th second

= u + a2

[2 × 10 – 1] (� n = 10)

or 25 = u + 19

2a

...(ii)

Subtracting equation (i) from (ii), we get

25 – 15 = ua

ua+�

��

− +��

��

192

92

or 10 = 19

292

102

a a a− =

∴ a = 2 m/s2. Ans.Substituting the value of ‘a’ in equation (i),

15 = u + 92

× 2 = u + 9

∴ u = 15 – 9 = 6 m/s. Ans.


NOTES

Engineering Mechanics Equation of Motions Due to Gravity. The acceleration due to gravity is ‘g’.Hence when a body falls, the equation of motions given by equations (4.5), (4.6) and(4.7) are modified by substituting ‘g’ in place of ‘a’. But when the body is movingvertically up the acceleration due to gravity is acting in the opposite direction. Inthat case the equations are modified by substituting (– g) in place of a. The value ofg is taken as 981 cm/s2 or 9.81 m/s2. The distance ‘S’ is replaced by ‘h’. Hence theequations of motions due to gravity in the downward directions and upward directionsbecomes as :

1. For downward motion 2. For upward motiona = + g a = – gv = u + gt v = u – gt

S = h = ut + 12 gt2 h = ut – 1

2gt2

v2 – u2 = 2gh. v2 – u2 = – 2gh.

Points to Be Remembered(i) If a body starts from rest, its initial velocity is zero, i.e., u = 0.

(ii) If a body comes to rest, its final velocity is zero, i.e., v = 0.(iii) If a body is projected vertically upwards, the final velocity of the body at the

highest point is zero, i.e., v = 0.(iv) If a body starts moving vertically downwards, its initial velocity is zero, i.e.,

u = 0.(v) Acceleration due to gravity is taken positive when a body is moving verti-

cally downwards. But if the body is moving vertically upwards, the acceleration due togravity is taken negative.

Problem 4. A tower is 90 m in height. A particle is dropped from the top of thetower and at the same time another particle is projected upward from the foot of thetower. Both the particle meet at a height of 30 m. Find the velocity, with which thesecond particle is projected upward.

Sol. Given :Height of tower, h = 90 m.Height at which both particle meet,

h1 = 30 m.Distance travelled by first particle,

S1 = h – h1 = 90 – 30 = 60 m.Distance travelled by the second particle,

S2 = h1 = 30 m.For the first particle, we have (particle is moving down).Initial velocity, u = 0Distance, S1 = 60 mAcceleration due to gravity, g = 9.81 m/s2.Let the two particles meet after an interval of ‘t’ seconds.Using the equation,

S1 = ut + 12

gt2 or 60 = 0 × t + 12

× 9.81 × t2

∴ t = 60 29 81

×.

= 3.497 s.

Fig. 4.1

90 m

30 m

Tower


NOTES

Dynamics of ParticlesFor the second particle, we have (particle is moving up)Time, t = 3.497 s (the time is same)Distance, S2 = 30 mAcceleration, g = – 9.81 m/s2

Let u = Initial velocity with which the second particle should be projected up.Using the equation,

S2 = ut + 12

gt2 or 30 = u × 3.497 + 12

(– 9.81) × 3.4972

= 3.497 u – 59.98

∴ u = 30 59.98

3.497+

= 25.73 m/s. Ans.

Problem 5. A body falling freely under the action of gravity passes two points20 m apart vertically in 0.4 seconds. From what height, above the higher point, didthe body start to fall ? Take g = 9.8 m/s2.

Sol. Given :Initial velocity, u = 0Let the body starts from A and passes the two points B

and C, 20 m apart in 0.4 seconds as shown in Fig. 4.2.Let distance AB = x mThen distance AC = (x + 20) mIf t = Time taken by the body to travel from A to B.Then (t + 0.4) = Time taken by the body to travel from A

to C.The distance ‘x’ travelled by the body in time ‘t’ is given

by

x = ut + 12 gt2

= 0 × t + 12 gt2 (� u = 0)

= 12 gt 2 ...(i)

The distance (x + 20) travelled by the body in time (t + 0.4) second is given by

(x + 20) = u × (t + 0.4) + 12 g(t + 0.4)2

= 0 × (t + 0.4) + 12 g(t + 0.4)2

= 12 g(t2 + 0.42 + 0.8t) ...(ii)

Subtracting equation (i) from equation (ii),

(x + 20) – x = 12 g(t2 + 0.42 + 0.8t) – 1

2 gt2

or 20 = 12 g[t2 + 0.16 + 0.8t – t2]

= 9.82

[0.8t + 0.16] or 20 2

9.8×

= 0.8t + 0.16

or 0.8t = 20 2

9.8×

– 0.16 = 4.08 – 0.16 = 3.92

∴ t = 3.920.80

= 4.9 s.

Fig. 4.2

X

20 m

A

B

C


NOTES

Engineering Mechanics Substituting the value of t in equation (i), we get

x = 12 g × (4.9)2 = 1

2 × 9.8 × (4.9)2 = 117.649 m. Ans.

Velocity and Acceleration of a Body Moving in a Straight Line byDifferentiation. Let the equation of motion of a body moving in a straight line isgiven in terms of displacement (s) and time (t) as s = t3 + 2t2 + 5t + 6 ...(i)

By differentiating the above equation with respect to time, we get

Velocity, i.e.,dsdt

= v

∴ v = dsdt

ddt

= (t3 + 2t2 + 5t + 6)

= 3t2 + 4t + 5 ...(ii)If the velocity at start is to be calculated, the value of t = 0 is to be substituted

in equation (ii).By differentiating the equation (ii) with respect to time, we get acceleration, i.e.,

dvdt

= a.

∴ Acceleration, a = dvdt

ddt

= (3t2 + 4t + 5)

= 6t + 4 ...(iii)The acceleration at start will be obtained by substituting t = 0, in equation

(iii).Problem 6. A particle moves along a straight line so that its displacement in

metre from a fixed point is given by, s = t3 + 3t2 + 4t + 5.Find :(i) Velocity at start and after 4 seconds

(ii) Acceleration at start and after 4 seconds.Sol. Given :

s = t3 + 3t2 + 4t + 5

Velocity, v = dsdt

dsdt

= (t3 + 3t2 + 4t + 5)

= 3t2 + 6t + 4 ...(i)(i) Velocity at start will be obtained, if t = 0 is substituted in equation (i),∴ v(at t = 0) = 3 × 02 + 6 × 0 + 4 = 0 + 0 + 4

= 4 m/s. Ans.Velocity after four seconds. Substitute t = 4 in equation (i)

v(at t = 4) = 3 × 42 + 6 × 4 + 4= 48 + 24 + 4 = 76 m/s. Ans.

(ii) Differentiating equation (i) with respect to time ‘t’, we get acceleration.

∴ Acceleration, a = dvdt

ddt

= (3t2 + 6t + 4)

= 6t + 6. ...(ii)Acceleration at start will be obtained, if t = 0 is substituted in equation (ii).∴ a(at t = 0) = 6 × 0 + 6 = 0 + 6

= 6 m/s2. Ans.


NOTES

Dynamics of ParticlesAcceleration after four seconds. Substitute t = 4 in equation (ii),∴ a(at t = 4) = 6 × 4 + 6 = 24 + 6

= 30 m/s2. Ans.Velocity and Displacement of a Body Moving in a Straight Line by In-

tegration. Let the equation of motion of a body moving in a straight line is given interms of acceleration (a) and time (t) as

a = t3 + 2t2 + 4t + 5 ...(i)

But a = dvdt

∴dvdt

= t3 + 2t2 + 4t + 5


∫ dv = ∫ (t3 + 2t2 + 4t + 5)dt

or v = t t t

t C4 3 2

1423

42

5+ + +�

��

��+ ...(ii)

where C1 is the constant of integration and its value is obtained from given condition.Let the condition is given that at t = 1, the velocity is 10 m/s.

Substituting this condition in equation (ii), we get

10 = 14

2 13

4 12

4 3 2

+ × + × + 5 × 1 + C1

= 14

23

+ + 2 + 5 + C1

= .25 + .67 + 7 + C1 = 7.92 + C1

∴ C1 = 10 – 7.92 = 2.08.

Substituting the value of C1 in equation (ii), we get the velocity at any time t,

v = t t t4 3 2

423

42

+ + + 5t + 2.08 ...(iii)

But v = dsdt

∴ dsdt

t t= +4 3

423

+ 2t2 + 5t + 2.08


dst t

t t = + + + +�

��

��4 3

2

423

2 5 2.08 dt

or s = t t t t5 4 3 2

4 52

3 423

52×

+×

+ + + 2.08t + C2 ...(iv)

where C2 is another constant of integration. The value of C2 is also obtained fromgiven condition. If at t = 1, the displacement (s) is given as equal to 10 m. Substitutingthis condition in equation (iv), we get

10 = 120

2 112

2 13

5 12

5 4 3 2

+×

+×

+× + 2.08 × 1 + C2


NOTES

Engineering Mechanicsor 10 =

120

16

23

+ + + 2.5 + 2.08 + C2

= .05 + .1667 + .667 + 2.5 + 2.08 + C2 = 5.4637 + C2

∴ C2 = 10 – 5.4637 = 4.5363.

Substituting the value of C2 in equation (iv), we get the displacement at anytime t as

s = t t t t5 4 3 2

20212

23

52

+ + + + 2.08t + 4.5363 ...(v)

If it is required to find the velocity and displacement at any time ‘t’, thenequation (iii) and (v) are used.

Problem 7. A particle moves along a straight line with an acceleration describedby the equation a = – 8s–2 where a is in m/s2 and s in m. When t = 1 s, s = 4 m andv = 2 m/s. Determine acceleration when t = 2 s.

Sol. Given :Acceleration, a = – 8s–2

When t = 1 s, s = 4 m and v = 2 m/sFind acceleration when t = 2 s.The acceleration in terms of velocity and displacement is given by equation

(4.3),

a = v dvds

But a = – 8s2 (given)

∴ – 8s–2 = v dvds

or vdv = – 8s–2 ds


v dv s dsv s = = −

+

+–

–2–2

–2

82

81

2 1

+ C1

or vs

2

28= + C1 ...(i)

where C1 is the constant of integration and its value is obtained from given conditioni.e., s = 4 m and v = 2 m/s when t = 1 s.

Substituting v = 2 m/s and s = 4 m in equation (i),

22

84

2

= + C1 or 2 = 2 + C1

or C1 = 0Substituting C1 = 0 in equation (i), we get

vs

2

28

= or v2 = 4s

or v = 4

s...(ii)

But v = dsdt

Substituting the value of v in equation (ii), we getdsdt s

= 4or s ds = 4dt


NOTES

Dynamics of ParticlesIntegrating, we get

s ds dt = 4 ors

12

1

12

1

+

+ = 4t + C2

or23

s3/2 = 4t + C2 ...(iii)

when t = 1 s, s = 4 m.

Substituting these values in equation (iii), we get

23

× 43/2 = 4 × 1 + C2 or163

= 4 + C2

or C2 = 163

416 12

343

− = =–

Substituting the value of C2 in equation (iii), we get

23

s3/2 = 4t 43

.

When t = 2 s, the above equation becomes as

43

s3/2 = 4 × 2 + 43

843

283

= + =

or s3/2 = 283

32

× = 14 or s = 142/3 = 5.81

Now acceleration, a = – 8s–2 = – 8 × 5.81–2 = 0.2373 m/s2. Ans.

��

The motion between two movingbodies is known as relative motion.

The motion of a moving body withrespect to another moving body, is knownas the relative motion of the first bodywith respect to second body.

Relative Motion between TwoBodies Moving in Straight Lines

Fig. 4.3 (a) shows the two bodiesA and B which are moving along parallellines in the same direction.

Let vA = Absolute velocity of AvB = Absolute velocity of B

Let vA > vB. To find the relativevelocity of A with respect to B, make thebody B stationary. This is possible if avelocity equal to vB is applied on the bodyB in the opposite direction as shown inFig. 4.3 (b). Then we shall have to applya velocity equal to vB on the body. A alsoin the opposite direction. The resultantof these two velocities acting on A will

(b)

vA

vB

(a)

B

A

vB B vB

vB A

vAB

vA

o b a

vA

vB

vAB

(c)

Fig. 4.3


NOTES

Engineering Mechanics represent the relative velocity of A with respect to B. As vA > vB, hence the resultantvelocity will be in the direction of vA and equal to (vA – vB). This resultant velocity isthe relative velocity of A with respect to B.

The relative velocity of A with respect to B in vector form is also given by,vAB = Vector difference of vA and vB

= v vA B

→ →− ...(i)

Graphically, the relative velocity of A with respect to B can also be determinedas [Refer to Fig. 4.3 (c)] given below :

(i) Take any point o. From o, draw oa parallel to velocity vA and take oa = vA tosome suitable scale.

(ii) From o, also draw ob parallel to velocity vB. Take ob = vB. Joint b to a.Then ba represents the relative velocity of A with respect to body B in

magnitude and direction. Hence relative velocity of A with respect to B may also bewritten in the vector form. [Refer to Fig. 4.3 (c)].

ba oa ob→ → →

= − .

Please note that the velocity of A with respect to B (i.e., vAB) in vector form iswritten as ba

→ and not ab

→. And mathematically it is written as vAB. Hence to find vAB

in magnitude and direction, start from point b and reach towards point a.Similarly the relative velocity of B with respect to A is given by

vBA = Vector difference of vB and vA

= v vB A

→ →− . ...(ii)

The relative velocity of B with respect to A in vector form is written as

ab ob oa→ → →

= − .From equations (i) and (ii), it is clear that the relative velocity of point A

with respect to B(vAB) and the relative velocity of point B with respect to A(vBA)are equal in magnitude but opposite in direction i.e.,

vAB = – vBA or ba ab→ →

= − .

Relative Motion between Two Bodies Moving Along Inclined LinesFig. 4.4 (a) shows the two bodies A and B

moving along inclined lines with absolutevelocities vA and vB respectively. To find therelative velocity of A with respect to B, make thebody B stationary. This is possible if a velocityequal to vB is applied on the body B in the op-posite direction as shown in Fig. 4.4 (b). Thenwe shall have to apply a velocity equal to vB onthe body A also in opposite direction.

The resultant gives the relative velocity ofA with respect to B as shown in Fig. 4.4 (b).

The vector oa in Fig. 4.4 (c) representsthe velocity vA in magnitude and direction,whereas the vector ob represents the velocityvB in magnitude and direction. Join points b

(b)

vB

vB

B

vB

vAB

A vA

A vA

B

vB

(a)


NOTES

Dynamics of Particlesand a. Then relative velocity of A with respect to B iswritten as vAB and is given by

vAB = Vector ba = v vA B

→ →−

and relative velocity of B with respect to A is given by,

vBA = Vector ab.

Problem 8. Two roads which are at an angleof 60°, intersect at a point O. A car 1 is moving with avelocity of 72 km/hr away from point O along x-axis atan instant of time. At the same time another car 2 ismoving away from point O with a velocity of 48 km/hralong the road 2.

Determine the velocity of car 1 with respect tocar 2.

Sol. Given :Velocity of car 1 along x-axis = 72 km/hr

= 72 1000

60 60×

× = 20 m/s

The car 1 is moving along x-axis, there is no component of velocity along y-axis.

Hence the velocity of car 1 in vector form is written asV1 = 20i(As velocity is along x-axis only, hence only i component exists)

Velocity of car 2 along road 2, which is at an angle of 60° with x-axis = 48 km/hr.

= 48 100060 60

××

= 13.33 m/s

The components of this velocity along x-axis = 13.33 cos 60° = 6.66 m/s andalong y-axis = 13.33 sin 60° = 11.54 m/s.

Hence the velocity of car 2 in vector form is written asV2 = (13.33 cos 60°)i + (13.33 sin 60°)j

= 6.66i + 11.54jNow the velocity of car 1 with respect to car 2 is given by

V12 = V1 – V2 = (20i) – (6.66i + 11.54j)= (20 – 6.66) i – 11.54j = 13.34i – 11.54j. Ans.

The velocity of car 1 with respect to car 2 is also known as the relative velocityof car 1 with respect to car 2.

The magnitude of this relative velocity

= 13.34 11.54 177.95 133.172 2+ = + = 17.64 m/s. Ans.

The direction of this relative velocity is given by,

tan θ = − 11.5413.34

( )( )

yx

-axis component-axis component

= – 0.865

Fig. 4.5

Road 2

Car 2

60°

Car 1O Road 1

vA

oa

vB vAB

(c)Fig. 4.4


NOTES

Engineering Mechanics As tan θ is negative, hence θ should be in second or fourth quadrant. If θ is insecond quadrant, then x-component should be – ve and y-component should be positive.But here y-component is negative, this means θ lies in fourth quadrant.

∴ θ = tan–1 (– 0.865) = 40.86° or (360 – 40.86) = 319.14°. Ans.Alternate method (Graphical Method)(i) Take any point o. From o draw a line oa parallel

to the velocity of car 1 or V1 in magnitude and direction.Cut oa = 20 m/s.

(ii) From o, also draw line ob parallel to the velocityof car 2 or V2 in magnitude and direction. Cut ob = 13.33m/s. Join b to a. Then vector ba is the relative velocityof car 1 with respect to car 2.

By measurement vector ba = 17.6 m/s and angleα = 40.86°.

∴ θ = 360 – 40.86 = 319.14°. Ans.Note. For graphical method both the velocities should be directed either away from

the origin (point o) or towards the origin.

Problem 9. If in the previous problem, the car 1 is at a distance of 100 m fromO along x-axis and is moving with a speed of 72 km/hr towards O at the time ofobservation and car 2 is moving with a speed of 48 km/hr. along road 2 and awayfrom the origin as shown in Fig. 4.7, then find the velocity of car 1 with respect tocar 2.

Sol. Given :Distance of car 1 from O along x-axis = 100 m

Speed of car 1 = 72 km/hr = 72 1000

60 60×

× = 20 m/s

∴ Velocity of car 1 = 20 m/sThe car 1 is moving along x-axis and towards

origin.Hence the velocity of car 1 in vector form is

written as V1 = – 20i [– ve sign is due to

opposite direction of car 1]Velocity of car 2 = 48 km/hr

= 48 100060 60

××

= 13.33 m/s

The velocity of car 2 in vector form is writtenas

V2 = 13.33 [cos 60i + sin 60j]= 6.66i + 11.54j

The velocity of car 1 with respect to car 2 is given by,V12 = V1 – V2 = – 20i – [6.66i + 11.54j]

= – 26.66i – 11.54j. Ans.Graphical Method(i) Take any point o. From o draw a line oa parallel to velocity V1 in magnitude

and direction. Take oa = 20 m/s.

a

q

q

b

o a

V=

13.3

3m

/s

2

60° a

V12

20 m/s = V1

Fig. 4.6

Fig. 4.7

V12

V2

V1a o

b

Fig. 4.8

1 x

60°

2

O100 m

y

y′

x′


NOTES

Dynamics of Particles(ii) From o, also draw a line ob parallel to velocity V2 in magnitude and direction.Take ob = 13.33 m/s. Now vector ba represents the velocity of 1 with respect to 2 inmagnitude and direction. Measure vector ba.

Problem 10. At an instant of time, a car 1 at a distance of 120 m from pointo along x-axis moving at a velocity of 72 km/hr decelerates at 1.5 m/s2 as it approachesthe point o. At the same instant, a car 2, is moving along a road which is at an angleof 60° with the road on which car 1 is moving. This car is moving away from o witha velocity of 48 km/hr and accelerates at 2 m/s2 as shown in Fig. 4.9. Find the velocityof car 1 with respect to car 2 after three seconds.

Sol. Given :Car 1. Initial velocity,

u1 = 72 km/hr

= 72 1000

60 60×

× = 20 m/s

= – 20i m/s (in vector notation)(– ve sign is due to car 1 is

moving towards o along x-axis)Deceleration = 1.5 m/s2 (towards o)

or Acceleration, a = 1.5 m/s2

(away from o along x-axis)= 1.5i m/s2 (in vector notation)

(+ ve sign is as the acceleration is actingalong x-axis away from o)

Time, t = 3 secondsLet v1 = Final velocity of car 1 after 3 secondsThen v1 = u1 + at = (– 20i) + (1.5i) × 3 (� u1 = – 20i ; a1 = 1.5i)

= – 20i + 4.5i = – 15.5i∴ V1 = – 15.5i [Final velocity of car 1, in vector notation]

Car 2. Initial velocity, u2 = 48 km/hr = 48 100060 60

×× = 13.33 m/s

This velocity is at an angle of 60° with x-axis. Hence in vector notation, it isgiven by

u2 = 13.33 (cos 60°)i + 13.33 (sin 60°)j= 6.67i + 11.54j

Acceleration, a = 2 m/s2. This acceleration is at an angle of 60° with x-axis.Hence this acceleration in vector form is given by,

a = 2(cos 60°)i + 2 sin (60°)j = 1.0i + 1.732jTime t = 3s

Let v2 = Final velocity of car 2 after 3 seconds.Then v2 = u2 + at

= (6.67i = 11.54j) + (1.0i + 1.732j) × 3= 6.67i + 11.54j + 3i + 5.196j = 9.67i + 16.736j

or V2 = 9.67i + 16.736jNow the velocity of car 1 with respect to the velocity of car 2 after 3 seconds

V12 = V1 – V2 = – 15.5i – [9.67i + 16.736j]= – 25.17i – 16.736j m/s. Ans.

Fig. 4.9

1 x

y

60°

2

o120 m


NOTES


The motion of a body along a circular path is known as circular motion. In circularmotion, the centre of rotation remains fixed. The examples of the bodies moving in acircular path are : shafts, flywheels, pulleys etc., rotating about their geometric axis.

Angular Velocity. It is defined as the rate of change of angular displacementof a body. Angular displacement is always measured in terms of angle covered bythe body from the initial position.

Let a body is moving along a circular pathas shown in Fig. 4.10. Let initially the body is at Aand after time ‘t’, the body is at B. Let ∠AOB = θ.

