Engineering Mechanics - Unit1- RA

UNIT - 1STATICS OF PARTICLES

1.1 Introduction

Mechanics is that branch of physical science that deals with the study of a body or bodies under the action of a force or system of forces. The application of principles of mechanics to solve our day to day engineering problems, is called ‘Engineering Mechanics’.

M e c h a n ic s

M e c h a n ic s o f F lu id s M e c h a n ic s o f S o lid s

R ig id b o d ym e c h a n ic s

D e fo rm a b le b o d ym e c h a n ic s

S ta tic s D y n a m ic s

K in e m a tic s K in e tic s

Figure 1.1: Classification of mechanics

In rigid body mechanics, all bodies are assumed to be rigid. In fact no body is rigid and all bodies deform when loaded. However, while studying the rigid bodies, the deformation of the bodies are not important and the concept of rigid body is justified. Mechanics of rigid body is further divided into statics and dynamics.

The branch of science which deals with the study of a body when the body is at rest is known as statics while the branch of science which deals with the study of a body when the body is in motion is known as dynamics. Dynamics is further divided into kinematics and kinetics. The study of a body in motion, when the forces which cause the motion are not considered is called kinematics and if the forces are also considered for the body in motion is called kinetics.

1.2 Laws of MechanicsThe study of mechanics is based on the following laws. These laws are called laws

of mechanics.

1.2.1 Newton’s Law of Motion

i) First Law

Every particle continues in a state of rest or uniform motion in a straight line unless it is disturbed by an external force to change its state of rest or motion.

ii) Second Law

If the resultant force acting on a particle is not zero, the acceleration of the particle will be proportional to the resultant force and act in the same direction.

iii) Third Law

To every action, there is always an equal and opposite reaction.

iv) Law of Gravitation

This law states that two particles of mass m1 and m

2 are attracted towards each other

along the line connecting them with a force whose magnitude ‘F’ is proportional to the product of their masses and inversely proportional to the distance (r) between them.

Mathematically,

1 22

m mF G

r

where, G = Universal Constant of Gravitation

1.2.2 Parallelogram Law of Force

This law states that two forces (F1, F

2) acting on a particle may be replaced by a

single force, called their resultant ‘R’ which can be obtained by drawing diagonal of the parallelogram, which has sides being equal to the given forces.

F 2

F 10

R

Figure 1.2

From figure,

where, = Angle between the forces

= Angle which the resultant force makes with one of the forces

1.2.3 Triangular Law of Force

It states that if two forces (F1, F

2) acting simultaneously on a particle may be

represented in magnitude and direction by the two sides of a triangle taken in order and their resultant ‘R’ may be represented in magnitude and direction by the third side of the triangle taken in opposite direction.

F 1

F 2R

F 1

F 2

(a) (b )

Figure 1.3

1.2.4 Lami’s Theorem

It states that, if three forces (F1, F

2, F

3) acting at a point be in equilibrium, each force

is proportional to the sine of the angle between the other two. This law is also known as sine law.

F 2

F 1

F 3

Figure 1.4

From Lami’s theorem,

31 2 FF F

sin sin sin

1.2.5 Principle of Transmissibility

This law states that the conditions of equilibrium or motion of a rigid body will remain unchanged if a force acting at a given point of the rigid body is replaced by a force of the same magnitude and same direction, but acting at a different point, provided that the two forces have the same line of action. In simple terms, along the line of action of the force, the effect of the force on a body is independent of the point of application.

=

b

b

aa

F

F

Figure 1.5

1.3 Force and Force SystemThe action of one body on the other may be described as a force. It can also be

defined as the agent which change or tends to change the position of rest or motion of the body. The force is a vector quantity and it can be identified by its magnitude, direction and point of application. The number of forces acting at a point is called force system.

1.3.1 Classification of Force System

Forces acting on a particle or rigid body are classified into following types.