Then angular displacement = ∠AOB = θ.Time taken = t.∴ Angular velocity

= Angular displacement

Time= θ

t

Mathematically, angular velocity = ddt

θ .

Angular velocity is denoted by the symbol, ω.

∴ ω = ddt

θ...(4.9)

It is measured in radians per second and is written as rad/s.Relation Between Linear Velocity and Angular Velocity. Consider the

body moving in a circle as shown in Fig. 4.10. The initial position of the body is at Aand after time ‘t’ the body is at B. The angle AOB is equal to θ.

Angular velocity = θt

.

Let V = linear velocity = Linear displacement

TimeBut linear displacement

= Arc AB = OA × θ = r × θ (� OA = radius of circle = r)

∴ V = r

t× θ

= r × Angular velocity �

θt

=��

��

Angular velocity

= r × ω ...(4.10)where ω = Angular velocity.

Angular Acceleration. It is defined as the rate of change of angular velocity.It is measured in radians per second per second and written as rad/s2. Mathematically,angular acceleration (α) is given as

α = Rate of change of angular velocity

= ddt

ddt

ddt

ω θ= �� ω θ=�

��

ddt

= d

dt

2

2θ ...(4.11)

Alsoddtω

= dd

ωθ

× ddt

θ =

dd

ωθ

× ω �

d

dt

θω=�

��

Fig. 4.10. Body moving in a circle

O

B

Aq


NOTES

Dynamics of ParticlesBut

ddtω

= angular acceleration = α

∴ α = ddtω

= ω ωθ

dd

...(4.12)

Relation Between Linear Acceleration and Angular Acceleration. Fromequation (4.10), we have V = ω × r.

Differentiating the above equation w.r.t. ‘t’

dVdt

ddt

= (ωr)

or = rddtω

(� r = constant)

ButdVdt

= linear acceleration = a

ddtω

= angular acceleration = α

Substituting these values in the above equation, we get a = rα ...(4.13)∴ Linear acceleration is equal to ‘r’ times the angular acceleration.

��

Consider a body, which is moving along a circular path.Let ω0 = Final angular velocity of the body in radians per second

ω = Final angular velocity in radians per second

t = Time in second during which angular velocity changes from ω0 to ωθ = Angular displacement or angle traversed in radians

α = Angular acceleration in radians/s2.

We know that angular acceleration (α) is the rate of change of angular velocity.Hence

α = Rate of change of angular velocity = Change of velocity

Time

= (Final angular velocity – Initial angular velocity)

Time

= ( – 0ω ω )

t...(4.14)

or αt = ω – ω0

∴ ω = ω0 + at ...(4.15)Equation for Angular Displacement (θ). We know average angular velocity

= (Initial Final angular velocity)

2+

= ω ω

ω0 +

.

∴ Angular displacement, θ = (Average angular velocity) × Time

= ( )

.ω ω0

2+

× t


NOTES

Engineering Mechanics Substituting the value of ω from equation (4.15) in the above equation, we get

θ =( ) ( )ω ω α ω α0 0 0

22

2+ +

× =+

×t

tt

t

= (ω0 + 12

αt)t = ω0t + 12 αt2

θ = ω0t + 12 αt2 ...(4.16)

Derivation of the Angular Displacement in Terms of Initial and FinalAngular Velocities. The final angular velocity is given by the equation (4.15)

ω = ω0 + αt or ω ω

α– 0 = t

Substituting the above value of t in the equation (4.16), we get

θ = ω0 × ω ωα

αω ω

α– –0 0

212

��

��

+ ��

��

= (ω – ω0) ωα α

ω ω00

12

+�

�

��( – )

= (ω – ω0) 2

20 0ω ω ω

α+�

�

��

– = (ω – ω0)

ω ωα

ω ωα

+��

��

=02

02

2 2–

or ω2 – ω02 = 2αθ ...(4.17)

Angular displacement in the nth second = ω0 + 2 1

2n –

.��

��

α

Relationship between r.p.m. (N) and Angular Velocity (ωωωωω)..... Let N = r.p.m.of a body ω = angular velocity of the body.

Now the number of revolution in one minute or 60 seconds = N.

∴ Number of revolution in one second = N60

.

But in one revolution, the body covers an angle equal to 360° or 2π radians.∴ Angle covered by the body in one second

= Angle covered in one revolution× Number of revolution per second

= 2π × N N60

260

= π .

But angle covered per second = Angular velocity = ω.

∴ ω = 260πN

...(4.18)

Problem 11. A body is rotating with an angular velocity of 5 radians/s. After4 seconds, the angular velocity of the body becomes 13 radians/s. Determine theangular acceleration of the body.

Sol. Given :Initial angular velocity, ω0 = 5 rad/sFinal angular velocity, ω = 13 rad/sTime, t = 4 sLet α = Angular acceleration of the body.Using equation (5.15),

ω = ω0 + αt or 13 = 5 + α × 4

∴ α =( – )13 5

484

= = 2.0 rad/s2. Ans.


NOTES

Dynamics of ParticlesProblem 12. The angle of rotation of a body is given as a function of time bythe equation,

θ = θ0 + at + bt2

where θ0 initial angular displacement, a and b are constants.Obtain general expressions for : (a) the angular velocity and (b) the angular

acceleration of the body.

If the initial angular velocity be 3π radian per second and after two seconds theangular velocity is 8π radian per second, determine the constants a and b.

Sol. Given :θ = θ0 + at + bt2.

Angular velocity (ω) is obtained by differentiating the above equation with respectto time (t).

∴ ω =ddt

ddt

θ = (θ0 + at + bt2) = a + 2bt ...(i)

� θθ

00 0is constant and hence

ddt

=��

��

(i) When t = 0, ω = 3πSubstituting these values in equation (i), we get

3π = a + 2b × 0 = a∴ a = 3π. Ans.(ii) When t = 2 seconds, ω = 8π radians.Substituting these values in ω = a + 2bt, we get

8π = a + 2b × 2 = a + 4b = 3π + 4b (� a = 3π)∴ 4b = 8π – 3π = 5π

or b =54π

= 1.25π. Ans.

(a) General expression for angular velocity

General expression for angular velocity is obtained by substituting the values ofa and b in equation (i).

∴ ω = a + 2bt

= 3π + 2 × 1.25π × t (� a = 3π, b = 1.25π)= 3π + 2.5πt. Ans.

(b) General expression for angular accelerationAngular acceleration (α) is obtained by differentiating the equation (i) with

respect to time (t).

∴ α =ddt

ddt

ω = (a + 2bt) = 2b � adadt

is constant, hence 0=��

��

= 2 × 1.25π (� b = 1.25π)= 2.50π. Ans.

��

When a body is at rest or moving in a straight line or rotating about an axis,the body obeys certain laws of motion. These laws are called Newton’s laws of motion.There are three laws of motion. These laws for linear motions and angular motionsare given in the next articles.


NOTES

Engineering Mechanics Newton’s Laws for Linear MotionFirst law. It states that a body continues in its state of rest or of uniform mo-

tion in a straight line unless it is compelled by an external force to change that state.Second law. It states that the rate of change of momentum of a body is pro-

portional to the external force applied on the body and takes place in the direction ofthe force.

Third law. It states that to every action, there is always an equal and oppo-site reaction.

Before discussing Newton’s laws of motion, let us define certain terms like mass,weight and momentum.

1. Mass. The quantity of matter contained in a body is known as mass of thebody. Mass is a scalar quantity. In C.G.S. units, the mass is expressed in gramme(gm) whereas in S.I. units the mass is expressed in kilogramme (kg).

2. Weight. Weight of a body is defined as the force, by which the body isattracted towards the centre of the earth. Mathematically weight of a body is givenby

Weight = Mass × Acceleration due to gravity = Mass × g ...(4.19)If mass is taken in kilogram (kg) and acceleration due to gravity in

metre per second square (m/s2), then weight is expressed in newton (N). But if massis taken in gramme (gm) and acceleration due to gravity in centimetre per secondsquare (cm/s2), then weight is expressed in dyne. The relation between newton (N)and dyne is given as

One Newton = 105 dyne.3. Momentum. The product of the mass of a body and its velocity is known as

momentum of the body. Momentum is a vector quantity. Mathematically, momentumis given by

Momentum = Mass × Velocity.Newton’s First Law of Motion. It consists of two parts. First part states

that a body continues in its state of rest unless it is compelled by an external forceto change that state. A book lying on a table remains at rest, unless it is lifted bysome external force.

Second part states that a body continues in its state of uniform motion in astraight line unless it is compelled by an external force to change that state. In actualpractice, we see that when a body is moving with a uniform velocity in a straightline, the body does not continue in its state of uniform motion but comes to rest aftersome time. This is due to frictional force acting on the body. For an ideal case (i.e.,when there is no frictional force acting on the body), the body will continue to movewith uniform velocity in a straight line, unless compelled by an external force tochange that state.

Newton’s Second Law of Motion. This law enables us to measure a force.Let a body of mass ‘m’ is moving with a velocity ‘u’ along a straight line. It is actedupon by a force F and the velocity of the body becomes v in time t. Then we have

u = Initial velocity of the body,

v = Final velocity of the body,m = Mass of the body,a = Uniform linear acceleration,


NOTES

Dynamics of ParticlesF = Force acting on the body, which changes the velocity u to v in time t,

t = Time in second to change the velocity from u to v. Initial momentumof the body

= Mass × Initial velocity = m × u

Final momentum of the body = m × v.

∴ Change of momentum

= Final momentum – Initial momentum = mv – mu = m(v – u)

Rate of change of momentum = Change of momentum

Time= m v u

t( – )

...(i)

But from equation (4.4),v u

t–

= a (i.e., linear acceleration)

Substituting the value of v u

t–�

�� in equation (i), we get

Rate of change of momentum = m × a.

But according to Newton’s second law of motion, the rate of change ofmomentum is directly proportional to the external force acting on the body.

∴ F ∞ m × a or F = k × m × a ...(ii)

where k is a constant of proportionality.

In equation (ii), k and m (mass of a body) are constants for a given body andhence force acting on a body is proportional to the acceleration produced by the force.This means that for a given body, greater force products greater acceleration while asmaller force produces smaller acceleration. The acceleration produced will be zeroif no force is applied on the body.

Two important conclusions are drawn from the first two Newton’s laws ofmotion :

(i) There will be no acceleration, if no external force is applied on the body.This means the body will continue in its state of existing uniform motion in a straightline.

(ii) Force applied on the body is proportional to the product of mass of the bodyand the acceleration produced by the force.

Unit of Force. Let us first define a ‘unit force’. A unit force can be suitablydefined so as to make the value of k in equation (ii) equal to one. A unit force (i.e.,Force = 1.0) is that which produces unit acceleration on an unit mass. Then bysubstituting F = 1.0, m = 1.0 and a = 1.0 in equation (ii) (i.e., F = k × m × a), we get

1 = k × 1 × 1 or k = 1.

Substituting the value of k = 1, in equation (ii), we get

F = m × a ...(4.20)

(i) If mass (m) = 1 kg and acceleration produced (a) = 1 m/s2 the unit of forceis known as newton (which is written as N). Thus newton is defined as that forcewhich acts on a body of mass one kg and produces an acceleration of 1 m/s2 in thedirection of force. Newton is the unit of force in S.I. system.

∴ 1 N = 1 kg × 1 m/s2 = 1 kg-m/s2.

(ii) If mass (m) = 1 gm and acceleration (a) = 1 cm/s2, then unit of force is knownas ‘dyne’. Thus a dyne may be defined as that force which acts on a body of mass one


NOTES

Engineering Mechanics gm and produces an acceleration of 1 cm/s2. Dyne is the unit of force in C.G.S. system.

∴ 1 dyne = 1 gm × 1 cm/s2 = gm cm/s2

By definition,1 N = 1 (kg) × 1 (m/s2) = 1 × 1000 (gm) × 1 × 100 (cm/s2)

= 105 (gm cm/s2)= 105 dyne (� 1 dyne = 1 gm × 1 cm/s2 = gm cm/s2)

Note. (i) The body will have acceleration if the external force is acting on the body inthe direction of motion of the body.

(ii) The body will have retardation if the external force is acting opposite to thedirection of motion of the body.

Problem 13. A force of unknown magnitude acts on a body of mass 150 kgand produces an acceleration of 3 m/s2 in the direction of force. Find the force.

Sol. Given :

Mass of the body, m = 150 kg

Acceleration, f = 3 m/s2

The force is given by equation (4.20),

Hence F = m × a = 150 (kg) × 3 (m/s2)

= 450 (kg-m/s2) = 450 N. Ans. (� kg-m/s2 = N)

Problem 14. A train of weight 1960 kN starts from rest and attains a speedof 120 km/hr in 5 minute. If the frictional resistance of the track is 10 N per kN ofthe train’s weight, find the average pull required. Take g = 9.8 m/s2.

Sol. Given :Weight of train, W = 1960 kN = 1960 × 1000 N

∴ Mass of train, m = Wg

= ×1960 10009.80

kg

Initial velocity of train, u = 0

Final velocity, v = 120 km/hr = 120 100060 60

××

m/s = 1003

m/s

Time in second, t = 5 × 60 = 300 seconds

Let a = Acceleration of train

Using the relation,

v = u + a × t or 1003

= 0 + a × 300 = 300 a

∴ a =1003

1300

19

× = m/s2

Let F* = Average pull required in newton

F1 = Frictional resistance in newton

F = Net force acting on the engine in the direction of motion.

Then F = F* – F1 ...(i)

But F1 = 10 N per kN of train’s weight

= 10 N × Weight of train in kN = 10 × 1960 = 19600 N.

Substituting this value in equation (i), we getF = (F* – 19600)


NOTES

Dynamics of ParticlesNow using equation (4.20), we getF = m × a

or (F* – 19600) =1960 1000 1

9× ×

9.80� m a= × =��

��

1960 1000 199.81

and

= 22222.22 N∴ F* = 22222.22 + 19600 = 41822.22 N = 41.822 kN. Ans.Motion on an Inclined Smooth Surface.

Fig. 4.11 shows a body of weight W, sliding down on asmooth inclined plane.

Let θ = Angle made by inclined plane withhorizontal

W = Weight of the bodya = Acceleration of the body

m = Mass of the body

= Wg

.

As the surface of the plane is smooth, hence the frictional force will be zero.Hence the forces acting on the body are its own weight W and reaction R of the plane.The resolved part of W perpendicular to the plane is W cos θ, which is balanced byR, while the resolved part parallel to the plane is W sin θ, which produces accelera-tion down the plane. This force is responsible for the movement of the body downthe plane.

∴ Net force acting on the body down the plane isF = W sin θ

Now using the equation (4.20), we haveF = m × a.

Substituting the values of F and m in the above equation, we get

W sin θ = Wg

× a

∴ a = g sin θ ...(4.21)If the body is moving up the plane, the corresponding acceleration will be – g

sin θ.Motion on an Inclined Rough Surface. Fig. 4.12 shows a body of weight

W, sliding down the rough inclined surface.Let a = Acceleration of the body

m = Mass of the body = Wg

θ = Inclination of the plane with horizontalµ = Co-efficient of friction

F1 = Force of friction.As the body is moving down the plane, the

force of friction will be acting up the plane as shown inFig. 4.12.

Now force of friction, F1 = µR= µ × W cos θ (� R = W cos θ)

Force acting down the plane, F2 = W sin θ

θ

θ

WW cos θ

R

W si

nθ

Fig. 4.11

θ

θ

W

W cos θ

R

Wsin

θ

F=

R

1

µ

Fig. 4.12


NOTES

Engineering Mechanics ∴ Net force acting on the body down the plane,F = F2 – F1 = W sin θ – µ W cos θ.

Now using the equation (4.20), F = m × a

or (W sin θ – µ W cos θ) = Wg

× a � mWg

=��

��

or W (sin θ – µ cos θ) = Wg

× a or a = W (sin θ – µ cos θ) × g

W

∴ a = g(sin θ – µ cos θ). ...(4.22)Problem 15. A body of weight 200 N is initially stationary on a 45° inclined

plane. What distance along the inclined plane must the body slide, before it reaches aspeed of 2 m/s. The co-efficient of friction between the body and the plane = 0.1.

Sol. Given :Weight of body, W = 200 N

∴ Mass of body, m = Wg

= 2009.81

kg

Angle of plane, θ = 45°Initial velocity, u = 0Final velocity, v = 2 m/sCo-efficient of friction, µ = 0.1.The acceleration of the body is given by

equation (4.22) as a = g[sin θ – µ cos θ] = 9.81[sin 45° – 0.1 cos 45°]

= 9.81[0.707 – 0.1 × .707] = 6.242 m/s2.Now using the relation,

v2 – u2 = 2as or 22 – 02 = 2 × 6.242 × s

∴ s = 2 2××2 6.242

= 0.32 m = 32 cm. Ans.

Problem 16. A truck weighing 6 kN just moves freely (engine is not running)at 36 kilometre per hour down a slope of 1 in 40, the road resistance at this speed justbeing sufficient to prevent any acceleration. Find the road resistance per kN weight oftruck.

What power will the engine have to exert to run up the same slope at double thespeed when the road resistance remains the same?

Sol. Given :Weight of truck, W = 6 kN = 6 × 1000 N

= 6000 NSpeed of truck, u = 36 km/hr

=36 1000

60 60×

× = 10 m/s

Slope of the road = 1 in 40

∴ sin θ =1

401st Case. Fig. 4.14 shows the position of the truck, when it is moving down

the plane. The road resistance (F1) is acting in the opposite direction of the motion ofthe truck. The truck is not having any acceleration and hence it is moving with aconstant velocity of 10 m/s. Hence the net force on the truck in the direction of motionshould be zero. But net force on the truck in the direction of motion (See Fig. 4.14).

Fig. 4.13

Fig. 4.14

q

R

200

sin45

°

F=

R

1

m

200 cos 45°

200 N45°

q

F1

W cos q

W

W sin q

q

R


NOTES

Dynamics of Particles

W cos θθ

F1W sin θ

P

W

Fig. 4.15

= W sin θ – F1

= 6000 × 140

– F1� W = = ��

��

60001

40N, sin θ

∴ According to the given condition,

6000 × 140

– F1 = 0

∴ F1 = 6000 × 140

600040

= = 150 N

∴ Frictional force per tonne weight of truck

=F1 150

6Weight of truck in kN= = 25 N. Ans.

2nd Case. Truck is moving up an inclined plane of slope 1 in 40 with doublespeed. Road resistance is same.

∴ Speed of truck = 2u = 2 × 10 = 20 m/s

sin θ = 140

.

Road resistance = F1 = 150 N.The truck is moving at constant speed of

20 m/s up the plane.Let P = Force exerted by engine up the

plane.As the truck is moving with uniform speed, the net force on the truck along

the plane should be zero (See Fig. 4.15).or P – W sin θ – F1 = 0

∴ P = W sin θ + F1 = 6000 × 140

+ 150 = 150 + 150 = 300 N

Power exerted by engine

= (Force exerted by engine in N) Speed of engine in m s1000

×

= ( )300 20

1000300 20

1000N × = ×m/s

= 6 kW. Ans.

Analysis of Lift Motion. Fig. 4.16 shows a lift(elevator or cage) carrying some weight and moving witha uniform acceleration.

Let W = Weight carried by the lift

m = Mass carried by lift = Wg

a = Uniform acceleration of the lift

T = Tension in cable supporting the lift. Thisis also called the reaction of the lift.

The lift may be moving upwards or movingdownwards.

Pulley

Force

W

T

Lift

Fig. 4.16


NOTES

Engineering Mechanics 1st Case. Let the lift is moving upwards asshown in Fig. 4.17(a). The weight carried by lift is actingdownwards while the tension in the cable is actingupwards. As the lift is moving up, the net force which isequal to (T – W) is acting upwards.

∴ Net force in upward direction = T – W.This net force produces an acceleration ‘a’Hence using,

Net force = Mass × Acceleration

or (T – W) =Wg��

× a � Mass =��

��

Wg

or T = W + Wg

× a = W 1 +��

��

ag

...(4.23)

2nd Case. As the lift is moving downwardsas shown in Fig. 4.17(b), the net force is acting down-wards. Hence in this case W is more than T (tension instring).

∴ Net force in downward direction = (W – T).This net force produces an acceleration ‘a’.Hence using, Net force = Mass × Acceleration

or (W – T) =Wg

× a

or T = W – Wg

× a

= W 1 –ag

��

��

...(4.24)

Problem 17. A lift carries a weight of 100 N and is moving with a uniformacceleration of 2.45 m/s2. Determine the tension in the cables supporting the lift, when :

(i) lift is moving upwards, and

(ii) lift is moving downwards. Take g = 9.80 m/s2.Sol. Given :Weight carried by lift, W = 100 N.Uniform acceleration, a = 2.45 m/s2.(i) Lift is moving upwardsLet T = Tension in the cables supporting the lift.

Using equation (4.23), T = W 1 +��

��

ag

= 100 1 +��

��

2.459.80

= 100 (1.25) = 125 N. Ans.(ii) Lift is moving downwards

Using equation (4.24), T = W 1 –ag

��

��

= 100 1 −��

��

2.459.80

= 100(1 – 0.25)

= 100 × .75 = 75 N. Ans.

Lift

Cable supportingthe lift

W

T

Lift

W

T

Fig. 4.17. (a) Lift is movingupwards

Fig. 4.17. (b) Lift is movingdownwards


NOTES

Dynamics of ParticlesProblem 18. An elevator weighing 5000 N is ascending with anacceleration of 3 m/s2. During this ascent its operator whose weight is 700 N is stand-ing on the scales placed on the floor. What is the scale reading ? What will be thetotal tension in the cables of the elevator during this motion ?

Sol. Given :Weight of the elevator, W1 = 5000 NAcceleration of elevator, a = 3 m/s2

Weight of the operator, W2 = 700 NWhen the operator is standing on the

scale, placed on the floor of the elevator, thereading of the scale will be equal to the reaction(R) offered by the floor on the operator.

Hence let R = Reaction offered by flooron operator. This is alsoequal to the reading ofscale.

T = Total tension in the cablesof elevator.

Consider the motion of operator. The operator is moving upwards along withthe elevator with an acceleration. a = 3 m/s2. The net force on the operator is actingupwards.

∴ Net upward force on operator= Reaction offered by floor on operator

– Weight of operator= (R – 700)

Mass of operator =Weight of operator

g g= 700

But, Net force = Mass × Acceleration

∴ (R – 700) =7009 8.

× 3 (� Acceleration = 3 m/s2)

∴ R = 700 + 7009 8.

× 3 = 700 + 214.28 = 914.28 N. Ans.

Total tension in the cables of elevator.Let T = Total tension in the cables of elevator

W = Total weight (i.e., weight of elevator + weight of operator)= 5000 + 700 = 5700 N.

As the elevator with the operator is moving upwards with an accelerationf = 3 m/s2, the net force will be acting on the elevator and operator in the upwarddirection.

∴ Net upward force on elevator and operator= Total tension in the cables – Total weight of elevator and operator= (T – 5700)

Mass of elevator and operator

=Total weight

g= 5700

9 80.

Ele

vato

r

T

Cable

Operator

R5000 N

Fig. 4.18


NOTES

Engineering Mechanics But, Net force = Mass × Acceleration

∴ (T – 5700) =57009 80.

× 3

∴ T = 5700 + 57009 80.

× 3 = 5700 + 1745 = 7445 N. Ans.

Analysis of the Motion of Two Bodies Con-nected by a String. Fig. 4.19 shows a light andinextensible string passing over a smooth andweightless pulley. Two bodies of weights W1 and W2are attached to the two ends of the string. Let W1 begreater than W2. As W1 > W2, the weight W1 will movedownwards, whereas the smaller weight (W2) willmove upwards. For an inextensible string, the upwardacceleration of the weight W2 will be equal to thedownward acceleration of the weight W1.

As the string is light and inextensible andpassing over a smooth pulley, the tension of the stringwill be the same on both sides of the pulley.

Let T = Tension in both strings,

a = Acceleration of the bodies.

Consider the motion of weight, W1. The weight W1 is moving downwards withan acceleration a. The forces acting on W1 are (i) its weight W1 acting downwardsand (ii) tension T acting upwards. As the weight W1 is moving downwards, hencenet force on the weight W1, is acting downwards.

∴ Net downward force = (W1 – T)

But net force = Mass × Acceleration

∴ (W1 – T) = Wg

1 × f � MassWeight

= =��

��g

Wg

1 ...(i)

Now consider the motion of weight, W2. The forces acting on W2 are : (i) itsweight W2 acting downwards and (ii) tension T acting upwards. But the weight W2is moving upwards, hence net force on weight W2, is acting upwards.

∴ Net upward force = (T – W2)But net upward force = Mass × Acceleration

or (T – W2) = Wg2 × a � Mass =

��

��

Wg2 ...(ii)

Adding equation (i) and (ii), we get

(W1 – W2) = ag

(W1 + W2)

or a = g W WW W( – )

( )1 2

1 2+...(4.25)

Equation (4.25) is used for finding the acceleration. If the value of thisacceleration is substituted either in equation (i) or in equation (ii) the value of tension(T) is obtained.

Hence substituting the value of ‘a’ in equation (ii), we get

(W1 – T) = Wg

gW WW W

W W WW W

1 1 2

1 2

1 1 2

1 2×

+=

+( – )( )

( – )( )

(Cancelling g)

Fig. 4.19

T

T

W2

W1

SmoothPulley

Inextensiblelight string


NOTES

Dynamics of Particlesor T = W1 –

W W WW W

WW WW W

1 1 2

1 21

1 2

1 21

( – )( )

–( – )( )+

=+

�

�

��

= W1 W W W W

W WW W

W W1 2 1 2

1 2

1 2

1 2

2+ ++

�

�

�� =

+–

( ) ( )...(4.26)

Problem 19. Two bodies of weight 50 N and 30 N are con-nected to the two ends of a light inextensible string. The string ispassing over a smooth pulley. Determine :

(i) The acceleration of the system, and(ii) Tension in the string. Take g = 9.80 m/s2.Sol. Given :Heavier weight, W1 = 50 NLighter weight, W2 = 30 NLet a = Acceleration of the system, and

T = Tension in the string.(i) Using the equation (4.25) for acceleration,

a =g W WW W( – )

( ). ( – )( )

1 2

1 2

9 80 50 3050 30+

=+

=9.8 20×

80 = 2.45 m/s2. Ans.

(ii) Using equation (4.26) for tension in the string,

T =2 2 50 30

50 302 50 30

801 2

1 2

W WW W( ) ( )+

=× ×

+=

× × = 37.5 N. Ans.

Problem 20. A pulley whose axis passes through the centre O, carriesload as shown in Fig. 4.21. Neglecting the inertia of the pulley and assuming that thecard is inextensible, determine the acceleration of the block A. How much weight shouldbe added to or taken away from the block A, if the acceleration of the block A is requiredto be g/3.0 downwards ?

Sol. Given :Bigger load, W1 = 700 NSmaller load, W2 = 500 NLet a = Acceleration of block A or the acceleration of

the system.Using equation (4.25),

a =g W W

W W( – )

( )1 2

1 2+

=g( – )( )

700 500700 500+ = g

2001200

= g6

. Ans.

How much weight should be added to or taken awayfrom the block A (i.e., from bigger load 700 N) whenacceleration of bigger load is g/3.0 downwards.

Let W1* = Total weight of block A when acceleration isg/3.0

a =g3

W2 = 500 N

Fig. 4.20

T

T

30 N

50 N

T

T

500 NB

700 A

A

O

D

Fig. 4.21


NOTES

Engineering MechanicsUsing equation (4.25), a =

g W WW W( * – )

( * )1 2

1 2+(� Here W1 = W1*)

or g g W

W3500

5001

1=

+( * – )

( * )

or13

=( * – )( * )WW

1

1

500500+

(Cancelling g to both sides)

or 3(W1* – 500) = W1* + 500 or 3W1* – 1500 = W1* + 500or 2W1* = 2000

∴ W1* =2000

2 = 1000 N.

As W1* is more than W1. Hence the weight must be added to the block A.∴ Weight added = W1* – W1 = 1000 – 700 = 300 N. Ans.

Analysis of the Motion of two Bodies Connected by a String when one Body isLying on a Horizontal Surface and other is Hanging Free

1. First case when the horizontal surface is smooth and the stringis passing over a smooth pulley. Fig.4.22 shows the two weights W1

and W2connected by a light inextensible string,passing over a smooth pulley. The weightW2 is placed on a smooth horizontal surface,whereas the weight W1 is hanging free.

The weight W1 is moving downwards,whereas the weight W2 is moving on smoothhorizontal surface. The velocity andacceleration of W1 will be same as that ofW2.

As the string is light and inextensibleand passing over a smooth pulley, the tensions of the string will be same on bothsides of the pulley.

Let T = Tension in the string

a = Acceleration of the weight W1 and also of W2

(i) Consider the motion of the hanging weight W1

The weight W1 is moving downwards with an acceleration a. The forces actingon W1 are : (i) its weight W1

acting downwards, and (ii) tension T acting upwards.

∴ Net downward force = (W1 – T).

Using, Net force = Mass × Acceleration

∴ (W1 – T) =Wg

1 × a � MassWeight

=��

��g

...(i)

(ii) Consider the motion of weight W2

The weight W2 is moving on the horizontal surface with an acceleration of ‘a’.As the weight W2 is moving in the horizontal direction, the only force causing themotion is T. The weight W2 is acting downwards and hence the component of thisweight in horizontal direction is W2 cos 90°, which is zero.

W2

Smooth horizontalsurface

T

T

W1

W2 Pulley

Fig. 4.22


NOTES

Dynamics of ParticlesUsing, Force = Mass × Acceleration

or T =Wg2 × a � Mass

Weight=

��

��g

...(ii)

Adding equations (i) and (ii),

W1 =Wg

1 × a + Wg2 × a =

ag

[W1 + W2]

∴ a = g W

W W×+

1

1 2( )...(4.27)

Equation (4.27) gives the acceleration of the system.To find the tension (T) in the string, substitute the value of ‘a’ in equation (ii).

∴ T =Wg

g WW W

W WW W

2 1

1 2

1 2

1 2×

×+

=+( ) ( )

. ...(4.28)

Equation (4.28) gives the tension in the string.2. Second case when the hori-

zontal surface is rough and string ispassing over a smooth pulley. Fig. 4.23shows the two weights W1

and W2 connectedby a light inextensible string, passing overa smooth pulley. The weight W1 is hangingfree, whereas the weight W2 is placed on arough horizontal surface. Hence in this caseforce of friction will be acting on the weightW2 in the opposite direction of the motionof weight W2 as shown in Fig. 4.23.

Let µ = Co-efficient of friction be-tween weight W2 andhorizontal surface

a = Acceleration of the systemT = Tension in the string

R2 = Normal reaction at the horizontal rough surface = W2∴ Force of friction = µR2

= µW2 (� R2 = W2) ...(i)(i) Consider the motion of the hanging weight, W1The weight W1 is moving downwards with an acceleration ‘a’. The net downward

force acting on weight W1 = (W1 – T).Using, Net force = Mass × Acceleration

W1 – T =Wg

1 × a � MassWeight

=��

��g

...(ii)

(ii) Consider the motion of the weight, W2.The weight W2 is moving on the rough horizontal surface towards right with

an acceleration ‘a’. The forces acting in the horizontal directions are : (i) tension (T)towards right, and (ii) force of friction = µR2 = µW2 towards left.

∴ Net horizontal force towards right = T – µW2.

W2

R2

T

T

W1

W2

Frictionalforce

( W )µ 2

Roughsurface

Fig. 4.23


NOTES

Engineering Mechanics Using, Net force = Mass × Acceleration

or T – µW2 =Wg2 × a ...(iii)

Adding equations (ii) and (iii),

W1 – µW2 =ag

(W1 + W2)

∴ a =g W W

W W( – )( )

1 2

1 2

µ+

m/s2 ...(4.29)

Equation (4.29) is used to find the acceleration.To find the tension T, substitute the value of ‘a’ in equation (ii).

∴ W1 – T =Wg

g W WW W

W W WW W

1 1 2

1 2

1 1 2

1 2×

+=

+( – )( )

( – )( )

µ µ

∴ T = W1 – W W W

W WW

W WW W

1 1 2

1 21

2 2

1 21

( – )( )

–( – )( )

µ µ+

=+

�

�

��

= W1 W W W W

W W1 2 1 2

1 2

++

�

�

��

– ( – )( )

µ

=W

W W1

1 2( )+ ( – )W W W W1 2 1 2+ + µ

=W W W

W WW W

W W1 2 2

1 2

1 2

1 2

1( )( )

( )( )

.+

+=

++

µ µ...(4.30)

Problem 21. Two bodies of weight 20 N and 10 N are connected to the two endsof a light inextensible string, passing over a smooth pulley. The weight of 20 N is placedon a smooth horizontal surface while the weight of 10 N is hanging free in air. Find :

(i) the acceleration of the system, and(ii) the tension in the string. Take g = 9.81 m/s2.Sol. Given :Weight placed on horizontal surface, W2 = 20 NWeight hanging free in air, W1 = 10 NAcceleration due to gravity, g = 9.81 m/s2

Let a = Acceleration of the system, andT = Tension in the string.

The horizontal surface is smooth. Hence the acceleration and tension are obtainedby using equations (4.27) and (4.28).

Using equation (4.27) for acceleration, we have

a = gW

W W1

1 2

9 81 1010 20( ).

( )+= ×

+ = 3.27 m/s2. Ans.

Using equation (4.28) for tension, we have

T = W W

W W1 2

1 2

10 2010 20

20030( ) ( )+

= ×+

= = 6.67 N. Ans.

Problem 22. Two bodies of weight 10 N and 1.5 N are connected to the two endsof a light inextensible string, passing over a smooth pulley. The weight 10 N is placedon a rough horizontal surface while the weight of 1.5 N is hanging vertically in air.Initially the friction between the weight 10 N and the table is just sufficient to preventmotion. If an additional weight of 0.5 N is added to the weight 1.5 N, determine :

(i) the acceleration of the two weights, and


NOTES

Dynamics of Particles(ii) tension in the string after adding ad-ditional weight of 0.5 N to the weight 1.5 N.Take g = 9.80 m/s2.

Sol. Given :Weight placed on rough horizontal surface,

W2 = 10 NWeight hanging free in air,

W1* = 1.5 NAdditional weight added

= 0.5 N∴ Total weight hanging in air in second case

W1 = 1.5 + 0.5 = 2.0 N.Let T = Tension in string when hanging weight is 1.5 N

R = Normal reaction between the weight 10 N and the table surfaceT1 = Tension in string when hanging weight is 2.0 Na = Acceleration of the system when hanging weight is 2.0 N

Initially, the friction between the weight 10 N and the table is just sufficient toprevent motion.

∴ Max. Frictional Force, F = T = 1.5 N.But frictional force, F = µR

or 1.5 = µ × 10 (� R = W2 = 10 N)

∴ µ = 1.510

= 0.15

where µ is the co-efficient of friction.When an additional weight 0.5 N is added to the hanging weight 1.5 N, the

system starts moving with an acceleration ‘a’.(i) Acceleration of the two weightsUsing equation (4.29), we have

a =g W W

W W( – )( )

. ( . – . )( . )

1 2

1 2

9 80 2 0 0 15 102 0 10

µ+

= ×+

=9 8 0 5

12. .×

= 0.408 m/s2. Ans.

(ii) Tension in the stringUsing equation (4.30), we have

T1 =W W

W W1 2

1 2

1 2 10 1 0 152 10

( )( )

( . )( )

++

= × ++

µ(� µ = 0.15)

=20 1 15

12× .

= 1.916 N. Ans.

3. Third case when the horizontal surface is rough and the string passesover a rough pulley. Fig. 4.25 shows the two weights W1 and W2 connected by astring passing over a rough pulley. The weight W1 is hanging free, whereas the weightW2 is placed on a rough horizontal surface. Hence in this case force of friction will beacting on the weight W2 in the opposite direction of the motion of weight W2 as shownin Fig. 4.25. As the string is passing over a rough pulley, the tensions on both sides ofthe string will not be same.

W2

R

T

T

Frictionalforce

1.5 N

10 N

Fig. 4.24


NOTES


W2

R2

T1

W1

T1

T2 T2W2

Frictional force( . W )µ1 2

Fig. 4.25

Let µ1 = Co-efficient of frictionbetween weight W2 and hori-zontal surface

µ2 = Co-efficient of frictionbetween string and surfaceof pulley

T1 = Tension in the string towhich weight W1 is attached

T2 = Tension in the string towhich weight W2 is attached.

R2 = Normal reaction at thehorizontal rough surface.

Here it is equal to W2.∴ Force of friction = µ1R2 = µ1W2 (� R2 = W2) ...(i)(i) Consider the motion of the hanging weight, W1

The weight W1 is moving downward with an acceleration ‘a’. The net downwardforce acting on weight W1 is equal to (W1 – T1).


∴ (W1 – T1) =Wg

1��

× a � MassWeight

=��

��g ...(ii)

(ii) Consider the motion of the weight, W2The weight W2 is moving on the rough horizontal surface towards right with an

acceleration ‘a’. The forces acting in the horizontal direction are : (i) Tension T2towards right and (ii) force of friction = µ1R2 = µ1W2 towards left.

∴ Net force towards right = (T2 – µ1W2)Now, Net force = Mass × Acceleration

(T2 – µ1W2) = Wg2�

��

× a ...(iii)

In equations (ii) and (iii), there are three unknowns i.e., T1, T2 and ‘a’. We requireone more equation. This equation is given by,

TT

1

2= eµ×θ = eµ2×θ ...(iv)

In equation (iv), the angle θ∗ is angle of lap and is used in radians. In this

particular case, from the geometry of the Fig. 4.26, the value of θ = 90° = π2

radians.

From the three equations [i.e., equations (ii), (iii) and (iv)], three unknown (i.e.,T1, T2 and acceleration ‘a’) can be obtained.

Problem 23. Two bodies of weight 1500 N and 1000 N are connected to the twoends of a string, passing over a rough pulley. The weight of 1500 N is placed on a roughhorizontal surface while the weight 1000 N is hanging free in air. If the co-efficient ofkinetic friction is 0.20 at all contact surfaces, find :

(i) the acceleration of the system, and(ii) tensions in the string.Sol. Given :Weight of body A, W2 = 1500 NWeight of body B, W1 = 1000 N


NOTES

Dynamics of ParticlesCo-efficient of kinetic friction forall contact surfaces,

µ = µ1 = µ2 = 0.2Let a = Acceleration of the

systemT1 = Tension in the string to

which weight 1000 N isattached

T2 = Tension in the string towhich weight 1500 N isattached

We know that

TT

1

2 = eµθ = e

0.22

× π = 1.3691

where µ = µ1 = µ2 = 0.2, θ = 90° = π2

radians

∴ T1 = 1.3691T2 ...(i)Now consider the motion of body B. The body B is moving downwards with an

acceleration ‘a’ Net force = Mass × Acceleration

(1000 – T1) =10009 81.

× a

or 1000 – 1.3691T2 =10009 81.

× a (� T1 = 1.3691T2) ...(ii)

Now consider the motion of body A.The body A is moving towards right with an acceleration ‘a’ Net force = Mass

× Acceleration

(T2 – µW2) =Wg2�

�� × a or T2 – 0.2 × 1500 =

15009 81. × a

or T2 – 300 =15009 81. × a ...(iii)

or T2 =15009 81. × a + 300 ...(iv)

Substituting the value of T2 in equation (ii), we get

1000 – 1.3691 15009 81

30010009 81. .

× +��

��

=a × a

or 1000 – 1.3691 × 15009 81. × a – 1.3691 × 300 =

10009 81. × a

or 1000 – 1.3691 × 300 =10009 81. × a +

1.3691 15009.81

× × a

or 1000 – 410.73 =1000 1.3691 1500

9.81+ ×�

��

× a

or 589.27 = 315.287 × a

∴ a =589 27

311 287..

= 1.89 m/s2. Ans.

Substituting the value of ‘a’ in equation (iv), we get

T2 =15009 81.

× 1.89 + 300 = 594.6 N. Ans.

Fig. 4.26

R2

W2

T2 T2

T1

T1

1000 N

1500 N

a

A

B a

µ. µ×

R = W= 0.2 1500

2 2


NOTES

Engineering Mechanics Substituting the value of T2 in equation (i), we getT1 = 1.3691 × 594.6 = N. Ans.

Analysis of the Motion of Two Bodies Connected by a String when oneBody is Lying on Inclined Plane and the other is Hanging Free in Air

1. First case when the inclinedsurface is smooth. Fig. 4.27 shows twobodies of weights W1 and W2 connected by alight inextensible string, which passes overa smooth and weightless pulley. The weightW2 is placed on an inclined smooth plane ofinclination θ with the horizontal, whereasthe weight W1 is hanging free in air.

As the inclined plane is smooth andhence the friction between the weight W2and the inclined plane will be neglected.When the weight W1 is moving downwards,the weight W2 will be moving upwards alongthe inclined plane. The velocity andacceleration of the weight W1 will be same as that of weight W2. Since the pulley issmooth and string is light and inextensible, the tension in the string on both sides ofthe pulley will be same.

Let a = Acceleration of the system i.e., acceleration of weight W1 as well asacceleration of weight W2,

T = Tension in the string,θ = Inclination of the inclined plane.

Consider the motion of weight W1. The weight W1 is moving downwards with anacceleration ‘a’. The forces acting on W1 are : (i) its weight W1 acting downwards, and(ii) tension T acting upwards.

∴ Net downwards force = W1 – TBut net downwards force = Mass × Acceleration.

∴ (W1 – T) = Wg

1 × a � mass =

��

��

Wg

1 ...(i)

Now consider the motion of Weight W2. The weight W2 is moving upwards alongthe inclined plane with an acceleration a. The forces acting on W2 along the plane areshown in Fig. 4.28. They are :

(i) W2 sin θ downwards, and(ii) Tension T upwards.∴ Net force acting on W2 along the

plane in the upwards direction= T – W2 sin θ.


∴ T – W2 sin θ = Wg2 × a

� MassWeight

=��

��g

...(ii)

W2 W1

TT

θ

Fig. 4.27

W1

T

θ W2

θ

R

Wsin

2

θ

Fig. 4.28


NOTES

Dynamics of ParticlesAdding equations (i) and (ii)

W1 – W2 sin θ =Wg

1 × a + Wg2 × a =

ag

(W1 + W2)

∴ a =g W W

W W( – sin )

( )1 2

1 2

θ+ m/s2 ...(4.31)

Hence equation (4.31) is used for finding the acceleration of the system.To find the value of tension T, in the string, the value of a from equation (4.31)

is substituted in equation (i).

∴ W1 – T =Wg

1 × g ( – sin )

( )W W

W W1 2

1 2

θ+

=W W W

W W1 1 2

1 2

( – sin )( )

θ+

(cancelling ‘g’)

or T = W1 – W W W

W WW

W WW W

1 1 2

1 21

1 2

1 21

( – sin )( )

–( – sin )θ θ

+=

+�

�

��

= W1

W W W WW W

W W WW W

1 2 1 2

1 2

1 2 2

1 2

+ ++

�

�

�� =

++

– sin [ sin ]( )

θ θ

=W W

W W1 2

1 2

1( sin )++

θ...(4.32)

Equation (4.32) is used for finding tension T in the string.2. Second case when the inclined surface is rough. As the surface of the

inclined plane is not smooth, hence a force of friction equal to µR will be acting on theweight W2 in the opposite direction of motion of weight W2 as shown in Fig. 4.29.