Force S ys tem

C oplanar (p lane) force system N on-cop lanar (space) fo rce sys tem

C oncurren tC oncurren tC olinear N on-concurren tN on-concurren t

Paralle lParalle l N onpara llelN onpara llel

If all forces acting on a particle or a rigid body lie in same plane, then it is called coplanar forces. In case of non-coplanar forces, all forces acting on a rigid body or a particle may not lie in the same plane but act in the different plane and have a common line of action is known as a collinear force system. Similarly, if all the forces act in the same plane and they intersect at a common point, then it is called coplanar concurrent force system. On other hand, if the entire forces act in the same plane but do not have a common point of application, then it is called coplanar non-concurrent force system. In case of coplanar non-concurrent force system, forces acting on a body may parallel or non-parallel to each other. If forces are parallel to each other, then it is called coplanar nonconcurrent parallel force system while if forces are non-parallel, then it is called coplanar nonconcurrent non-parallel force system.

In case of non-coplanar concurrent force system, all the forces act in different planes but they intersect at a common point. However, the non-coplanar non-concurrent force system, all forces act in different planes but do not intersect at a common point. Some of the classification of forces are shown in figure 1.6

o

P L A N E

P L A N E

a) C ollinear fo rce sys tem b) C oncurren t fo rce system

F 1F 2

F 3

F 3F 1

F 4

PL A N E

c) Para llel fo rce sys tem d) N on-concurren t non-parallelforce system

F 2

F 1

SPA C E

e) N on coplanar para llel fo rce sys tem

y

x

z

SPA C E

f) N on cop lanar noncurren t fo rce sys tem

Figure 1.6: Classification of force system.

1.3.2 Resultant of Concurrent Forces in Plane

When a number of forces act on a system, it is necessary to find out the resultant of the forces acting on a particle or rigid body since the state of rest or motion of a body can be determined by the resultant. Therefore, the resultant of a force system is defined as a single force having the same effect as that of the total effect of the force system.

The resultant of plane concurrent forces can be done by using two methods.

a) Graphical method

b) Analytical method

a) Graphical Method

There are three graphical methods are available to find the resultant of concurrent forces.

i. Parallelogram law of forces

ii. Triangular law of forces

iii.Polygon law of forces

i. Parallelogram Law of Forces

O

a d

b c

F 2

F 1

R

Figure 1.7: Parallelogram law of forces.

Let us consider two forces (F1, F

2) acting at a point O. The resultant of these two

forces can be obtained by constructing a parallelogram such that the forces F1 and F

2

represent the two adjacent sides of the parallelogram are shown in figure 2.7. In the above figure, the angle ‘’ represent the angle made by the resultant with the force F

1. However,

this method can be used only to determine the resultant of two forces only.

ii. Triangular Law of Forces

F 1

F 2R

F 1

0

F 2

(a) (b )

Figure 1.8: Triangular law of forces.

In this method, the resultant of two forces (F1, F

2) acting at a point can be determined

by constructing a triangle such that the forces F1 and F

2 are represented by two sides of the

triangle taken in an order. The closing side of the triangle represents the resultant of two forces as shown in figure 1.8. The major limitation of this method is that it connot be used for finding the resultant of more than two forces.

iii. Polygon Law of Forces

The most advantage of this method is that it can be used for finding resultant of more than two forces. In this method the resultant of concurrent forces acting at a point can be obtained by drawing a polygon in a systematic order and the closing side of the polygon taken in an opposite direction represents the resultant of concurrent forces.

To illustrate, let us consider four forces (F1, F

2, F

3, F

4) acting at a point as shown in

figure 1.9. The resultant of these forces is the closing side of the polygon.

F 2

F 2F 3

F 1

F 4

R

F 1

O

F 4

F 3

(a) (b )

Figure 1.9: Polygon law of forces

b) Analytical Method

Resultant of concurrent forces can also be found out by the following analytical methods.

i. Parallelogram Law of Forces

F 2

O C D

BA

F 1

R

Figure 1.10: Parallelogram law of forces.

Figure 1.10 shows the law of parallelogram of forces where the resultant of two forces (F

1, F

2) inclined at an angle ‘’ and acting at point ‘O’ is given by the diagonal of a

parallelogram passing through O. The resultant ‘R’ is inclined an angle with force F1.