Let µ = Co-efficient of frictionbetween the weight W2and inclined surface,

θ = Angle of the inclination ofthe plane,

a = Acceleration of the system,T = Tension in the string,R = Normal reaction acting on

W2.The forces acting on the weight

W2 are shown in Fig. 4.29. Equating theforces normal to the plane, we get

R = W2 cos θ.∴ Frictional force, F = µR

= µW2 cos θ ...(i)Consider the motion of weight W1. The weight W1 is moving downwards with an

acceleration a.The net downward force acting on W1 = (W1 – T).

Mass of weight W1 =Wg

1 .


∴ (W1 – T) =Wg

1 × a ...(ii)

W1

TT

θ W2

R

Wsin

2

θ F =Rµ

W cos2 θ

Fig. 4.29


NOTES

Engineering Mechanics Now consider the motion of weight W2. The weight W2 is moving upwards alongthe inclined plane with an acceleration a. The net upward force along the inclinedplane acting on weight W2

= T – W2 sin θ – µR= T – W2 sin θ – µW2 cos θ (� From (i), µR = µW2 cos θ)

Mass of weight, W2 =Wg2

Using, Net force = Mass × Acceleration

∴ T – W2 sin θ – µW2 cos θ =Wg2

× a ...(iii)

Adding equations (ii) and (iii), we get

W1 – W2 sin θ – µW2 cos θ =Wg

1 × a + Wg2 × a =

ag

[W1 + W2]

∴ a =g W W W

W W( – sin – cos )

( )1 2 2

1 2

θ µ θ+

...(4.33)

Equation (4.33) is used for finding acceleration of the system. To find tension Tin the string, the value of ‘a’ from equation (4.33) is substituted in equation (ii).

∴ W1 – T =Wg

g W W WW W

1 1 2 2

1 2×

+( – sin – cos )

( )θ µ θ

=W W W W

W W1 1 2 2

1 2

( – sin – cos )

( )

θ µ θ+

∴ T = W1 – W W W W

W W1 1 2 2

1 2

( – sin – cos )

( )

θ µ θ+

= W1 1 1 2 2

1 2–

( – sin – cos )( )

W W WW W

θ µ θ+

�

�

��

= W1 W W W W W

W W1 2 1 2 2

1 2

+ + ++

�

�

��

– sin cos

( )

θ µ θ

=W

W W1

1 2( )+ (W2 + W2 sin θ + µW2 cos θ)

=W W

W W1 2

1 2( )+ (1 + sin θ + µ cos θ) ...(4.34)

Equation (4.34) is used for finding tension in the string.Problem 24. Two bodies of weights 40 N and 15 N are connected to the two ends

of a light inextensible string, which passes over a smooth pulley. The weight 40 N isplaced on a smooth inclined plane, while the weight 15 N is hanging free in air. If theangle of the plane is 15°, determine :

(i) acceleration of the system, and(ii) tension in the string. Take g = 9.80 m/s2.Sol. Given :Weight placed on inclined plane, W2 = 40 NWeight hanging free in air, W1 = 15 NAngle of inclination, θ = 15°Acceleration due to gravity, g = 9.80 m/s2

Let a = Acceleration of the systemT = Tension in the string.


NOTES

Dynamics of ParticlesThe inclined surface is smooth. Hence the acceleration and tension are obtainedby using equations (4.31) and (4.32).

(i) Using equation (4.31) for acceleration,

a =g W W

W W( – sin )

( )1 2

1 2

θ+

=9 80 15 40 15

15 409 80 15 0 2588

55. ( – sin )

( ). ( – . )° 40

+= ×

=9 80 4 684

55. .×

= 0.828 m/s2. Ans.

(ii) Using equation (4.32) for tension,

T =W W

W W1 2

1 2

1( sin )( )

++

θ

=15 40(1 .2588)

(15 40)15 40 1.2588

55× +

+= × ×

= 13.732 N. Ans.

��

The work energy equation states that the total work done by all the forces actingon a particle moving from one position to other, is equal to change of the kinetic energyof the particle. Mathematically it is written as :

(ΣFx) . S = 12

m (v2 – u2)

where ΣFx = Sum of all the forces acting on the particlev = Final velocity of particleu = Initial velocity of particle.

To understand fully the work energy equation, let us first define work and energy.Work and Energy. Work is defined as the product of force and displacement

of the body on which force is acting. The force and displacement should be in thesame direction. Energy is defined as the capacity to do work. The energy exists inmany forms like, mechanical, electrical, heat, chemical and light etc. But inengineering mechanics, we only consider mechanical energy. This article deals withwork and energy.

Work. As defined above, work is the product of force and distance. The distanceshould be in the direction of the force. If a force P is acting on a body and the bodymoves a distance of S in the direction of the force, as shown in Fig. 4.30, then thework done on the body is given by :

S S

P q

P

(a) (b)

Fig. 4.30

Work done = Force × Distance= P × S ...(4.35)


NOTES

Engineering Mechanics But if the force acting on the body and the distance moved are not in the samedirection as shown in Fig. 4.30, then the work done on the body is given by :

Work done = Component of force in the direction of motion× Distance

= P cos θ × S ...(4.36)where θ is the angle made by the force with the direction of motion.

If θ = 90°, cos θ = 0 and hence work done will be zero. Hence if the force anddisplacement are at right angles, then work done will be zero. If a man of weight W ismoving on a horizontal road and the distance moved by man is S, then the work doneby the weight of man will be zero. This is due to the fact that W and S are at rightangle.

Units of Work. Work is the product of force and distance. In S.I. system, forceis expressed in Newton (N) whereas the distance is expressed in metre (m). Hencethe unit of work is N m.

The work done by a force of 1 N on a body which moves a distance of 1 m, iscalled 1 Nm. This is also known as one Joule (J). Hence one Joule is the work done bya force of 1 N on a body whose displacement is 1 m.

Problem 25. Find the work done in drawing a body :(i) Weighing 500 N through a distance of 5 m along a horizontal surface by a

horizontal force of 250 N.(ii) Weighing 500 N through a distance of 5 m along a horizontal surface by a

force of 200 N whose line of action makes an angle of 30° with the horizontal.Sol. Given :Weight, W = 500 NDistance, S = 5 mForce, P = 250 N(i) The forces acting on the body are shown in

Fig. 4.31.Work done = Force × Distance

= 250 × 5 = 1250 Nm. Ans.(Note that here the force applied and not the weight of the body is taken into account).

(ii) W = 500 N, P = 200 N, θ = 30°[Refer to Fig. 4.32 (b)]

Here the work done is given by equation (4.32).∴ Work done = P cos θ × Distance

= (200 × cos 30°) × 5= 200 × 0.866 × 5= 866 Nm. Ans.

Problem 26. Determine the work done by electric motor in winding up a uniformcable which hangs from a hoisting drum if its free length is 10 m and weighs 500 N.The drum is rotated by the motor.

Sol. Given :Free length of cable, L = 10 mWeight of 10 m length cable = 500 N

Weight of cable per m length =50010

= 50 N

W = 500 N

P = 250 N

Fig. 4.31

W = 500 NP = 200 N

θ

Fig. 4.32


NOTES

Dynamics of ParticlesRefer to Fig. 4.33.Consider an element of length dx at a distance

of x m from free end. This element is to be raised to aheight of (10 – x) m. The work done on this element isequal to the product of its weight and the heightthrough which it is raised.

Weight of element= Weight of cable per m length

× length of element= 50 × dx = 50dx

Work done on the element,dW = Weight of element

× Height through which it is raised= (50dx) × (10 – x) = 50 × (10 – x)dx

Total work done will be obtained by integratingthe above equation from 0 to 10 m.

∴ W =0

10

50 (10 – x) dx

= 50 102

2

0

10

xx−

�

�

�� = 50 10 10

102

2

× −�

�

��

= 50 [100 – 50] = 50 × 50 = 2500 Nm. Ans.Power. The rate of doing work is known as power. Hence, power can be

obtained by dividing the total work done by time. Or in other words, the power isthe work done per second. The unit of power in S.I. system is Nm/s or Watt. It isdenoted by W. The power in kilowatts is obtained by dividing watt by 1000.

∴ Power = Work done per second

=Force Distance

Time×

= Force × Distance

Time = Force × Velocity ...(4.37)

The force and velocity should be in the same direction.Problem 27. A train of weight 2000 kN is pulled by an engine on a level track at

a constant speed of 36 kilometre per hour. The resistance due to friction is 10 N per kNof the trains weight. Find the power of the engine.

Sol. Given :Weight of train, W = 2000 kNSpeed of train, v = 36 km/hr

=36 1000×

×60 60 = 10 m/s

Resistance due to friction,F = 10 N per kN weight of train

= 10 × weight of train in kN = 10 × 2000 = 20,000 NLet P = Force exerted by the engine to pull the train at constant speed.∴ Net force in the direction of motion = P – F = (P – 20,000) NAs engine is moving with uniform speed, the acceleration of engine will be zero.

The net force, which is equal to mass × Acceleration, will also be zero.∴ Net force = 0

or P – 20,000 = 0or P = 20,000 N

dx

(10 – x)

10 m

x

Fig. 4.33

FForce of friction(or resistance)

P

Fig. 4.34


NOTES

Engineering Mechanics Now the power is given by equation (4.37), asPower = Force exerted by engine × Velocity

= 20,000 × 10 Nm/s= 200,000 W (� Watt, W = Nm/s)= 200 kW. Ans.

Problem 28. A train of weight 2000 kN moves down a slope of 1 in 150 at18 km/hr and engine develops a power of 35 kW. If the train is pulled up at the samespeed, what power will be required to pull the train.

Sol. Given :Weight, W = 2000 kN = 2000 × 1000 NSlope = 1 in 150

∴ sin θ =1

150

Speed of train, v = 18 km/hr = 18 100060 60

×× = 5 m/s

Power developed by engine = 35 kW = 35 × 1000 W = 35000 W.1st Case. Train is moving down the plane

as shown in Fig. 4.35 with a constant velocityand hence it will not have any acceleration. Sothe net force acting on the train in the directionof motion should be zero. But the forces actingon the train are :

(i) W sin θ in the direction of motion ;(ii) Force of friction (F) in the opposite

direction of motion ;(iii) Force (P) exerted by engine in the

direction of motion.Now net force acting on the engine in the direction of motion

= W sin θ – F + P = (2000 × 1000) × 1

150 – F + P

= 13333.3 – F + PThe power developed by engine

= (Force exerted by engine) × Velocity= P × 5 or 35,000 = 5 × P

∴ P =35 000

5,

= 7000 N

Substituting the value of P in equation (i), we getNet force = 13333.3 – F + 7000But net force in the direction of motion is zero.∴ 13333.3 – F + 7000 = 0 or 13333.3 + 7000 = F or 20333.3 = F∴ F = 20333.3 N ...(ii)2nd Case. The train is moving up the same plane with the same speed as shown

in Fig. 4.36.As the plane is same and hence frictional force will be same.Let P* = Force exerted by engine while moving upNet force in the direction of motion

= P* – W sin θ – F

Direction

of motion

W sin θ F = Force

of

frictio

n

Wθ

θW cos θP

Fig. 4.35


NOTES

Dynamics of Particles= P* – (2000 × 1000) ×

1150– 20333.3

= P* – 13333.3 – 20333.3= P* – 33666.6.

The train is moving with constant speedand hence acceleration is zero. This means the netforce will also be zero.

∴ P* – 33666.6 = 0or P* = 33666.6 N

∴ Power developed by engine= Force exerted by engine × Speed= P* × v = 33666.6 × 5 Nm/s= 168333 W = 168.333 kW. Ans.

Work Done by a Torque. Fig. 4.37 shows a circular body, which can rotateabout the point A. A tangential force P is applied at point B and the body movesthrough a small angle θ.

Let R = Radius.Now the distance moved by the force P in

rotating the body through an angle θ is equal to thelength of arc BC.

But length of arc BC = R × θ.As the body has moved a distance R × θ, the

work will be done by the force on the body.∴ Work done

= Force × Distance moved= P × R × θ = ( P × R) × θ

But P × R is equal to torque (T).∴ Work done = T × θ ...(4.38)In the above equation, θ should be taken in

radians.Power Developed by a TorqueLet T = torque

N = r.p.m. (or revolution per minute)Consider a body acted upon by a torque (T) and rotating at N r.p.m.Now angle turned in one revolution = 2π or 360°.∴ Angle turned in N revolution = 2π × N.But N revolutions are performed in one minute.∴ Angle turned in one minute = 2πN

or Angle turned in one second =260πN

Now using equation (4.38), we getWork done = T × θ

or Work done per second = T × Angle turned in one second

= T × 260πN

Nm/s

Direction

of motion

W sin θ F = Force

of

frictio

n

θ

W cos θ

P*

Fig. 4.36

Aθ

RB

PC

P

Fig. 4.37


NOTES

Engineering Mechanics But, power = Work done per second

= T × 260

260

π πN NT= W. ...(4.39)

Problem 29. A tangential force of 1800 N is acting on a shaft of diameter 10 mm.Find the work done by the force for one revolution of the shaft.

Sol. Given :Force, P = 1800 N

Dia. of shaft, D = 10 mm = 10

1000 = 0.01 m

∴ Radius of shaft, R =0.01

2 = 0.005 m

Angle turned for one revolution = 2π radians∴ θ = 2π radiansTorque, T = P × R = 1800 × 0.005 = 9.0 NmWork done is given by equation (4.38).∴ Work done = T × θ

= 9 × 2π Nm = 56.54 Nm. Ans.Energy. The capacity of doing work is known as energy. It is the product of

power and time. The energy is expressed in Nm. It exists in many forms i.e.,mechanical, electrical, heat, chemical, light etc. In engineering mechanics, we are onlyconcerned with mechanical energy and the same will be dealt with.

Mechanical energy consists of the following two types :1. Potential energy (or position energy or datum energy)2. Kinetic energy.

Depending upon the state of rest or that of motion, a body may possess anyoneor both of the above energies.

1. Potential energy. Potential energy is also known as position energy or datumenergy. It is the energy by virtue of position of a body with respect to any givenreference or datum. A weight W lying on the top of a tower of height h possesses apotential energy of W × h with respect to the ground, as the weight W is capable ofdoing W × h work if it falls on the ground.

A compressed spring has potential energy, because it can do work in recoveringits original shape. Similarly, compressed air also possesses potential energy becauseit is capable of doing work when allowed to expand.

2. Kinetic energy. The energy possessed by a body by virtue of its velocity (orits motion) is known as kinetic energy. It is represented by K.E.

The expression for the kinetic energy is derived as follows : Consider a body ofmass m starting from rest. Let it be subjected to an accelerating force F and aftercovering a distance S, its velocity becomes v.

∴ Initial velocity, u = 0Now work done on the body = Force × Distance

= F × S ...(i)But Force = Mass × Acceleration∴ F = m × aSubstituting the value of F in equation (i),

Work done = m × a × S= m × (a × S) ...(ii)


NOTES

Dynamics of ParticlesBut from equation of motion, we havev2 – u2 = 2a × S or v2 – 02 = 2a × S (� u = 0)

or a × S =v2

2Substituting the value of a × S in equation (ii),

Work done = m × v2

2But work done on the body is equal to K.E. possessed by the body.

∴ K.E. = 12 mv2.

Derivation of Work Energy Equation. The work energy principle states thatthe total work done by all forces is equal to change of kinetic energy. This is provedas :

We know that, F = m × a. ...(i)where F = Resultant of all forces acting on a body,

m = Mass of the body, anda = Acceleration in the direction of resultant force

= v dvds

Substituting the value of a in equation (i), we get

F = m × vdvds

��

�� or F . ds = m × v dv ...(iii)

But F . ds is the work done by the resultant force F in displacing the body by asmall distance ds. The total work done by the resultant force F in displacing the bodyby a distance S is obtained by integrating the above equation (ii).

Hence integrating equation (ii) on both sides, we get

0

S

u

vF ds m v dv = × ×.

where u is the initial velocity when distance is zero and v is the final velocity whendistance is S .

∴ F . S = m . v m

u

v2

2 2

�

�

�� = [v2 – u2] =

mv mu2 2

2 2−

or Work done by resultant force = change in kinetic energyThe above equation is known as work energy equation.Note. All the forces acting on a body don’t do work when the body is displaced. These

forces are : (i) the forces acting on a fixed point such as support reactions of pin ball and socketconnections about which the body rotates, (ii) the forces acting perpendicular to direction ofmotion such as normal reaction of the surface over which the body is moving and weight ofbody if body moves in a horizontal direction.

Problem 30. A body of mass 2 kg is moving with a velocity of 50 m/s. What willbe the kinetic energy of the body ?

Sol. Given :Mass, m = 2 kgVelocity, v = 50 m/s

K.E. is given by 12 mv2 .


NOTES


∴ K.E. = 12 mv2 = 1

2 × 2 × 502 kg

ms

2��

= 2500 kg . ms2

. m

= 2500 Nm. � kgms

N2 =��

��

. Ans.

Problem 31. A bullet of mass 81 gm and moving with a velocity of 300 m/s isfired into a block of wood and it penetrates to a depth of 10 cm. If the bullet movingwith the same velocity, were fired into a similar piece of wood 5 cm thick, with whatvelocity would it emerge ? Find also the force of resistance, assuming it to be uniform.

Sol. Given :Mass of bullet, m = 81 gm = 0.081 kgInitial velocity of bullet, u = 300 m/sFirst CasePenetration of bullet, S = 10 cm = 0.1 mFinal velocity of bullet,v = 0Let P = Force of resistance in Newton (N)The force of resistance will be acting in the

opposite direction of motion of bullet. The work done bythe force of resistance

= Force of resistance × Depth of penetration= (– P) × S = (– P) × 0.1 Nm

= – 0.1P (– ve sign is due to opposite direction of force)

Change of K.E. of bullet = 12 mv2 – 1

2 mu2 = 0 – 1

2 × 0.081 × 3002 Nm = – 3645 Nm

But the change of K.E. of the bullet is equal to work doneChange of K.E of bullet = Work done by force of resistance

or – 3645 = – 0.1P

∴ P =36450.1

= 36450 N

Second Case. Velocity of bullet after 5 cm penetrationThe force of resistance will be same i.e., P = 36450 N.Now depth of penetration,

S = 5 cm = 0.05 m∴ Work done by force of resistance

= – P × S = – 36450 × 0.05 Nm(– ve sign is due to opposite direction of force)

= – 1822.5 NmLet v1 = Velocity of bullet after 5 cm penetrationChange of K.E. of bullet after penetration

= 12 mv1

2 – 12 mu2

= 12 × 0.081 × v1

2 – 12 × 0.081 × 3002 ( )� u = 300 m/s

But the change of K.E. is equal to work done

∴ 12 × .081 × v1

2 – 12 × 0.081 × 3002 = – 1822.5

or 12 × 0.081 × v1

2 – 3645 = – 1822.5

v = 300 m/s

0.1 m

Fig. 4.38


NOTES

Dynamics of Particlesor 1

2 × 0.081 v1

2 = 3645 – 1822.5 or 12 × 0.081 v1

2 = 1822.5

or v1 = 1822.5 20.081

1/2×��

��

= 212.132 m/s.

Law of Conservation of Energy. It statesthat the energy can neither be created nor destroyedthough it can be transformed from one form toanother form. The second statement of this law is :

“The total energy possessed by a body remainsconstant provided no energy is added to or takenfrom it.”

Fig. 4.39 shows a body resting on the top of atower.

Let W = Weight of body

m = Mass of body = Wg

h = Height of towerLet us find the potential energy and K.E. of

the body at the top of the tower.Potential energy of the body with respect to ground

= W × h Nm ...(i)K.E. of the body

= 12 mv2 = 0 (� v = 0)

∴ Total energy at the top of tower= Potential energy + Kinetic energy= Wh + 0 = Wh

Suppose the body falls down by a height h1 to the position 1 as shown in Fig. 4.39.Let v1 is the velocity at position 1.

Using equation v2 – u2 = 2gh, we getv1

2 – 02 = 2gh1 (� u = 0, h = h1)v1

2 = 2gh1

∴ K.E. of the body at position 1

= 12 mv1

2 = 12 m × 2gh1 (� v1

2 = 2gh1)

= mgh1 = Wh1 (� W = mg)Potential energy at position 1

= W × Height of body at 1 w.r.t. ground = W × (h – h1)∴ Total energy at position 1

= Potential energy + Kinetic energy= W (h – h1 ) + Wh1 = Wh – Wh1 + Wh1 = Wh

And it is the same energy, the body was possessing at the top of the tower.Similarly, it can be proved that total energy of the body on the ground will be

equal to ‘Wh’.Problem 32. A truck of weight 20 kN is travelling at 45 km/hr on a level road.

It is brought to rest in 20 metre, what is the average force of resistance acting on thetruck.

1

h1

h

(h – h )1

v1

Ground

Body

Fig. 4.39


NOTES

Engineering Mechanics Sol. Given :Weight of truck, W = 20 kN = 20 × 1000 = 20,000 N

Speed of truck, u = 45 km/hr = 45 1000

60 60×

× m/s = 12.5 m/s

∴ Initial velocity of truck, u = 12.5 m/sFinal velocity of truck, v = 0

Change of K.E. of the truck, =12

mv2 – 12

mu2 = 0 – 12

mu2

= – 12

mu2, where m = Wg

= 20,0009.81

kg

= – 12

× 20,000

9.81��

�� × 12.52 Nm ...(i)

The truck is brought to rest, in a distance of 20 m.∴ Distance = 20 mLet P = Average force of resistance which is acting in the opposite direction of

motion.Work done by force of resistance = (– P) × Distance.

= – P × 20 = – 20P ...(ii)But work done is equal to change of K.E.∴ Equating the two equations (i) and (ii),

– 20 P = – 12

20000× ��

��9.81

× 12.52

or P =

12

12 5

20

2× ��

��

×200009.81

. = 7963.8 N. Ans.

Problem 33. A 10 gm bullet is shot horizontally in a wood block of mass 1 kg.The bullet gets embedded in the block and the block is displaced on a rough horizontaltable (µ = 0.2) through 1 m .What was the velocity of bullet ?

Sol. Given :Mass of bullet, m1 = 10 gm = 0.01 kgMass of block, m2 = 1 kgDistance travelled by block and bullet, S = 1 mValue of µ = 0.2Let u1 = Initial velocity of bullet

u2 = Initial velocity of block which is equal to zeroV = Common velocity of bullet and block after impactM = Total mass of bullet and block = 0.01 + 1 = 1.01 kg

From the law of conservation of momentum, we getTotal momentum before impact = Total momentum after impact

or (m1u1 + m2u2) = M × Vor (0.01 × u1 + 1 × 0) = 1.01 × V or 0.01 u1 = 1.01 V

∴ V =0.011.01

u1 ...(i)

Now force of friction = µR (R is normal reaction )= µW (Here R = W )= µ × M × g (� W = M × g)= 0.2 × 1.01 × 9.81 N


NOTES

Dynamics of ParticlesWork done by force of friction = Force of friction × Distance moved= – 0.2 × 1.01 × 9.81 × 1 Nm= – 1.9816 Nm ...(ii)

(– ve sign is due to opposite direction of force of friction)

Initial velocity of block and bullet, Vi = V = 0.011.01

u1

After travelling a distance of 1 m, the block and bullet comes to rest.Hence final velocity of block of bullet, Vf = 0

Change of K.E. of bullet and block =12

M[vf2 – vi

2] = 12

× 1.01 0 12

− ��

��

�

�

��

0.011.01

u

= – 12

(1.01) × 0.011.01

u12

��

��

�

�

��

...(iii)

But the work done by force of friction is equal to change of K.E. of block andbullet.

Equating the work done by force of friction to the change of K.E. of bullet andblock i.e., equating equations (ii) and (iii), we get

– 1.9816 = – 12

× 1.01 × 0.011.01

u12

��

��

0.011.01

u12

��

��

=1.9816 2

1.010.011.01

1.9816 21.01

× ��

��

= ×u1 = 1.98 m/s

or 0.01 u1 = 1.01 × 1.98

or u1 =1.01 1.98

0.01×

= −199.98 200 m/s. Ans.~

Energy Lost by a Body Falling on Another Body and to Calculate theResistance Offered the Ground. Fig. 5.66 shows a body of mass M falling onanother body of mass m, from a height h. The mass m penetrates into the groundthrough a distance S before coming to rest.

Let v = velocity of mass M before striking themass m

= 2gh

V = Velocity of both masses M and mafter impact (i.e., common velocity)

The velocity of mass m before impact is zero.Applying law of conservation of momentum :

Total momentum of M and m before impact = Totalmomentum of both masses after impactor M × v + m × 0 = (M + m) × V

or M × 2gh = (M + m) × V

∴ V =M gh

M m×

+2

( )...(i)

Resistance offered by the groundThe K.E. of the system (i.e., K.E. of both masses) just after the impact

= 12(Total mass of system) × V2 = 1

2 [M + m] × V2 Nm

m

h

S

M

Fig. 4.40


NOTES

Engineering Mechanics and potential energy to the system just after the impact= (Total weight of system)

× Vertical depth of penetration= [(M × m) × g] × S Nm

Total energy of the system= K.E. of system + P.E. of system

= 12 [M + m] × V2 + (M + m) × g × S

This total energy is lost in penetrating the mass m into the ground.∴ Work done during penetration of mass m into ground

= Resistance of ground × Depth of penetration = R × SNow total energy lost is equal to work done during penetration.

∴ 12[M + m] × V2 + (M + m) × g × S = R × S

From the above equation, the resistance R can be calculated.Problem 34. A hammer of mass 1500 kg drops from a height of 60 cm on a pile

of mass 750 kg. Find :(i) the common velocity after impact assuming plastic impact.

(ii) the average resistance of the ground if the pile comes to rest after penetrating5 cm into the ground.

Sol. Given :Mass of hammer, M = 1500 kgHeight from which hammer drops,

h = 60 cm = 0.6 mMass of pile, m = 750 kgDepth of penetration, S = 5 cm = 0.05 m.(i) Let V = common velocity after impactPlastic impact means that after impact, both (hammer and pile) masses will

move together with a common velocity.v = velocity of the hammer before impact

The velocity of hammer after falling through a height of 0.6 m from rest, isgiven by

v = 2gh = × ×2 9.81 0.6 = 3.43 m/s

The velocity of pile before impact is zero.Total momentum of hammer and pile just before impact

= M × v + m × 0 = M × v = 1500 × 3.43 = 5145 kg-m/sTotal momentum of hammer and pile just after impact

= (Total mass of hammer and pile) × Common velocity= (M + m) × V = (1500 + 750) × V = 2250 V kg-m/s

From the principle of conservation of momentum, we knowTotal momentum before impact

= Total momentum after impactor 5145 = 2250 × V

∴ V =51452250

= 2.29 m/s. Ans.

(ii) Let R = Average resistance of ground to the penetration.Now first of all calculate the total energy of the system (i.e., hammer and pile)

just after the impact. This total energy is the sum of potential energy and kineticenergy.


NOTES

Dynamics of ParticlesK.E. of the system (hammer and pile) just after the impact

= 12(M + m) V2 = 1

2(1500 + 750) × 2.292 = 5899.6 Nm

Loss of potential energy of system in moving a distance S

= Total weight of system × S = [(M + m) × g] × S

= (1500 + 750) × 9.81 × 0.05 (� S = 0.05 m)

= 1103.625 Nm

∴ Total energy loss of the system

= Loss of K.E. of system + Loss of P.E.

= 5899.6 + 1103.625 = 7003.225 Nm

This total energy lost is used in penetrating the pile into the ground.

The work done during penetration of the file

= Resistance of ground × Depth of penetration = R × 0.05

Equating the work done during penetration to the total energy lost, we get

∴ Work done = Total energy lost

R × 0.05 = 7003.225

∴ R =7003.225

0.05 = 140064.5 N. Ans.

��

Let F = Net force acting on a rigid body in the direction of motion, throughthe C.G. of the body

m = Mass of the rigid body, anda = Acceleration of the body.

Then we have

F m a mdvdt

= × = ×

� Acceleration is rate of change of velocity i e advdt

. . , =�

�

��

or F dt m dv× = ×Integrating the above equation, we get

F dt m dvt

t

v

v

× = × 1

2

1

2

= m(v2 – v1)

If time-interval is very small, then F dtt

t

× 1

2

is known as impulse. Hence impulse

is the product of force and time when time is very small. Momentum is the product ofmass and velocity.

∴ Impulse = F dt m v v mv mvt

t

× = = 1

2

2 1 2 1( – ) –


NOTES

Engineering Mechanics or Impulse = Change of momentum

[� Momentum = Mass × Velocity]

= Final momentum – Initial momentum

or Initial momentum + Impulse = Final momentum

The above equation gives the relation between impulse and momentum of arigid body. The impulse-momentum equation is based on integrating the equation ofmotion with respect to time. The question of motion relates force, velocity and time.

Note. (i) The impulse-momentum approach is particularly convenient in situations whenforces act for very small interval of time as in an impact or sudden blow.

(ii) In satellite motion, a combination of impulse-momentum method and work-energymethod is used.

(iii) The unit of impulse is Ns (i.e., Newton second) in S.I. system.

��

If the balls of different materials are allowed to fall on a marble floor, theywill rebound to different heights due to their elasticity. Elasticity is the property ofbodies by virtue of which they rebound after impact. The body, which rebounds to agreater height is known as more elastic, than a body which rebounds to a lesserheight. Inelastic body is one which does not rebound at all.

Whenever two elastic bodies collide with each other, they tend to compress eachother. Immediately after this, the two bodies attempt to regain its original shape, dueto their elasticity. This process, of regaining the original shape, is called restitution.

The important terms, used in collision, are defined as :1. Time of compression. The time taken by two bodies in compression, after

the instant of collision, is known as time of compression.2. Time of restitution. The time taken by two bodies to regain the original

shape, after compression, is known as time of restitution.3. Time of collision. The sum of time of compression and time of restitution is

known as time of collision or period of collision or period of impact.4. Law of conservation of momentum. It states that if the resultant of the

external forces acting on a system is zero, the momentum of the system remainsconstant. This means that the total momentum of the system before collision is equalto the total momentum of the system after collision. The system may consist of onebody or two bodies or more.

Types of Impacts. Impact means the collision of two bodies which occurs ina very small interval of time and during which the two bodies exert very large forceon each other. The important types of impacts are :

(i) Direct impact(ii) Indirect (oblique) impact.Direct Impact of Two Bodies. The two bodies A and B are moving in a

horizontal line before collision with velocities u1 and u2 in the same direction i.e.,along x-axis as shown in Fig. 4.41 (a). If u1 > u2, the body A will strike the body B andcollision will take place. Let C is the point of collision of the two bodies as shown inFig. 4.41 (b). The point C is also known as the point of contact. The line joining thecentres of these two bodies and passing through the point of contact is known as line ofimpact. Hence here the line O1-C-O2 is called line of impact.


NOTES

Dynamics of ParticlesThe collision between two bodies is known as direct impact if the two bodiesbefore impact, are moving along the line of impact.

m1

m2A

O1 O2

u1

B

u2

v1

v2

Point of contact orpoint of collision

Line ofimpact

O1O2

Body ABody B

C

(a) (b)

Fig. 4.41

The two bodies shown in Fig. 4.41 is having a direct impact.

Let m1 = Mass of the body A

u1 = Initial velocity of body A, i.e., the velocity of body A before collision

along x-axis.

v1 = Final velocity of body A (after collision) along x-axis.

m2, u2 and v2 are the mass of body B, velocity of body B before collision andvelocity of the body B after collision along x-axis respectively.

The momentum of the body A before collision

= Mass × Velocity = m1u1 kg m/s.

The momentum of the body B before collision

= m2 × u2 = m2u2 kg m/s.

∴ Total initial momentum (i.e., momentum before collision)

= m1u1 + m2u2 kg m/s ...(i)

Similarly, total final momentum (i.e., momentum after collision)

= Mass of body A × Final velocity of A + Mass of body B

× Final velocity of body B

= m1v1 + m2v2

But according to the law of conservation of momentum,

Total initial momentum = Total final momentum

or m1u1 + m2u2 = m1v1 + m2v2.

Indirect Impact of Two Bodies. Fig. 4.42 (a) shows the two bodies A and B,moving with a velocity of u1 and u2 in different direction. The velocity of body A makesan angle θ1 with horizontal direction whereas the velocity of body B makes an angleθ2 with horizontal direction. As the two bodies are moving in different direction, thecollision between two bodies takes place as shown in Fig. 4.42 (b). The point C is thepoint of contact. And line joining the centres O1, O2 with point C is known as line ofimpact.


NOTES


θ1

θ1

O1

O1

O2

O2

A

Bu1

θ2

θ2

φ1φ2

v2

v1

u2

Line ofimpact

A

B

Direction of motionof body A before collision

Direction of motionof body B before collision

(a)

(b)

Fig. 4.42

If the two bodies, before impact are not moving along the line of impact, thecollision is known as indirect impact.

Here the two bodies A and B are not moving along the line of impact beforecollision, hence the collision is known as indirect impact (or oblique impact).

Let m1 = Mass of body Au1 = Initial velocity of Aθ1 = Angle made by body A with horizontal before collisionv1 = Velocity of body A after collisionφ1 = Angle made by body A with horizontal after collision.

m2, u2, θ2, v2 and φ2= Corresponding values of body B.Now according to the law of conservation of momentum the total initial

momentum in any direction must be equal to final momentum in that direction.Now initial momentum (i.e., momentum before collision) of body A in x-direction

= Mass of A × Initial velocity of A in x-direction= m1 × u1 cos θ1

Initial momentum of body B in x-direction= Mass of B × Initial velocity of B in x-direction= m2 × u2 cos θ2

∴ Total initial momentum in x-direction= m1 × u1 cos θ1 + m2 × u2 cos θ2 ...(i)

Final momentum of body A (i.e., momentum after collision) in x-direction= m1 × Final velocity of A in x-direction = m1 × v1 cos φ1

Similarly, final momentum of B in x-direction= m2 × Final velocity of B in x-direction = m2 × v2 cos φ2

∴ Total final momentum in x-direction= m1 × v1 cos φ1 + m2 × v2 cos φ2 ...(ii)

Equating the initial momentum to final momentum given by equations (i) and(ii), we get

m1 × u1 cos θ1+ m2 × u2 cos θ2 = m1 × v1 cos φ1 + m2 × v2 cos φ2.


NOTES

Dynamics of ParticlesCo-efficient of Restitution. It is defined as the ratio of velocity of separation(of the two moving bodies which collides with each other ) to their velocity of approach.It is also defined as the ratio of the relative velocities of colliding bodies after impactto their relative velocity before impact. It is denoted by symbol ‘e’. The relative ve-locities are measured along the line of impact, which is the common normal to thecolliding surfaces.

According to Newton’s Law of collision of elastic bodies, ‘‘the velocity of separation,of the two moving bodies which collide with each other, bears a constant ratio to theirvelocity of approach’’. And the constant of proportionality is known as co-efficient ofrestitution.

Fig. 4.43 shows two bodies A and B.Let u1 = Velocity of A before collision along x-axis

v1 = Velocity of A after collision along x-axisu2 = Velocity of B before collision along x-axisv2 = Velocity of B after collision along x-axis.

u1

u2

v1

v2

A AB B

(a) Before Collision (b) After Collision

Fig. 4.43

The body A will collide with body B if velocity of A is more than that of B. Hencevelocity of approach (or relative velocity of colliding bodies before impact)

= Initial velocity of A – Initial velocity of B = (u1 – u2)After collision, the separation of the two bodies will take place if final velocity of

B is more than that of A.Hence velocity of separation (or relative velocity of colliding bodies after impact)

= Final velocity of B – Final velocity of A = (v2 – v1)Now according to Newton’s Law of collision of elastic bodies,Velocity of separation ∝ Velocity of approach

or (v2 – v1 ) ∝ (u1 – u2 )or (v2 – v1 ) = e × (u1 – u2 )where e is the constant of proportionality and is known as co-efficient of restitution.

∴ e = ( )( )v vu u

2 1

1 2

−−

...(4.40)

For most of bodies, the value of e lies between 0 and 1. For perfectly elasticbodies e = 1 and for perfectly plastic bodies e = 0.

To determine the velocities after the impact (i.e., velocity v1 and v2) the equation(4.40) is not sufficient to determine the two unknowns. One more equation is needed.This equation is law of conservation of momentum i.e., total initial momentum is equalto total final momentum.

Note 1. The velocity of approach or separation of the two bodies, which are moving inthe same direction before or after the impact, is the difference of their velocities.

2. For the two bodies moving in the opposite direction, the velocity of approach orseparation is the sum of their velocities.


NOTES

Engineering Mechanics 3. Newton’s law of collision of elastic bodies, also holds good for indirect impact, i.e.,

(v2 cos φ2– v1 cos φ1) = e ( u1 cos θ1 – u2 cos θ2 ) ...(4.41)

4. If the velocity in a direction is + ve then the velocity in the opposite direction will be– ve.

Problem 35. Ball A of mass 1 kg moving with a velocity of 2 m/s, strikes directlyon a ball B of mass 2 kg at rest. The ball A, after striking, comes to rest. Find thevelocity of ball B after striking and co-efficient of restitution.

Sol. Given :Mass of ball A, m1 = 1 kgInitial velocity of ball A, u1 = 2 m/sMass of ball B, m2 = 2 kgInitial velocity of ball B, u2 = 0Final velocity of ball A, v1 = 0This is a case of direct impact.Let v2 = Velocity of ball B after impact, and

e = Co-efficient of restitution.Total initial momentum = m1u1 + m2u2 = 1 × 2 + 2 × 0 = 2 kg m/s.Total final momentum = m1v1 + m2v2 = 1 × 0 + 2 × v2 = 2v2 kg m/s.According to the law of conservation of momentum,

Total initial momentum = Total final momentum∴ 2 = 2 × v2

∴ v2 =22

= 1 m/s. Ans.

To find the co-efficient of restitution, the equation (4.40) is used

∴ e =( )( )

( )( )

v vu u

2 1

1 2

1 02 0

−−

= −−

= 12

. Ans.

Problem 36. Three perfectly elastic balls A, B and C of masses 2 kg, 6 kg and12 kg are moving in the same direction with velocities 12 m/s, 4 m/s and 2 m/srespectively. If the ball A strikes with the ball B, which in turns, strikes with the ball C,prove that the balls A and B will be brought to rest by the impact.

Sol. Given :For perfectly elastic balls, e = 1Mass of first ball, m1 = 2 kgMass of second ball, m2 = 6 kgMass of third ball, m3 = 12 kgInitial velocity of first ball, u1 = 12 m/sInitial velocity of second ball, u2 = 4 m/sInitial velocity of third ball, u3 = 2 m/s(a) First of all consider the impact of first and second ballLet v1 = Final velocity of first ball after impact

v2 = Final velocity of second ball after impactTotal momentum of 1st and 2nd ball before impact

= Total momentum of 1st and 2nd ball after impact∴ m1u1 + m2u2 = m1v1 + m2v2

or 2 × 12 + 6 × 4 = 2 × v1 + 6 × v2

or 48 = 2v1 + 6v2 or v1 + 3v2 = 24 ...(i)Now using equation (4.40),


NOTES

Dynamics of Particlese =

( )( )v vu u

2 1

1 2

−−

or (v2 – v1) = e (u1 – u2) = 1( 12 – 4 ) (� e = 1)or (v2 – v1) = 8 ...(ii)

Adding equations (i) and (ii), we get4v2 = 24 + 8 = 32

∴ v2 =324

= 8 m/s.

Substituting this value of v2 in equation (ii),8 – v1 = 8 or v1 = 0.

The velocity of ball A after impact is zero. Hence the ball A will be brought torest by the impact of A and B. Ans.

The velocity of ball B after impact with A will be (v2 = 8 m/s).(b) Now consider the impact of second and third ball. The second ball is now

moving with a velocity of 8 m/s and strikes the third ball which is moving with avelocity of 2 m/s.

Now Initial velocity of second ball, u2* = 8 m/sInitial velocity of third ball, u3 = 2 m/sLet v2* = New final velocity of second ball after striking the third ball.

v3 = Final velocity of third ballLaw of conservation of momentum gives :Total momentum of 2nd and 3rd ball before impact

= Total momentum of 2nd and 3rd ball after impactor m2 × u2* + m3u3 = m2 × v2* + m3 × v3or 6 × 8 + 12 × 2 = 6 × v2* + 12 × v3

or 48 + 24 = 6v2* + 12v3 or 72 = 6v2* + 12v3or 6v2* + 12v3 = 72 or v2* + 2v3 = 12 ...(iii)


e =Velocity of separationVelocity of approach

=−

−( *)( * )v vu u

3 2

2 3

or (v3 – v2*) = e (u2* – u3) = 1(8 – 2) = 6∴ v3 – v2* = 6 ...(iv)Adding equations (iii) and (iv), we get

3v3 = 12 + 6 = 18

∴ v3 =183

= 6 m/s.

Substituting the value of v3 in equation (iv),6 – v2* = 6 or v2* = 6 – 6 = 0.

Hence the velocity of second ball, after impact with third ball, is zero. Hence thesecond ball will also be brought to rest after impact with third ball C. Ans.

Problems on Indirect Impact

Problem 37. A ball of mass 1 kg, moving with a velocity of 6 m/s, strikes on aball of mass 2 kg moving with a velocity of 2 m/s. At the instant of impact, the velocitiesof the two balls are parallel and inclined at 30° to the line joining their centres. If co-

efficient of restitution is 12

, find :


NOTES

Engineering Mechanics (i) the velocity and direction in which the 1 kg ball will move after impact.(ii) the velocity and direction in which the 2 kg ball will move after impact.Sol. Given :Mass of first ball,

m1 = 1 kgMass of second ball,

m2 = 2 kgInitial velocity of first ball,

u1 = 6 m/sInitial velocity of second ball,

u2 = 2 m/sAngle made by first ball with

line of impact, θ1 = 30°Angle made by second ball

with line of impact, θ2 = 30°

Co-efficient of restitution, e = 12

Let v1 = Velocity of first ball after impact

v2 = Velocity of second ball after impact

φ1 = Angle made by first ball after impact with line of impact

φ2 = Angle made by second ball after impact with line of impact.

Velocity of each ball normal to the line of impact remains unchanged. This meansthat the components of velocities of each ball normal to line of impact before and afterimpact is same.

∴ For ball A,

Initial normal component = Final normal component

or 6 sin 30° = v1 sin φ1 or v1 sin φ1 = 6 sin 30° = 3 ...(i)

For ball B,

Final normal component = Initial normal component

v2 sin φ2 = 2 sin 30° = 1 ...(ii)

According to law of conservation of momentum,

Total initial momentum along the line of impact

= Total final momentum along the line of impact.

or m1 × u1 cos θ1 + m2 × u2 cos θ2 = m1 × v1 cos φ1 + m2 × v2 cos φ2

or 1× 6 cos 30° + 2 × 2 cos 30° = 1 × v1 cos φ1 + 2 × v2 cos φ1

or 6 × 0.866 + 4 × 0.866 = v1 cos φ2 + 2v2 cos φ2

or 8.66 = v1 cos φ1 + 2v2 cos φ2

or v1 cos φ1 + 2v2 cos φ2 = 8.66 ...(iii)

The co-efficient of restitution for this case is given by the equation

(v2 cos φ2 – v1 cos φ1) = e (u1 cos θ1 – u2 cos θ2)

= 12 [6 × cos 30° – 2 × cos 30°] = 2 × cos 30° = 1.732

∴ v2 cos φ2 – v1 cos φ1 = 1.732 ...(iv)

AB

30° 30°φ1 φ2

2 m/s6 m/s

1 kg

2 kg

v1

v2

Line ofimpact

Fig. 4.44


NOTES

Dynamics of ParticlesAdding equations (iii) and (iv),3v2 cos φ2 = 8.66 + 1.732 = 10.392

or v2 cos φ2 = 10.3923

= 3.464 ...(v)

Substituting the value of v2 cos φ2 in equation (iii),v1 cos φ1 × 2 × 3.464 = 8.66

or v1 cos φ1 = 8.66 – 2 × 3.464 = 1.732 ...(vi)Dividing equation (i) by equation (vi), we get

tan φ1 = 31.732

= 1.732

∴ φ1 = tan–1 1.732 = 60°. Ans.Dividing equation (ii) by equation (v), we get

tan φ2 = 13.464

= 0.2886

∴ φ2 = tan–1 0.2886 = 16.1°. Ans.Substituting the value of φ2 in equation (v),

v2 cos 16.1° = 3.464

∴ v2 = 3.464cos 16.1°

= 3.6 m/s. Ans.