To get the resultant analytically, first draw BD perpendicular to OC.

From BOD, R2 = BD2 + OD2

= BD2 + (OC + CD)2 ( OD = OC + CD)

= BD2 + OC2 + CD2 + 2 OC · CD

= BC2 + OC2 + 2 OC · CB cos ( BD2 + CD2 = BC2)

( CD = CB cos = F

22 + F

12 + 2 F

1 F

2 cos

Resultant,2 2

1 2 1 2R F F 2F F cos

2

1 2

BD CB sin F sintan

OC CD OC CB cos F F cos

ii. Triangle Law of Forces

F 1C D

B

O

F 2

R

Figure 1.11: Triangle law of forces.

To get the resultant analytically, first draw BD perpendicular to OC.

Then, from BOD

R2 = BD2 + OD2 = BD2 + (OC + CD)2

= BD2 + OC2 + CD2 + 2 OC · CD

= BC2 + OC2 + 2 OC · CB cos = F

22 + F

12 + 2 F

1 F

2 cos

Resultant,2 2

1 2 1 2R F F 2F F cos

2 21 2 1 2F F 2F F cos 180

2

1 2

BD CB sin F sintan

OC CD OC CB cos F F cos

1 2

1 2

F sintan

F F cos

iii) Polygon Law of Forces

The law of polygon of forces can be proved by using the law of triangles as shown in figure 1.8. The resultant of forces F

1 and F

2 is R

1, R

1 and F

3 is R

2 and then that of R

2 and

F4 is R. Thus, the resultant of force system is R.

F 3

F 4

F 1

O

F 2F 3

F 4

F 1

F 2

R 1 R 1

R

O(a) (b )

Figure 1.12: Polygon law of forces.

1.3.3 Resolution and Composition of Several Concurrent Forces in Plane

A force can be splitted up into two or more forces along any desired axis or lines such that the resultant of these forces is a given force. Therefore, resolution is nothing but

12

3 4

x

x

x

x

x

yy

y y

0

F 3

F4

F 1F2

F 2 si

n 2

– F 2 cos 2 F 1 cos 1

F 4 cos 4

– F 4

sin

4

– F 3 cos 3

– F

3 co

s 3

F 1 si

n 1

y

Figure 1.13: Rectangular components of a force along x and y axis.

Resolving a given force along any desired axis for finding its effect. Generally, a force can be resolved along x and y axis as in case of plane force. While in case of space force, it can be resolved along x, y and z directions or axes. Figure 2.13 shows a resolution of plane force along x and y directions. When a force is resolved along any two perpendicular axis, the components of the force are called rectangular components. Composition of concurrent forces is nothing but algebraic sum of the force components along x and y axes.The resultant is the output of composition of forces. This method is used to find the resultant of several concurrent forces acting on a particle.

In the above figure, each force acting at O can be replaced by its rectangular components.

Let ∑Fx=Algebraic sum of the force components along x - axis.

1 1 2 2 3 3 4 4F cos F cos F cos F cos

∑Fy=Algebraic sum of the force components along y - axis.

1 1 2 2 3 3 4 4F sin F sin F sin F sin

The resultant of and is given by,

R = Resultant 22

x yF F

The angle made by the resultant with x - axis is given by

y

x

Ftan

F

1.3.4 Equilibrium of a Particle in Plane

A particle at rest or moving with a constant velocity with respect to an inertial reference is said to be in equilibrium. For the equilibrium, net force or resultant force acting on a particle must be equal to zero. The equilibrium of a particle obey’s Newton’s first law.

Let R = Resultant of given concurrent forces.

22

x yF F

where,

∑Fx = Algebraic sum of the force components along x - axis.

∑Fy = Algebraic sum of the force components along y - axis.

For equilibrium condition, ∑Fx = 0 and ∑Fy = 0

1.3.5 Resultant and Equilibriant

As we know, resultant is the force having the same effect as that of the total effect of the force system. It is obvious that the particle will move in the direction of resultant acting. If a single force ‘E’ whose magnitude is equal to resultant ‘R’ but opposite in direction applied on a particle, the particle will be in equilibrium. The force ‘E’ is known as an equilibriant. Therefore, the equilibriant of a force system is a single force having its magnitude equal to that of the resultant of the system but opposite direction along the same line of action.