Substituting the value of φ1 in equation (vi),v1 cos 60° = 1.732

∴ v1 = 1.732cos 60°

= 3.464 m/s. Ans.

Loss of Kinetic Energy During Impact. When the two bodies collide witheach other, the loss of kinetic energy takes place due to impact. This loss of energymay be obtained by finding out the kinetic energy of the two bodies before and afterthe impact. The loss of kinetic energy during impact is equal to the difference of thetwo kinetic energies.

Consider two bodies A and B having a direct impact.Let m1 = Mass of the first body,

u1 = Velocity of first body before impactv1 = Velocity of first body after impact

m2, u2 and v2 = corresponding values of mass, initial velocity and final velocityof second body

EL = Loss of energye = Co-efficient of restitution.

Kinetic energy of first body before impact

= 12 × m1 × u1

2

Kinetic energy of second body before impact

= 12 m2u2

2

∴ Total K.E. of the two bodies before impact

= 12 m1u1

2 + 12 m2u2

2 ...(i)

Similarly, total K.E. of the two bodies after impact

= 12 m1v1

2 + 12 m2v2

2 ...(ii)


NOTES

Engineering Mechanics ∴ Loss of K.E., during impact= Total K.E. before impact – Total K.E. after impact

= ( 12 m1u1

2 + 12 m2u2

2 ) – ( 12 m1v1

2 + 12 m2v2

2 ) .

Problem 38. A vehicle of mass 600 kg and moving with a velocity of 12 m/sstrikes another vehicle of mass 400 kg, moving at 9 m/s in the same direction. Both thevehicles get coupled together due to impact. Find the common velocity with which thetwo vehicles will move. Also find the loss of kinetic energy due to impact.

Sol. Given :Mass of first vehicle, m1 = 600 kgInitial velocity of first vehicle, u1 = 12 m/sMass of second vehicle, m2 = 400 kgInitial velocity of second vehicle, u2 = 9 m/sWhen the two vehicles get coupled, then total mass of the two vehicles,

M = m1 + m2 = 600 + 400 = 1000 kg(i) Let V = Common velocity of the two vehicles after impactTotal momentum before impact

= m1u1 + m2u2 = 600 × 12 + 400 × 9= 7200 + 3600 = 10800 kg m/s ...(i)

Total momentum after impact= (m1 + m2) × V = (600 + 400) × V = 1000 V kg m/s ...(ii)

According to law of conservation of momentum,Total momentum before impact = Total momentum after impact

or 10800 = 1000 × V

∴ V =108001000

= 10.8 m/s. Ans.

(ii) Let EL = Loss of kinetic energy due to impactTotal K.E before impact

= 12 m1u1

2 + 12 m2u2

2 = 12 × 600 × 122 + 1

2 × 400 × 92

=118800

2 Nm

Total K.E after impact = 12 MV2 =

12

× 1000 × 10.82 = 116640

2 Nm

∴ Loss of K.E. = Total K.E. before impact – Total K.E. after impact

=118800

2116640

22160

2− = = 1080 Nm. Ans.

Impact of a Body on a Fixed Plane. There are two types of impact of a bodyon a fixed plane. These are :

(i) Direct impact of a body on a fixed plane, and(ii) Indirect impact of a body on a fixed plane.Direct Impact of a Body on a Fixed Plane. The fixed plane is at rest before

impact and after impact. The mass of the fixed plane is very large. Consider a bodyhaving a direct impact on a fixed plane.

Let u = Initial velocity of the body,v = Final velocity of the bodye = Co-efficient of restitution.

The velocity of approach between body and fixed plane= Initial velocity of body – velocity of fixed plane = u – 0 = u.


NOTES

Dynamics of ParticlesThe velocity of separation= Final velocity of body – Velocity of fixed plane = v – 0 = v.

∴ Co-efficient of restitution is given by,

e =Velocityof separationVelocityof approach

= vu

or v = e × u ...(4.42)

Important Cases :(a) A body is falling on a floor from some heightLet H = Height from which the body is falling

u = Velocity with which the body strikes the floorThis velocity will be obtained by using the equation(Final vel.)2 – (Initial vel.)2 = 2g × HInitial velocity of the body = 0Final velocity of the body = u∴ u2 – 0 = 2gH

or u = 2gH . ...(4.43)(b) A body is first projected upwards and then strikes the floor. When the body is

projected upwards with some initial velocity, the body will reach the greatest heightand then will return to the ground with the same velocity with which it was projected.

Problem 39. A ball of mass 500 gm is dropped on a horiontal floor from aheight of 18 m. The ball rebounds due to impact with horizontal floor to a height of 8 m.Find the co-efficient of restitution between the floor and the ball.

Sol. Given :Mass of the ball, m = 500 gm = 0.5 kgHeight from which the ball is dropped,

H = 18 mThe velocity with which the ball strikes the floor is given by equation (4.41),

∴ u = 2 2 9 81 18 36 9 81 6 9 81gH = × × = × = ×. . . m/s.Height to which the ball rebounds = 8 m

or h = 8 mThis means that after the impact, the ball rises to a height of 8 m. At the highest

point, the velocity will be zero. The velocity of the ball after impact, will be obtained byusing the relation

(Final vel.)2 – (Initial vel. )2 = 2ghLet v = velocity of the ball after impact

Then v = 2 2 8 4gh = × × = ×9.81 9.81The co-efficient of restitution is given by equation (4.42), as

v = e × u or 4 × 9.81 = e × 6 9.81

∴ e =4 9.81

6 9.81

××

=23

. Ans.

Indirect Impact of a body on a fixedplane. Fig. 4.45 shows a body striking a fixedplane at certain angle. This type of impact isknown as indirect impact. The point C is knownas point of contact. The line joining the centreof the body with the point of contact is knownas line of impact.

Let u = Velocity of body before impactv = Velocity of body after impact

30°q f

45°O

C

Line ofImpact

u

v

Fig. 4.45


NOTES

Engineering Mechanics θ = Angle made by the initial velocity with line of impactφ = Angle made by the final velocity with line of impacte = Co-efficient of restitution.

The component of initial velocity along the line of impact will cause the directimpact with the fixed plane. The components of the initial velocity and final velocitynormal to the line of impact will remain unchanged.

Here velocity of approach = u cos θand velocity of separation = v cos φ

But, velocity of separation= e × velocity of approach

or v cos φ = e × u cos θ. ...(4.44)Problem 40. A ball is thrown against a wall with a velocity 10 m/s making an

angle of 30° with the wall as shown in Fig. 4.46. If the co-efficient of restitution= 0.5, find :

(i) direction of the ball after impact, and(ii) velocity of the ball after impact.Sol. Given :Initial velocity of ball, u = 10 m/s.Angle made by initial velocity with wall = 30°.∴ Angle made by initial velocity with line of

impact,θ = 90 – 30 = 60°

Co-efficient of restitution, e = 0.5Let v = Final velocity of the ball,

and θ = Angle made by final velocity withline of impact.

The components of the initial velocity and finalvelocity at right angles to the line of impact is same.

∴ u sin θ = v sin φor 10 sin 60° = v sin φor 10 × 0.866 = v sin φor v sin φ = 8.66 ...(i)

Now using equation (4.44),v cos φ = e × u cos θ = 0.5 × 10 × cos 60° = 2.5 ...(ii)

Divide equation (i) by equation (ii),vv

sincos

φφ =

8.662.5

∴ tan φ = 8.662.5

= 3.464

∴ φ = tan–1 3.464 = 73.9°. Ans.Substituting the value of φ in equation (i), we get

v sin 73. 9° = 8.66

∴ v = 8.66

sin (73.9) = 9.013 m/s. Ans.

Line ofimpact f

q

30°

u

Wall

v

OC

Fig. 4.46


NOTES

Dynamics of Particles��

1. The motion of a body in a straight line is called linear motion.2. The rate of change of displacement of a body is known as velocity. It is given by

v = st

ordsdt

.

3. If the equation of motion of a body moving in a straight line is given in terms ofdisplacement (s) and time (t), the velocity of the body is obtained by differenti-ating the displacement with respect to time. And acceleration is obtained bydifferentiating the velocity with respect to time.

4. If the equation of motion of a body moving in a straight line is given in terms ofacceleration (f) and time (t), then velocity is obtained by integrating the accel-eration. The constant of integration for velocity is obtained from the givencondition in terms of velocity and time.The displacement is obtained by integrating the velocity. There will be anotherconstant of integration. This constant of integration is also obtained from givencondition in terms of displacement and time.

5. The quantity of matter contained in a body known as mass of the body.6. The weight of a body is defined as the force by which the body is attracted to-

wards the centre of earth. Weight of the body is given byWeight = Mass × g.

7. Momentum of a body is the product of the mass and its velocity.8. Kinetic energy due to rotation of a body is given by

K.E. = 12Iω2.

9. If a body is having motion of translation as well as motion of rotation, then totalkinetic energy is given by,

Total K.E. = 12mv2 + 1

2Iω2.

10. Work is the product of force and distance. Energy is the capacity to do work.The work done is given by

Work done = P × S ... When force and distance are in the same direction= P cos θ × S When force acts at an angle θ with the

direction of displacement.11. The units of work is N m (or Joule). Hence one Joule is the work done by a vorce

of 1 N when displacement is 1 m.12. Law of conservation of energy states, “Energy can neither be created nor

destroyed, though it can be transformed from one form into another form”.13. The resistance offered by the ground against penetration of body into ground is

obtained by equating the work done against penetration to the total energy lostby the system. Total energy means the loss of K.E. + loss of P.E. of the system.

14. According to the law of conservation of momentum, the total momentum of asystem before collision is equal to the total momentum of the system aftercollision.


NOTES

Engineering Mechanics 15. Line of impact is obtained by joining the centres of the two bodies with the pointof contact.

16. The collision between two bodies, which are moving along the line of impactbefore collision, is known as direct impact.

17. The collision between two bodies, which are moving at some angle to the line ofimpact before collision, is known as indirect impact.

18. The loss of K.E. during impact in terms of masses, initial velocities of two bodiesand co-efficient of restitution is given by

E1 = m m u u

m m1 2 1 2

2

1 22( )

( )−

+ [1– e2].

��

• Velocity. It is the rate of change of displacement of body moving in a straightline.

• Acceleration. It is defined as the rate of change of velocity of a body.• Angular velocity. It is defined as the rate of change of angular displacement

of a body, denoted by ω (omega).• Angular acceleration. It is the rate of change of angular velocity denoted by α

(alpha).• Mass. The quantity of matter contained in a body.• Weight. It is defined as the force by which the body is attracted towards the

centre of the earth.• Momentum. It is the product of mass of a body and its velocity (v).

��

1. Define the terms : velocity and acceleration.

2. (a) What do you mean by linear motion ? Give some examples of linear motion.

(b) Distinguish between uniform motion and uniformly accelerated motion.

3. Acceleration can be expressed as

ad s

dt=

2

2 and audvds

=

Derive the above two expressions.

4. Derive the following equations of motion of a body moving in a straight linewith uniform acceleration :

(i) v = u + at (ii) s = ut + 12at2 (iii) v2 – u2 = 2as.

5. Derive a relation for the distance travelled, by a body in the nth second.

6. Define the terms : Mass of a body and weight of a body. What is the relationshipbetween the two ?

7. Explain the terms : Momentum of a body and angular momentum of a body.


NOTES

Dynamics of Particles8. Define and explain the Newton Laws of motion for linear motion and rotationalmotion.

9. Deduce the relation,

Kinetic energy = 12 Iw2

where I = Moment of inertia about the axis of rotation, andω = Angular velocity.

10. A body is having motion of translation as well as motion of rotation. How willyou determine the total kinetic energy of the body ?

11. Define the following terms :Work, energy, momentum, impulse and moment of momentum.

12. What are the units of work ? What is the relation between work done and power ?Also mention the unit of power.

13. Prove that the K.E. of a moving body is expressed as :

K.E. = 12 mv2 where m = mass of body and v = velocity of body.

14. State the law of conservation of energy. Give proof of this law taking mechanicalenergy only into account.

15. Define and explain the following terms :Collision of elastic bodies, law of conservation of momentum and law ofconservation of energy.

16. (a) What do you mean by direct impact and indirect impact ?(b) Define the co-efficient of restitution.

17. A ball strikes directly on a similar* ball which is at rest. The first ball comes torest by the impact. If half of the initial kinetic energy is lost by impact, thenprove that the co-efficient of restitution will be equal to 0.707.

18. A body of mass m is allowed to fall from a height ‘H’ on a floor. After the impact,the body rises to a height of ‘h’ (or body rebounds to a height of h), then prove

that the co-efficient of restitution will be equal to h H/ .

19. What is the difference between the impact of the two bodies and the impact of abody on a fixed plane ?

20. A heavy elastic ball drops from the ceiling of a room, and after rebounding twicefrom the floor reaches a height equal to one-half that of ceiling. Prove that the co-efficient of restitution is (1/2)1/4.

21. Find an expression for co-efficient of restitution in case of a body which is havingindirect impact on a fixed plane.

22. Prove that the two elastic bodies of equal masses exchange velocities in the caseof direct central impact.

23. A body is moving with a velocity of 3 m/s. After five seconds the velocity of the bodybecomes 13 m/s. Find the acceleration of the body. [Ans. 2 m/s2]

24. A car is moving with a velocity of 20 m/s. The car is brought to rest by applyingbrakes in 4 seconds. Determine : (i) the retardation and (ii) distance travelledby the car after applying brakes. [Ans. – 5 m/s2, 40 m/s]


NOTES

Engineering Mechanics 25. A bullet, moving at the rate of 200 m/s, in fired into a log of wood. The bulletpenetrates to a depth of 50 cm. If the bullet moving with the same velocity isfired into a similar piece of wood 25 cm thick, with what velocity would it emerge.Take the resistance to be uniform in both the case. [Ans. 141.4 m/s]

26. A body moves along a straight line and its acceleration (a) which varies withtime (t) is given by : a = 4 – 5t. After 4 seconds from start of observations itsvelocity is observed to be 16 m/s. After 8 seconds from start of observations, thebody was 70 m from the origin. (i) Determine its acceleration, velocity anddistance from the origin at the start of observations, (ii) determine the timeafter start of observation in which the velocity becomes zero and its distancefrom the origin. [Ans. (i) 4 m/s2, 40 m/s, 48.67 m, (ii) 8.4 s, 31.87 m]

27. The velocity of a particle moving along x-axis is defined by v = kx3 – 4x2 + 6xwhere v is in m/s and x in m and k is a constant. If k = 2, compute the accelerationwhen x = 2 m. Also find the smallest value of k that will make acceleration= 16 m/s2 at x = 3 m.

[Hint. a = v dvds

or =v dvdx

. But v = kx3 – 4x2 + 6x and dvdx

= 3kx2 – 8x + 6

∴ a = (kx3 – 4x2 + 6x) (3kx2 – 8x + 6)When k = 2 and x = 2 then a = (2.23 – 4.22 + 6.2) (3.2.22 – 8.2 + 6)or a = (16 – 16 + 12) (24 – 16 + 6) = 12 × 14 = 168 m/s2. Ans.]

28. Find the force acting on a body of mass 100 kg and producing an acceleration of2 m/s2 in its direction. [Ans. 200 N]

29. A force of 450 N acts on a body having a mass of 150 kg for 5 seconds. If theinitial velocity of the body is 10 m/s, determine :

(i) acceleration produced in the direction of force, and

(ii) distance moved by the body in 4 seconds. [Ans. (i) 3 m/s2, (ii) 64 m]

30. The weight of a body on earth is 490 N. If the acceleration due to gravity onearth = 9.8 m/s2, what will be the weight of the body on :

(i) the moon where gravitational acceleration is 1.5 m/s2, and

(ii) the sun, where gravitational acceleration is 300 m/s2.

[Ans. (i) 75 N, (ii) 15000 N]

31. A force of 300 N acts on a body of mass 150 kg for 30 seconds. If the initialvelocity of the body is 25 m/s, determine the final velocity of the body, when theforce :

(i) acts in the direction of motion, and

(ii) acts in the opposite direction of motion. [Ans. (i) 85 m/s, (ii) – 35 m/s]

32. A body of mass 20 kg falls on the muddy ground from a height of 39.2 m. Thebody penetrates into the ground. Find the distance through which the body willpenetrate into the ground, if the resistance by the ground to penetration isconstant and equal to 980 N. Take g = 9.8 m/s2. [Ans. 9.8 m]

33. A lift carries a weight of 110 N and is moving with a uniform acceleration of3 m/s2. Determine the tension in the cables supporting the lift, when

(i) lift is moving upwards, and

(ii) lift is moving downwards. Take g = 9.80 m/s2. [Ans. 143.67 N, 76.34 N]


NOTES

Dynamics of Particles34. A lift has an upward acceleration of 1.5 m/s2. What pressure will a man weighing500 N exert on the floor of the lift ? What pressure would he exert if the lift hadan acceleration of 1.5 m/s2 downward ? Take g = 9.8 m/s2.

[Ans. 576.5 N, 423.5 N]

35. An elevator weighs 2000 N and is moving vertically downwards with a uniformacceleration. Write the equation for the elevator cable tension. Startingfrom rest it travels a distance of 30 m during an interval of 12 seconds. Findthe cable tension during this time. Neglect all other resistances to motion. Whatare the limits of cable tension ?

[Ans. 1915.2 N, at f = 0, T = 2000 and at f = 9.81, T = 0]

36. An elevator weighing 6000 N is ascending with an acceleration of 2 m/s2. Duringthis ascent its operator whose weight is 600 N is standing on the scales placedon the floor. What is the scale reading ? What will be the total tension in thecables of the elevator during this motion ? [Ans. 722.3 N, 7945.5 N]

37. Find the work done in drawing a body :

(i) Weighing 1000 N through a distance 10 m along a horizontal surface by ahorizontal force of 400 N.

(ii) Weighing 1000 N through a distance 10 m along a horizontal surface by aforce of 400 N whose line of action makes an angle of 30° with horizontal.

[Ans. 4000 N m, 3464 N m]

38. A body of weight 2000 N moves on a level horizontal rough road for a distanceof 200 m. The resistance of the road is 10 N per 1000 N weight of the body. Findthe work done by the resistance on the body. [Ans. 4000 N m]

39. A hammer of mass 1500 kg drops from a height of 60 cm on a pile of mass750 kg. Find the depth of penetration of the pile into the ground, if the averageresistance of the ground is 140 kN. Assume the impact between the hammerand pile to be plastic. [Ans. 5.005 cm]

40. A hammer of mass 200 kg falls through a height of 4 m on a pile of negligiblemass. If it drives the pile 50 cm into the ground, find the average resistance ofthe ground for penetration. [Ans. 17658 N]

41. A bullet of mass 30 gm is fired into a body of mass 10 kg , which is suspended bya string 0.8 m long. Due to this impact, the body swings through an angle of 30°.Find the velocity of bullet. [Ans. 483 m/s]

42. A bullet of mass 10 gm moving with a velocity of 100 m/s is fired into a body ofmass 1 kg which is suspended by a string 1 m long. Due to the impact, the bodyswings through some angle. Find the angle of swing when the bullet getsembedded in the body. [Ans. 18.2°]

43. If in problem 63, the bullet, instead of getting embedded into the body, escapesfrom the other end of the body with a velocity of 20 m/s, find the angle throughwhich the body will swing. [Ans. 14.7°]

44. If in problem 63, the bullet rebounds from the surface of the body with a velocityof 20 m/s, find the angle through which the body will swing. [Ans. 22.09°]

45. Ball A of mass 2 kg moving with a velocity of 4 m/s, strikes directly on a ball Bof mass 4 kg at rest. The ball A, after striking, comes to rest. Find the velocity ofball B after striking and co-efficient of restitution. [Ans. 2 m/s, 0.5]


NOTES

Engineering Mechanics 46. A ball of mass 1 kg moving with a velocity of 2 m/s, strikes directly on anotherball of mass 2 kg at rest. If the co-efficient of restitution between the two ballsis 0.5, find the velocities of the two balls after impact. [Ans. 0, 1 m/s]

47. A vehicle of mass 600 kg and moving with a velocity of 12 m/s strikes anothervehicle of mass 400 kg moving at 9 m/s in the same direction. Due to the impactboth the vehicles get coupled and move together. Find the common velocity withwhich the two vehicles move after impact. [Ans. 10.8 m/s]

48. A body of mass 100 kg, moving with a velocity of 9 m/s, collides directly with astationary body of mass 50 kg. If the two bodies become coupled so that theymove on together after the impact, what is their common velocity.