1.3.6 Free Body Diagram (FBD)

In dealing with any equilibrium problems in engineering mechanics it is necessary to consider all the forces acting on the body. The problem becomes much simple if each body is considered in isolation. Such a body which has been so separated or isolated form the surrounding bodies is called free body. In other words, it is a diagram of a body or a group of bodies which is isolated from its surroundings and on which all the external forces such as weight, applied forces, reactions and friction are acting. The free body diagram of some of the figures are given below.

B ody under equ ilib rium

String

A

C B

F.B .D

R C

T B A

W B

B

CBA

B ody under equ ilib rium

W C

R BR A

C

F.B .D

CA

D

B

E W A

W B

R B A

R B A

R F

R D

R CA

B ody under equ ilib rium F.B .C

B

W C

T C A

T C B

CB

C

A

B ody under equ ilib rium F.B .D

AP

W AR A

P

F

B ody under equilib rium F.B .D

Figure 1.14: Free body diagram of some figure.

SOLVED PROBLEMS1. If two forces F

1= 30 kN and F

2= 25 kN act on a particle as shown

figure 1.15.

F 1 = 30 kN

50°

O

F 2= 25 k

N

Figure 1.15

Find the resultant of these two forces by following methods.

a) Parallelogram law of forces

b) Triangle law of forces.

Solution:

a) Parallelogram Law of Forces

We know,

Resultant,2 2

1 2 1 2R F F 2F F cos

2 230 25 2 30 25 cos50º

= 49.89 kN

Inclination of resultant force with F1,

2

1 2

F sin 25 sin 50ºtan 0.4157

F F cos 30 25 cos50º

= tan–1 (0.4157) = 22º 341

b) Triangle Law of Forces

We know, 2 2

1 2 1 2R F F 2F F cos

Resultnat force,

where, = 180 – = 180º – 50º = 130º

2 2R 30 25 2 30 25 cos130º

= 49.89 kN

R

OF = 30 kN1

= 50°

= 130°

F =

25 kN

2

Figure 1.16

By applying law of cosines,

R 25

sin130º sin

25 sin130º 25 sin130º

sinR 49.89

= 0.3839

= sin–1 (0.3839) = 22º 341

2. Two forces 60 N and 70 N act on a screw at an angle of 30º and 80º fromthe base respectively as shown in figure 1.17. Determine the magnitudeand direction of their resultant.

Solution:

Apply parallelogram law of forces.

F = 70 kN2

F = 60 kN1

O

80°30°

Figure 1.17

Resultant,2 2

1 2 1 2R F F 2F F cos

where, = Angle between the two forces

= 80º – 30º = 50º

2 2R 60 70 2 60 70 cos50º

= 117.89 kN

Direction of resultant,

2

1 2

F sin 70 sin 50ºtan

F F cos 60 70 cos50º

= 0.5107

= tan–1 (0.5107) = 27º 31

Inclination of the resultant from the base

= + 30º = 27º 31 + 30º

= 27º 31

3. A cart is pulled by two ropes as shown in figure 1.18. The tension in ropePQ is 4 kN. The resultant acts along the axis of the cart. Determine thetension in rope PR and the magnitude of the resultant of two forces.

T = 4 kN

1

T 2

35°

25°

Q

R

P

Figure 1.18

Solution:

Let T1

= Tension in rope PQ = 4 kN (given)

T2

= Tension in rope PR

Resolve the forces along x - axis.

H PQcos35º PR cos 25º

= T1 cos 35º + T

2 cos 25º

= 4 cos 35º + T2 cos 25º

Similarlly, resolve all the forces along y - axis.