[Ans. 6 m/s]49. A bullet of mass 100 gm is fired into a freely suspended target of mass 5 kg. On

impact, the target along with the bullet moves with a velocity of 5 m/s in thedirection of firing. Find the velocity of bullet. [Ans. 255 m/s]

50. A ball of mass 30 kg moving with a velocity of 4 m/s strikes directly another ballof mass 15 kg moving in the opposite direction with a velocity of 12 m/s. If theco-efficient of restitution is equal to 5/6, then determine the velocity of each ballafter impact. [Ans. v1 = – 3.33 m/s, v2 = 10 m/s]

��

• Ashok Gupta, ‘‘Interactive Engineering Mechanics – Statics – A Virtual Tutor(CDROM)’’, Pearson Education Asia Pvt., Ltd., (2002).

• Dr. I.S. Gujral, ‘‘Engineering Mechanics’’, Laxmi Publications (P) Ltd.• Dr. R.K. Bansal, ‘‘A Textbook of Engineering Mechanics’’, Laxmi Publications

(P) Ltd.


NOTES

Friction and Elements ofRigid Body Dynamics

U N I T

5FRICTION AND ELEMENTSOF RIGID BODY DYNAMICS

STRUCTURE

5.1 Introduction5.2 Limiting Force of Friction and Definitions of Certain Terms5.3 Laws of Coulomb Friction5.4 Simple Contact Friction5.5 Rolling Resistance5.6 Belt Friction5.7 Transmission of Power through Belts5.8 Motion of Translation5.9 Motion of Rotation5.10 Plane Motion of a Rigid Body• Summary• Glossary• Review Questions• Further Readings

��

After going through this unit, you should be able to :• define frictional force, law of Columb friction and simple contact friction.• elaborate the concepts of rolling resistance and belt friction.• determine translation and rotation of rigid bodies and general plane motion.

��

When a solid body slides over a stationary solid body, a force is exerted at thesurface of contact by the stationary body on the moving body. This force is called theforce of friction or frictional force and is always acting in the direction opposite tothe direction of motion. The property of the bodies by virtue of which a force is exertedby a stationary body on the moving body to resist the motion of the moving body is


NOTES

Engineering Mechanics called friction. Friction acts parallel to the surface of contact and depends upon thenature of surface of contact.

��

��

For defining the terms like co-efficient of friction (µ) and angle of friction (φ),consider a solid body placed on a horizontal plane surface as shown in Fig. 5.1.

Let W = Weight of body acting through C.G. downward,R = Normal reaction of body acting through C.G. upward,P = Force acting on the body through C.G. and parallel to the horizontal

surface.If P is small, the body will not move as the force of friction acting on the body

in the direction opposite to P will be more than P. But if the magnitude of P goes onincreasing, a stage comes, when the solid body ison the point of motion. At this stage, the force offriction acting on the body is called limiting forceof friction. The limiting force of friction is denotedby F.

Resolving the forces on the body verticallyand horizontally, we get

R = WF = P.

If the magnitude of P is further increased the body will start moving. The forceof friction, acting on the body when the body is moving, is called kinetic friction.

Co-efficient of Friction (µ). It is defined as the ratio of the limiting force offriction (F) to the normal reaction (R) between two bodies. It is denoted by the symbolµ. Thus

µ = Limiting force of friction

Normal reaction= F

R .

∴ F = µR ...(5.1)Angle of Friction (φ). It is defined as the angle

made by the resultant of the normal reaction (R) andthe limiting force of friction (F) with the normal reac-tion (R). It is denoted by φ. Fig. 5.2 shows a solid bodyresting on a rough horizontal plane.

Let S = Resultant of the normal reaction (R) andlimiting force of friction (F)

Then angle of friction = φ= Angle between S and R

From Fig. 5.2, we have

tan φ = FR

RR

= µ[� F = µR from (5.1)]

= µ = Co-efficient of friction ...(5.2)Thus the tangent of the angle of friction is equal to the co-efficient of friction.

Fig. 5.1. Solid body onhorizontal surface

Horizontalplane

FForce offriction R

C. G.

Solidbody

W

P

Fig. 5.2. Angle of friction

C. G.

W

P

R S

F

Fφ


NOTES


A block of weight W is placed on a rough horizontal plane surface as shown inFig. 5.3 and a force P is applied at an angle θ with the horizontal such that the blockjust tends to move.

Let R = Normal reaction µ = Co-efficient of frictionF = Force of friction

= µRIn this case the normal reaction R will not be

equal to weight of the body. The normal reaction isobtained by resolving the forces on the blockhorizontally and vertically. The force P is resolved intwo components i.e., P cos θ in the horizontal directionand P sin θ in the vertical direction.

Resolving forces on the block horizontally, weget

F = P cos θor µR = P cos θ ...(i) (� F = µR)

Resolving forces on the block vertically, we get R + P sin θ = W

∴ R = W – P sin θ ...(ii)From equation (ii), it is clear that normal reaction is not equal to the weight

of the block.If in equation (ii), the values of W, P and θ are

known, then value of normal reaction (R) can beobtained. This value of R can be substituted inequation (i) to determine the value of co-efficient offriction µ.

Note. (i) The force of friction is always equal to µR(i.e., F = µR).

(ii) The normal reaction (R) is not equal to the weightof the body always.

Cone of Friction. It is defined as the right cir-cular cone with vertex at the point of contact of thetwo bodies (or surfaces), axis in the direction of normalreaction (R) and semi-vertical angle equal to angle offriction (φ). Fig. 5.4 shows the cone of friction in which,

O = Point of contact between two bodiesR = Normal reaction and also axis of the cone of frictionφ = Angle of friction.Types of Friction. The friction is divided into following two types depending

upon the nature of the two surfaces in contact :1. Static friction, and2. Dynamic friction.If the two surfaces, which are in contact, are at rest, the force experienced by

one surface is called static friction. But if one surface starts moving and the other isat rest, the force experienced by the moving surface is called dynamic friction. If

Fig. 5.3

Fig. 5.4. Cone of friction

F = Rm

WP

R

q

RAxis

Cone offriction

Point ofcontact

ff

P


NOTES

Engineering Mechanics between the two surfaces, no lubrication (oil or grease) is used, the friction, that existsbetween two surface is called ‘Solid Friction’ or ‘Dry Friction’. Solid friction or Dryfriction is also known as Coulomb friction.

��

The friction, that exists between two surfaces which are not lubricated, isknown as solid friction. The two surfaces may be at rest or one of the surface is movingand other surface is at rest. The following are the laws of solid friction :

1. The force of friction acts in the opposite direction in which surface is havingtendency to move.

2. The force of friction is equal to the force applied to the surface, so long as thesurface is at rest.

3. When the surface is on the point of motion, the force of friction is maximumand this maximum frictional force is called the limiting friction force.

4. The limiting frictional force bears a constant ratio to the normal reactionbetween two surfaces.

5. The limiting frictional force does not depend upon the shape and areas of thesurfaces in contact.

6. The ratio between limiting friction and normal reaction is slightly less whenthe two surfaces are in motion.

7. The force of friction is independent of the velocity of sliding.The above laws of solid friction are also called laws of static and dynamic friction.

��

The problems on simple contact friction consists of :(i) Simple contact friction on a horizontal surface,

(ii) Simple contact friction on an inclined plane, and(iii) Simple contact friction on Ladder surface.Problem 1. A body of weight 100 Newtons is placed on a rough horizontal plane.

Determine the co-efficient of friction if a horizontal force of 60 Newtons just causesthe body to slide over the horizontal plane.

Sol. Given :Weight of body, W = 100 NHorizontal force applied,

P = 60 N∴ Limiting force of friction,

F = P = 60 NLet µ = Co-efficient of friction.The normal reaction of the body is given as

R = W = 100 NUsing equation (5.1),

F = µR

or µ = FR

= 60100

= 0.6. Ans.

Fig. 5.5

W = 100 N

P = 60 N

R

F


NOTES


Angle of Repose. The angle of repose is defined as the maximum inclinationof a plane at which a body remains in equilibrium over the inclined plane by theassistance of friction only.

Consider a body of weight W, resting on a rough inclined plane as shown inFig. 5.6.

Let R = Normal reaction acting at right angle to the inclined plane.α = Inclination of the plane with the horizontal.F = Frictional force acting upward along the plane.

Let the angle of inclination (α) be graduallyincreased, till the body just starts sliding downthe plane. This angle of inclined plane, at whicha body just begins to slide down the plane, iscalled angle of repose.

Resolving the forces along the plane, we get

W sin α = F ...(i)

Resolving the forces normal to the plane,we get

W cos α = R ...(ii)

Dividing equation (i) by equation (ii),

WW

FR

sincos

αα

= or tan α = FR

...(iii)

But from equation (5.2), we know

tan φ = FR

...(iv)

where φ = angle of friction.Hence from equations (iii) and (iv), we have

tan α = tan φor α = φor Angle of repose = Angle of friction.

Equilibrium of a Body Lying on a Rough Inclined Plane. We have alreadystudied that if the inclination of the plane, with the horizontal, is less than the angleof friction, the body will remain in equilibrium without any external force. If the bodyis to be moved upwards or downwards in this condition an external force is required.But if the inclination of the plane is more than the angle of friction, the body willnot remain in equilibrium. The body will move downward and an upward externalforce will be required to keep the body in equilibrium.

Such problems are solved by resolving the forces along the plane andperpendicular to the planes. The force of friction (F), which is always equal to µR isacting opposite to the direction of motion of the body.

Problem 2. Prove that the angle of friction (φ) is equal to the angle made byan inclined plane with the horizontal when a solid body, placed on the inclined plane,is about to slide down.

Sol. A solid body of weight, W is placed on an inclined plane AC as shown inFig. 5.7.

Fig. 5.6

R

W

α

α

W sin α

F

W cos α


NOTES

Engineering Mechanics Let α = Angle of the inclined plane AC with horizontal plane AB, such thatbody just starts moving downward.

The body is in equilibrium under the action of following forces :1. Weight of the body (W) acting vertically downwards.2. Normal reaction (R), acting perpendicular to the inclined plane, AC.3. The force of friction, F = µR, acting up the plane as the body is about to

slide down the plane.The weight, W can be resolved in two

component one along the plane and other per-pendicular to the plane. The components are Wsin α and W cos α respectively.

As the body is in equilibrium, the forcesalong and perpendicular to the inclined planeare :

W sin α = F = µRW cos α = R

Dividing WW

RR

sincos

αα

µ= = µ

or tan α = µBut from equation (5.2), we have

tan φ = µwhere φ = angle of friction

∴ tan α = tan φ = µor α = φ.

The above relation shows that the angle of friction is equal to angle of theinclined plane when a solid body, placed on the inclined plane is about to slide down.

Problem 3. A body of weight 500 N is pulled up an inclined plane, by a forceof 350 N. The inclination of the plane is 30° to the horizontal and the force is appliedparallel to the plane. Determine the co-efficient of friction.

Sol. Given :Weight of body, W = 500 NForce applied, P = 350 NInclination, α = 30°Let µ = Co-efficient of Q friction

R = Normal reactionF = Force of friction = µR.

The body is in equilibrium under the actionof the forces shown in Fig. 5.8.

Resolving the forces along the plane,500 sin 30° + F = 350

or 500 sin 30° + µR = 350 (� F = µR) ...(i)Resolving forces normal to the plane,

R = 500 cos 30° = 500 × .866 = 433 N

Fig. 5.7

R

W

α

W sin α

C

W cos αα

A B

F = Rµ

Solid body

Direction

of motion

Fig. 5.8. Body moving up

500 N

500 sin 30°

500 cos 30° N

R350 N

F30°

30°


NOTES


Substituting the value of R in equation (i), we get 500 sin 30° + µ × 433 = 350

or 500 × 0.5 + 433 µ = 350or 433 µ = 350 – 500 × 0.5 = 350 – 250 = 100

∴ µ = 100433

= 0.23. Ans.

Problem 4. A body of weight 450 N is pulled up along an inclined plane havinginclination 30° to the horizontal at a steady speed. Find the force required if the co-efficient of friction between the body and the plane is 0.25 and force is applied parallelto the inclined plane. If the distance travelled by the body is 10 m along the plane,find the work done on the body.

Sol. Given :Weight of body, W = 450 NInclination of plane, α = 30°Co-efficient of friction, µ = 10.25Distance travelled by body = 10 mLet the force required = P.The body is in equilibrium under the action

of forces shown in Fig. 5.9.Resolving forces along the plane,

P = W sin 30° + µR = 450 × 0.5 + 0.25 × RP = 225 + 0.25 R

Resolving forces normal to the plane, R = W cos 30° = 450 × 0.866 = 389.7 N

Substituting the value of R in equation (i), P = 225 + 0.25 × 389.7 = 322.425 N. Ans.

Work done on the body = Force × Distance travelled in the direction of force = 322.525 × 10 Nm = 3224.25 Nm = 3224.25 J (where J = Joules = Nm). Ans.

Analysis of Ladder Friction. Fig 5.10 shows a ladder AC resting on theground and leaning against a wall.

Let RA = Reaction at A RC = Reaction at C FA = Force of friction at A

= µRA FC = Force of friction at C

= µRC.Due to the self weight of the ladder or when some

man stands on the ladder, the upper end A of the laddertends to slip downwards, and hence the force of frictionbetween the ladder and the vertical wall FA = µRA will beacting upwards as shown in Fig. 5.10. Similarly, the lowerend C of the ladder will tend to move towards right andhence a force of friction between ladder and floor FC = µRCwill be acting towards left.

For the equilibrium of the system, the algebraic sum of the horizontal andvertical components of the forces must be zero. Also the moments of all the forces aboutany point must be zero.

Note. If the vertical wall is smooth, there will be no force of friction between the ladderand the vertical wall.

Fig. 5.9. Body moving up

W

W sin 30°

W cos 30°

RP

F =Rµ 30°

30°

Fig. 5.10

RA

F = RA A�

Ladder

Ver

tical

wal

l

A

B CRC

F = RC C�

Floor


NOTES

Engineering Mechanics Problem 5. A uniform ladder of length 10 m and weighing 20 N is placedagainst a smooth vertical wall with its lower end 8 m from the wall. In this positionthe ladder is just to slip. Determine :

(i) the co-efficient of friction between the ladder and the floor, and(ii) frictional force acting on the ladder at the point of contact between ladder

and floor.Sol. Given :Weight of ladder, W = 20 NLength of ladder, AC = 10 mDistance of lower end of ladder from wall,

i.e., BC = 8 m.

In right-angled triangle ABC,

AB = AC BC2 2−

= 10 8 100 642 2− = −

= 36 = 6 m.As vertical wall is smooth, hence there will

be no force of friction between ladder and wall.In the position of the ladder shown in Fig. 5.11, the ladder is just to slip. Hence

the lower end C of the ladder will tend to move towards right and hence a force offriction between ladder and floor (i.e., FA = µRC) will be acting towards left.

Let RA = Reaction at ARC = Reaction at C

µ = Co-efficient of friction between ladder and floor at CFC = Force of friction at C = µRC

Ladder is uniform and the weight of ladder (W = 20 N) is acting at the middlepoint of AC to G. The line of action of W will pass through the middle point of BC.Hence distance

CD = 12

× BC = 82

= 4.0 m.

Resolving the force vertically, we get RC = 20 N

Resolving the forces horizontally, we get RA = FC = µ × RC = µ × 20 = 20 µ ...(i)

Taking the moments of all forces about point C,Clockwise moments = Anti-clockwise moments

RA × AB = 20 × CD(The moments of RC and FC about point C is zero)

or 20 µ × 6 = 20 × 4 [� RA = 20 µ from equation (i)]

∴ µ = 20 420 6

×× = 0.67. Ans.

(ii) Frictional force acting at C is given as FC = µ × RC = 0.67 × 20 = 13.40 N. Ans.

Fig. 5.11

G

20 N

DB FC

A

RA

10 m

C

8 m RC


NOTES


To prove that ladder is in equilibrium for the given positionLet FA′ = Force of friction when ladder is in equilibrium

RB′ = Normal reaction at B when ladder is in equilibriumRA′ = Normal reaction A when ladder is in equilibrium

The ladder will be in equilibrium if FA′ is less than FA.From Fig. 5.12, ABC is a right-angled triangle. G

is the middle point of AB and GD is normal to AC. HenceD is the middle point of AC.

∴ AD = CD = 2.5 mAlso AB2 = AC2 + BC2

∴ BC = AB AC2 2− = 13 52 2−

= 169 25−

= 144 = 12 m.The ladder will be in equilibrium, if the moments

of all forces (acting on the ladder) about any point is zero.Taking moments of all forces shown in Fig. 5.13

about A,25 × AD = RB′ × BC = RB′ × 12

or 25 × 2.5 = RB′ × 12 (� AD = 2.5)

∴ RB′ = 2 5 2 5

12. .×

= 5.21 N

Equating horizontal forces, FA′ = RB′ = 5.21 N

Hence for equilibrium, the force of friction required as 5.21N. But maximum amount of force of friction available is 7.5 Nwhich is more than the required amount. Hence the ladder willremain in equilibrium in the given position.

Problem 6. The weight of 14 m long bar as shown in Fig. 5.13 (a) is 600 Nand it may be considered to be concentrated at a point 6 m from the bottom. It restsagainst a smooth vertical wall at A and on a rough horizontal floor at B. The co-efficient of static friction between the bar and the floor is 1/3. Establish by calculationsif the bar would stand in the 60° position as shown.

Sol. Given : Length, AB = 14 m Weight, W = 600 N

The weight is acting at D, which is 6 m frompoint B.

∴ Length BD = 6 m and length AD = 8 mVertical wall is smooth and hence FA = 0Co-efficient of friction, µ = 1/3Angle CBA = 60°Let us find :(i) actual force of friction available at point B, and

(ii) force of friction required for equilibrium atpoint B.

Fig. 5.12

RB′B

G

C

25 kgf

D F = RA A′ µ ′A

RA

Fig. 5.13

A

Sm

ooth

60°

C B F = RB Bµ

600 N

8 m

6 mD

14 m

E

RB


NOTES

Engineering Mechanics The force of friction required for equilibrium will be obtained from momentequation i.e., ΣM = 0. The actual force of friction available will be obtained fromsummation of forces in horizontal and vertical directions i.e., ΣFx = 0 and ΣFy = 0.

Let RA = Normal reaction at A RB = Normal reaction at B FB = Force of friction at B = µRB

The forces acting on the ladder are shown in Fig. 5.13.Resolving forces horizontally, RA = FB ...(i)Resolving forces vertically, RB = 600 N ...(ii)

Now FB = µ × RB = 13

× 600 = 200 N

∴ Actual force of friction available at point B= FB = 200 N

To find the force of friction required for equilibrium, take the moments of allforces about point B.

∴ 600 × BE = RA × AC ...(iii)

But BE = BD cos 60° = 6 × 12

= 3 m

and AC = AB sin 60° = 14 × 0.866 = 12.124 mSubstituting these values in equation (iii), we get

600 × 3 = RA × 12.124

∴ RA = 600 312 124

×. = 148.46 N

But from equation (i), FB = RA = 148.46 N

∴ Force of friction required for equilibrium = 148.46 N.As the actual force of friction available at B is 200 N, which is more than the

force of friction required for equilibrium, the ladder will stand stable in the 60°position as shown in the given figure. Ans.

��

Fig. 5.14 shows a hard roller moving withoutslipping on a horizontal surface while supporting aload W at the centre. If the roller is moving withuniform velocity, due to horizontal force P, thensome sort of resistance must be present so that thenet force in the direction of uniform motion is zero[� Net force = Mass × Acc. As due to uniformmotion of roller, acceleration is zero, hence net forceis zero. But net force = P – resisting force]. Theresistance to the motion of the roller in such casesis known as rolling resistance.

Hence rolling resistance is the resistance tothe motion of the roller moving without slipping ona horizontal surface while supporting a load W atthe centre.

B

W

Direction of Motion

Hardroller

F

A

RN

Fig. 5.14


NOTES


The rolling resistance can be understoodif we imagine the surface to be yielding as shownin Fig. 5.15. The same result may be if the rolleris assumed to be yielding and moving over a rigidsurface or both the roller and surface areyielding. The ground in front of the roller isdepressed, causing the normal reaction (RN) toact ahead of the line of action of the weight Was shown in Fig. 5.15. Now for equilibrium of theroller, the three forces i.e., W, P and RN must beconcurrent. As W and P intersect at B, hence RNwill also pass through B, making an angle φ withthe vertical direction as shown in Fig. 5.15.

Now for equilibrium of the roller,ΣFx = 0 or P = RN sin φ …(i)

and ΣFy = 0 or W = RN cos φ …(ii)To eliminate RN from equations (i) and (ii), divide (ii) by equation (i),

∴ PW

= sincos

φφ

= tan φ …(iii)

Generally the angle φ is very small, hence tan φ ~− sin φ. But sin φ from Fig.

5.15, is equal to aR

where R = Radius of roller.

∴ tan φ ~− sin φ = ar

Substituting the value of tan φ in equation (iii), we get

PW

aR

=

or P = W a

R×

…(5.3)

The distance ‘a’ in the above equation is known as co-efficient of rolling

resistance. For a given material of given radius of roller, the ratio PW

is constant.

This means with the increase of W, the value of P also increases. Also for a givenmaterial of constant radius of roller, the value of ‘a’ (i.e., value of co-efficient of rollingresistance) is constant.

From equation (5.3), it is also clear that for a given load W, the force requiredto maintain uniform motion of the roller decreases as the radius of roller increases.

Problem 7. A railroad freight car is having a weight of 80 metric tons. Thediameter of the wheels is 0.75 m and co-efficient of rolling resistance betweenwheel and track is 0.025 mm. Find the horizontal force required to maintain uniformspeed.

What will be the horizontal force for the truck and trailer of weight 80 metrictons if the diameter of the tires is 1.2 m and co-efficient of rolling resistance betweenthe truck tires and road is 0.625 mm ? In which case, the horizontal force is minimum.

B

W

A

RN

P

Direction of motion

a

φ

Fig. 5.15


NOTES

Engineering Mechanics Sol. Given :Railroad freight car Truck and trailer

Weight, W1 = 80 metric tons, W2 = 80 metric tons = 80 × 1000 kgf = 80 × 1000 × 9.81 N

(� 1 ton = 1000 kgf = 1000 × 9.81 N) = 80 × 1000 × 9.81 N D2 = 1.2 m = 1.2 × 103 mm

Dia., D1 = 0.75 m = 0.75 × 103 mm R2 = 1.2 1000

2×

= 1.2 × 500 mm

∴ Radius, R1 = 0 75 1000

2. ×

= 0.75 × 500 mm a2 = 0.625 mm.

Co-efficient of rolling resistance,a1 = 0.025 mm.


P1 = W a

R1 1

1

×P2 =

W aR

2 2

2

×

= ( . ) .

.80 1000 9 81 0 025

0 75 500× × ×

× = (80 1000 9.81) 0.625

1.2 500× × ×

×= 52.32 N Ans. = 817.5 N Ans.

In case of railroad freight car, the horizontal force required to maintain uniformspeed, is minimum. Ans.

��

A belt is passing over a pulley and hence the belt is in contact with the surfaceof the pulley. If the surface of the pulley is perfectly smooth, the tension in the belton both sides of the pulley will be same (i.e., the tension throughout the belt will beconstant). Also for the perfectly smooth surface, there will be no frictional resistanceand hence no driving torque will be developed.

But if the surface of the pulley is rough, the tension in the belt will not beconstant. The tension will vary throughout the length of the belt which is in contactwith pulley. This variation in tension is due to frictional resistance. The frictionalresistance depends on the co-efficient of friction (i.e., value of µ) between the beltand pulley surface. It will be shown in the next articles that

TT

1

2 = eµθ

where T1 = Tension in the belt on tight side,T2 = Tension in the belt on slack side, µ = Co-efficient of friction, and θ = Angle of contact in radians.

Ratio of Belt Tensions. Fig. 5.16 shows a driver pulley A and driven pulleyB rotating in the clockwise direction. Fig. 5.17 shows only the driven pulley B.Consider the driven pulley B.

Let T1 = Tension in the belt on the tight sideT2 = Tension in the belt on the slack side


NOTES


θ = Angle of contact, i.e., the angle subtended by the arc EF at the centreof the driven pulley.

µ = Co-efficient of friction between the belt and pulley.

Driver pulleySlack side

T2 T2

Driven pulley

A B

T1

T1

Tight side

Fig. 5.16

The ratio of the two tensions may befound by considering an elemental piece of thebelt MN subtending an angle δθ at the centreof the pulley B as shown in Fig. 5.16. Thevarious forces which keep the elemental pieceMN in equilibrium are :

(i) Tension T in the belt at M acting tan-gentially,

(ii) Tension T + δT in the belt at N actingtangentially,

(iii) Normal reaction R acting radiallyoutward at P, where P is the middlepoint of MN,

(iv) Frictional force F = µR acting at right angles to R and in the opposite direc-tion of the motion of pulley.

Now angle PBM = δθ2

. Also angle TPF = δθ2

.

Resolving all the forces acting on the belt MN in the horizontal direction, weget

R = T sin δθ2

+ (T + δT) sin δθ2

Since the angle δθ is very small, sin δθ2

can be written as δθ2

. Hence the aboveequation becomes as

R = T × δθ2

+ (T + δT) × δθ2

= T × δθ2

+ T × δθ2

+ δT × δθ2

= T × δθ + δ δθT ×

2= T × δθ ...(i)

Neglecting the small quantityTδ δθ×�

��2

Fig. 5.17

E

Bq

dq

TT2

F = Rm

dq2

M

P

R

N

F(T + T)d

T1

Drivenpulley

Directionof rotation


NOTES

Engineering Mechanics Now resolving all the forces vertically, we get

F = (T + δT) cos δθ2

– T cos δθ2

Since δθ is very small, hence cos δθ2

reduces to unity i.e., 1. Hence the aboveequation becomes as

F = (T + δT) – T = δT

or µR = δT (� F = µR)

or R = δµT

...(ii)

Equating the two values of R given by equations (i) and (ii), we get

T × δθ = δµT

orδTT

= µ . δθ

Integrating the above equation between the limits T2 and T1, we get

δ µ θ µ θTT

d dT

T

2

1

� � �= =.

or loge TT

1

2 = µ × θ

orTT

1

2 = eµ × θ ...(5.4)

In equation (5.4), θ should be taken in radians. Here θ is known as angle ofcontact. For an open belt or for a crossed belt the angle of contact is determined asgiven below.

Angle of Contact for Open Belt Drive. With an open belt drive, the beltwill begin to slip on the smaller pulley, since the angle of lap is smaller on this pulleythan on the large pulley. The angle θ should be taken as the minimum angle of contact.Hence in equation (5.4), the angle of contact of lap (θ) at the smaller pulley must betaken into consideration.

Angle of contact, θ = (180 – 2α) ...(5.5)But the value of α is given by,

sin α = r r

x1 2−

...(5.6)

where r1 = Radius of larger pulley,r2 = Radius of smaller pulley, and x = Distance between the centres of two pulleys.

Angle of Contact for Crossed Belt Drive. For a crossed belt drive, the angleof lap on both the pulleys is same.

∴ Angle of contact, θ = (180 + 2α) ...(5.7)

The value of α is given by, sin α = r r

x1 2+

...(5.8)

where r1 = Radius of larger pulley,r2 = Radius of smaller pulley, and x = Distance between the centres of the two pulleys.


NOTES

Friction and Elements ofRigid Body Dynamics��

Let T1 = Tension in the tight side of the beltT2 = Tension in the slack side of the belt

v = Velocity of the belt in metre/s.The effective tension or force acting at the circumference of the driven pulley

is the difference between the two tensions (i.e., T1 – T2).∴ Effective driving force = (T1 – T2)∴ Work done per second = Force × Velocity

= (T1 – T2) × v Nm

∴ Power transmitted = ( )T T1 2

1000− × ν

kW ...(5.9)

or P = (T1 – T2) × v Watts ...(5.10)Equation (5.9) gives the power in kW whereas equation (5.10) gives the power

in watts. In case of equation (5.10), the tensions T1 and T4 are taken in Newtons.Torque exerted on the driving pulley = (T1 – T2) × r1 ...(5.11)

and Torque exerted on the driven pulley = (T1 – T2) × r2 ...(5.12)Problem 8. A belt is running over a pulley of diameter 120 cm at 200 r.p.m.

The angle of contact is 165° and co-efficient of friction between the belt and pulley is0.3. If the maximum tension in the belt is 3000 N, find the power transmitted by thebelt.

Sol. Given :Dia. of pulley, d = 120 cm = 1.2 mSpeed of pulley, N = 200 r.p.m.

Angle of contact, θ = 165° = 165 × π

180 radians.

� 1180

° =��

��

πrad.

Co-efficient of friction, µ = 0.3Max. tension, T1 = 3000 N

Velocity of belt, v = π πd N60

1.2= × × 20060

= 12.56 m/s.

Let T2 = Tension on the slack side of the belt.Using equation (5.4), we get

TT

1

2 = eµ × θ = e0.3 × 165 × π/180 = e0.8635 = 2.3714

or300

2T = 2.3714 ∴ T2 =

30002 3714.

= 1265 N

Power transmitted is given by equation (5.14) as,

P = ( )T T1 2

1000− × ν

kW

= ( ) .3000 1265 12 65

1000− ×

= 21.79 kW. Ans.


NOTES


If a body moves in such a way that all its particle movein parallel planes and travel the same distance, then the bodyis said to have motion of translation. When a rigid body is intranslation, all the points of the particles of the body have thesame velocity and same acceleration at any particular instant.The motion of the rigid link AB, from its initial position ABto A1B1, shown in Fig. 5.18 is an example of motion oftranslation.

��

If a body rotates about a fixed point in such a way thatall its particle move in circular path, the body is said to havethe motion of rotation. The fixed point about which the bodyrotates is called the point of rotation and the axis, passingthrough the fixed point, is known as axis of rotation. The par-ticles, lying on the axis of rotation, have zero velocity and zeroacceleration. The motion of the link AB, from its initial posi-tion AB to AB1, shown in Fig. 5.19 is an example of motionof rotation.

Combined Motion of Translation and Rotation.Consider a link AB, which moves from its initial position ABto A1B1 in a short interval of time as shown in Fig. 5.20. Thelink has neither entirely motion of translation nor entirelyrotation, but a combination of the two. The motion of the linkfrom the position AB to the position A1B1 may be regardedas to consists of :1st Case

(i) A motion of entirely translation from the position AB to the position A1B* sothat A1B* is parallel to AB as shown in Fig. 5.21, and

(ii) A motion of entirely rotation about A1 from the position A1B* to the positionA1B1.2nd Case

(i) A motion of entirely rotation about A from the position AB to the positionAB* as shown in Fig. 5.22, and

(ii) A motion of entirely translation from the position AB* to the position A1B1.

B

AA1

B1

*B

A A1

B

*BB1

Fig. 5.21 Fig. 5.22

B B1

A A1

A

B

B1

Fig. 5.18

Fig. 5.19

A

B

A1

B1

Fig. 5.20


NOTES


3rd CaseThe combined motion of translation and rotation of the link from its initial

position AB to the position A1B1, may be assumed to be a motion of entirely rotationabout a certain point. This point is known as instantaneous centre of rotation.

Instantaneous Centre. It is the point aboutwhich the rotation of a link (having combined motionof translation and (rotation) is taking place. Theposition of the instantaneous centre of rotation is de-termined as given below :

Let the link AB in a short interval of timechanges its position from AB to A1B1. The point A ofthe link has moved to point A1 where the point B ofthe link has moved to the point B1 as shown inFig. 5.23. Draw the right bisectors of chord AA1 andchord BB1. Let CD is the right bisectors of AA1whereas EF is the right bisectors of BB1. Let these twobisectors meet at the point O. Then the point O is theinstantaneous centre of rotation of the link AB. Thismeans the link AB as a whole has rotated about O.

Let VA = Linear velocity of point AVB = Linear velocity of point B

ω = Angular velocity of link AB about O.This means the angular velocity of point A and

point B about O will be ω.Now, we know that v = r × ωLinear velocity = Angular velocity × r∴ Linear velocity of point A is given by

VA = Angular velocity of point A about O

× Distance of A from centre of rotation (i.e., from O)= ω × OA

∴ ω = VOA

A ...(i)

Similarly, the linear velocity of point B is given byVB = Angular velocity of point B about O

× Distance of B from centre of rotation (i.e., from O)= ω × BO

∴ ω = VBO

B ...(ii)

Equating equations (i) and (ii), we get

VOA

A = VBO

B or VV

A

B =

OABO ...(5.13)

The direction of the velocity at A will be at right angle to OA whereas thevelocity at B will be at right angle to BO.

Thus, if the directions of velocities at A and B are known, then the instantaneouscentre of AB is obtained by drawing perpendiculars to the directions of the velocities

E

F

B B1

C

DA

O

A1

Fig. 5.23


NOTES

Engineering Mechanics at A and B. The point, where these two perpendiculars, meet is the instantaneouscentre.

If the directions of velocities at A and B are parallel but unequal in magnitude,then instantaneous centre of AB is obtained by determining the point of intersectionof the line AB with the line joining the extremities of the vectors VA and VB as shownin Fig. 5.24 (a) and (b). In Fig. 5.24 (a), the instantaneous centre O is outside thelink AB whereas in Fig. 5.24 (b) the instantaneous centre O is on the link AB. Henceinstantaneous centre may lie on the link or outside the link.

(Point O is outside the link AB) (Point O is on the link AB)

Fig. 5.24 (a) Fig. 5.24 (b)

If the directions of velocities at A and Bare parallel and also are equal in magnitude,then the instantaneous centre is at infinityand all the points of the link AB have the samevelocity as shown in Fig. 5.24 (c). The twoparallel lines will meet at infinity.

Note.(i) The instantaneous centre is at rest and

has zero velocity.

(ii) The instantaneous centre may be outsidethe body or within the body.

(iii) If any point on the body or outside the body has zero velocity, then that point will beinstantaneous centre for the body at that instant.

Problem 9. A link AB is moving in a vertical plane. At a certain instant, whenthe link is inclined at 30° to the horizontal, the point A is moving horizontally at 4 m/s,while B is moving vertically upwards. Find the velocity of B.

Sol. Given :Inclination of link AB with horizontal = 30°Horizontal velocity of A, VA = 4 m/sLet VB

= Velocity of B in the vertically upwarddirection.

Since the actual directions of motion of A and Bare known, the position of the instantaneous centre canbe easily determined by drawing perpendiculars to thedirections of motions at A and B as shown in Fig. 5.25.

Fig. 5.24 (c)

Fig. 5.25


NOTES


AO is the perpendicular to the direction of motion of A whereas BO is theperpendicular to the direction of motion of B. The point O, at which these twoperpendiculars meet, is the instantaneous centre of the link AB.

Now using equation (5.13), we getVV

A

B =

OABO

= tan 30° � In ∆ AOBAOBO

, tan 30° =��

��

or4

VB =

1

3∴ VB = 4 × 3 = 4 × 1.732 = 6.928 m/s. Ans.

��! �� "

A plane motion of a rigid body is the combination of its rectilinear motion andits rotation about an axis perpendicular to the plane of motion. Or in other words, ageneral plane motion can be regarded equivalent to a combination of translation ofthe centre of gravity and rotation about an axis passing through the centre of mass.Therefore, the equations of motion will be :

Fx = max ...(5.14)

Fy = may ...(5.15)

for translation of the centre of mass,

and MG = IG . α ...(5.16)

for rotation about an axis passing through centre of mass of the body.

Problem 10. A cricket ball hits the bat at point A. The mass of the bat is mand G is its centre of gravity, as shown in Fig. 5.26 (a). Find the position of point Awhen the horizontal force exerted on the hands of the batsman at O is zero.

(a) Ball hitting the bat at A (b) Free-body diagram of bat

Fig. 5.26

Sol. Let the horizontal force exerted by the ball on the bat be F which acts atA. Let ax be horizontal acceleration of the bat, and let Rx and Ry be reactions in x-andy-direction acting at O.


NOTES

Engineering Mechanics The equations of motion for the bat are :

ΣFx = max gives

Rx – F = max ...(i)

ΣM0)ext. = I0 . α)inertia, gives

– Fh = (IG + mr 2 ) . α

or, – Fh = (mkG2 + mr 2 ) α ...(ii)

or – F = m k r

hG( )2 2+ α

...(iii)

∴ Eqn. (i) gives Rx = max – m k r

hG( )2 2+ α

But, ax = α . r

∴ We get Rx = mαr – m k r

hG( )2 2+ α

...(iv)

when, Rx = 0, we get

mαr = m k r

hG( )2 2+ α

or, h = k r

rG

2 2+...(v)

Note. When the ball hits, the bat at point A such that Rx = 0, then point A is calledthe point of percussion.


NOTES

Friction and Elements ofRigid Body Dynamics��

1. Force of friction always acts in the direction opposite to the direction of motion.2. The maximum value of frictional force acting on a body, when the body is on the

point of motion, is called limiting force of friction. It is denoted by F.3. The force of friction, acting on a body when the body is moving, is called dynamic

friction.4. The ratio of the limiting force of friction (F) to the normal reaction (R) between

two bodies is known as co-efficient of friction. It is denoted by µ. Mathemati-

cally, µ = FR

.

5. If a body is placed on a rough inclined plane and the angle of inclination of theplane is gradually increased, till the body just starts sliding down the plane.The angle of the inclined plane, at which the body just begins to slide down theplane, is called angle of repose.

6. Angle of repose is equal to angle of friction.7. Ratio of the tensions on the two sides of a belt is given by

TT

1

2 = eµ × θ

where T1 = Tension on the tight side,T2 = Tension on the slack side,

µ = Co-efficient of friction between belt and pulley,and θ = Angle of contact in radians.

= (180 – 2α) ... For an open belt= (180 + 2α) ... For a crossed belt

where α = sin–1 r r

x1 2−��

�� ... For an open belt

= sin–1 r r

x1 2+��

�� ... For a crossed belt.

8. Power transmitted by a belt is given by

P = ( )T T v1 2

75× ×

h.p. ...in M.K.S. Units where T1 and T2 are in kgf

= (T1 – T2) ⋅ v Watts ... in S.I. Units where T1 and T2 are in N.

��

• Friction. It is a force, exerted by a stationary body on the moving body to resistthe motion of moving body.

• Angle of friction. An angle made by the resultant of the normal reaction (R)and the limiting force of friction (F) with the normal reaction (R). It is denotedby φ.

φ = FR


NOTES

Engineering Mechanics • Static friction. If the two surfaces are in contact, in resting position, the forceexperienced by one surface is called static friction.

• Coulomb. If between the two surfaces, no lubrication is used, the friction thatexists between two surfaces is coulomb friction.

• Angle of repose. It is the maximum inclination of a plane at which a bodyremain in equilibrium over the inclinded plane by the assistance of friction only.

• Rolling resistance. It is the resistance to the motion of the roller moving withoutslipping on a horizontal surface while supporting a load W at the centre.

��

1. Define the terms : Friction, limiting force of friction, co-efficient of friction andangle of friction.

2. Explain the difference between co-efficient of friction and angle of friction.3. (a) State the laws of static and dynamic friction.

(b) State the laws of solid friction.4. Prove that the angle of friction is equal to the angle of the inclined plane, when

a solid body of weight W placed on the inclined plane, is about to slide down.5. What do you mean by ‘angle of repose’ ? Prove that angle of repose is equal to

the angle of friction.6. A body of weight W is placed on a rough inclined plane having inclination α to

the horizontal. A force P is applied to the body in such a way that it makes anangle θ to the inclined plane. Prove that the force required to drag the bodywhen the body is on the point of motion up the plane is given by

P = W sin ( )

cos ( )α φ

θ φ+

−

where α = Inclination of the inclined plane with horizontal θ = Inclination of the force with the inclined plane φ = Angle of friction.

7. Prove that in question 10, (i) the force P will be minimum if the angle ofinclination of the force with the inclined plane is equal to the angle of frictionand (ii) least force is given by

Pmin = W sin (α + φ).8. Distinguish between slip and creep in a belt drive. Derive an expression for the

ratio of tensions in the tight and slack sides in terms of µ and θ, when the belt isjust on the point of slipping.

9. Derive the expression for optimum speed of flat belt for the transmission ofmaximum power considering the effect of centrifugal tension.

10. A body of weight 150 N is placed on a rough horizontal plane. If the co-efficientof friction between the body and the horizontal plane is 0.4, determine thehorizontal force required to just slide the body on the plane. [Ans. 60 N]

11. The force required to pull a body of weight 40 N on a rough horizontal plane is15 N. Determine the co-efficient of friction if the force is applied at an angle of20° with the horizontal. [Ans. 0.404]


NOTES


12. A body of weight 60 N is placed on a rough horizontal plane. To just move thebody on the horizontal plane, a push of 18 N inclined at 20° to the horizontalplane is required. Find the co-efficient of friction. [Ans. 0.255]

13. A pull of 60 N inclined at 25° to the horizontal plane, is required just to move abody placed on a rough horizontal plane. But the push required to move thebody is 75 N. If the push is inclined at 25° to the horizontal, find the weight ofthe body and co-efficient of friction. [Ans. 253.83 N, .238]

14. A uniform ladder of weight 250 N and of length 5 m rests on a horizontal groundand leans against a rough vertical wall. The co-efficient of friction between theladder and floor is 0.3 and between the ladder and vertical wall is 0.2. When aweight of 900 N is placed on the ladder at a distance of 2 m from the top of theladder ; the ladder is at the point of sliding.

15. With the help of a belt, an engine running at 150 r.p.m., drives a line shaft. Thediameter of the pulley on the engine is 70 cm and the diameter of the pulley onthe line shaft is 35 cm. A 80 cm diameter pulley on the line shaft drives a 20 cmdiameter pulley keyed to a dynamo shaft. Find the speed of the dynamo shaftwhen (i) there is no slip (ii) there is a slip of 3% at each drive.

[Ans. (i) 1200 r.p.m. (ii) 1129 r.p.m.]16. Two parallel shafts 12 metres apart are to be connected by a belt running over

pulleys of diameters 480 cm and 80 cm respectively. Determine the length ofthe belt required : (i) if the belt is open, and (ii) belt is crossed.

[Ans. (i) 33.13 m, (ii) 33.45 m]17. A shaft which rotates at a constant speed of 160 r.p.m. is connected by belling to

a parallel shaft 72 cm apart which has to run at 60, 80 and 100 r.p.m. Thesmallest pulley on the driver shaft is 4 cm in radius. Determine the remainingradii of the two stepped pulleys for : (i) a crossed belt, and (ii) an open belt.[Ans. (i) r2 = 10.67 cm, r3 = 4.89 cm, r4 = 9.78 cm, r5 = 5.64 and r6 = 9.02 cm ;

(ii) r2 = 10.69, r3 = 5, r4 = 10, r5 = 6 and r6 = 9.6 cm]18. A shaft rotating at 200 r.p.m. drives another shaft at 300 r.p.m., and transmits

8 H.P. through a belt. The belt is 10 cm wide and 1 cm thick. The distancebetween the shafts is 4 m. The smaller pulley is 50 cm in diameter. Calculatethe stress in (i) open-belt, and (ii) crossed-belt. Take µ = 0.3. Neglect centrifugaltension. [Ans. (i) 12.68 kgf/cm2, (ii) 11.847 kgf/cm2]

��

• Rajasekaran, S, Sankarasubramanian, G., ‘‘Fundamentals of EngineeringMechanics’’, Vikas Publishing House Pvt. Ltd., (2000).

• Hibbeller, R.C., ‘‘Engineering Mechanics’’, Vol. 1 Statics, Vol. 2 Dynamics,Pearson Education Asia Pvt. Ltd. (2000).

• Dr. R.K. Bansal, ‘‘A Textbook of Engineering Mechanics’’, Laxmi Publications(P) Ltd.

Documents

DocumentEM