1 2v T sin 35º T sin 25º = 4 sin 35º – T

2 sin 25º

Given that the resultant acts along the axis of the cart.

i.e., H R and v 0

4 sin 35º – T2 sin 25º = 0

2

4 sin 35ºT 5.428 kN

sin 25º

Then, Resultant R = 4 cos 35º + 5.428 cos 25º

= 8.196 kN

4. Determine the magnitude and direction of the resultant of the forces acting on the bolt as shown in figure 1.19.

30N30°

30°

20°

20N

10N

Figure 1.19

Solution:

Resolve all the forces along x - axis.

∑Fx = 10 cos 30º + 20 cos 70º – 30 cos 60º

= 0.50 N

Resolve all the forces along y - axis.

∑Fy=10 sin 30º + 20 sin 70º + 30 sin 60º

= 49.77 N

Resultant, R 22

x yF F

= 2 20.50 49.77

= 49.8 N

Inclination of resultant,

y1 1

x

F 49.77tan tan

F 0.50

= 89.25º

5. The force system shown in figure 1.20 has a resultant of 100 N pointing upalong the y - axis. Compute the values of F and required to give thisresultant.

4 5 °

x

y

F

O2 0 0N

1 2 0N

Figure 1.20

Solution:

Resolve all the forces along x and y axis.

xF = F cos + 120 cos 45º – 200

yF = F sin – 120 sin 45º

Given that the resultant of 100 N pointing up along y - axis.

i.e., ∑Fx = R

and ∑Fy = 0

F cos+ 120 cos 45º – 200 = 0

F cos

F sin– 120 sin 45º= 100

F sin = 184.84

Then,

F sin 184.84

F cos 115.14

tan = 1.6054

= tan–1 (1.6054) = 58º 41

1

1150.14F 217.6 N

cos 58º 4

6. Find the magnitude and direction of resultant of four concurrent forcesacting on a plane as shown in figure 1.21.

6Py

P

3P

30°

45° 60°o

4P

Figure 1.21

Solution:

Resolve all the forces along x and y direction.

i.e., ∑Fx = P cos 60º + 3P – 6P cos 45º + 4P cos 60º

= 1.257 P

∑Fy = P sin 60º + 6 P sin 45º – 4 P sin 60º

= 1.644 P

Resultant, R 22

x yF F

2 21.257 P 1.644 P

= 2.06 P

Direction of resultant force,

y1 1

x

F 1.644 Ptan tan

F 1.257 P

= 52º 351

7. Find the magnitude and direction of the resultant from the following cases.

a) 30 N inclined at 25º towards North of East

b) 25 N towards North

c) 35 N inclined at 40º towards south of west

d) 30 N towards North west.

30°

30N

30N

O

35N

EW40°

45°

25NN

S

Figure 1.22

Solution:

Let us consider x and y axes along the East and North directions.

Resolve all the forces along x and y direction

∑Fx = 30 cos 30º – 30 cos 45º – 35 cos 40º

= –22.04 N

∑Fy = 30 sin 30º + 25 + 30 sin 45º – 35 sin 40º

= 38.71 N

Resultant, R 22

x yF F

2 222.04 38.71

= 44.54 N

y

x

F 38.71tan

F 22.04

= 60.34º

The resultant lies in the second quadrant.

8. Find the resultant of the system of forces shown in figure 1.23.

2 260.64 70.49 92.98 N

y

x

F 70.49tan

F 60.64

15N

50N

4

3

70N12N

15

8

5

12

10Ny - ax is

Figure 1.23

Solution:

From figure 1.23

cos = 4/5, or tan = 3/4, sin = 3/5

cos = 12/13, sin = 5/13

cos = 15/17, sin = 8/17

Resolve all forces along x and y axis

∑Fx = 50 + 15 cos + 10 cos – 12 cos

= 50 + 15 × 4/5 + 10 × 12/13 – 12 × 15/17

= 60.64 N

∑Fy = 15 sin – 10 sin – 12 sin– 70

= 15 × 3/5 – 10 × 5/13 – 12 × 8/17 – 70

= –70.49 N

Resultant, R 22

x yF F

= 49.29º

The resultant lies in the fourth quadrant.

9. A system of five forces of 4 kN, 5 kN, 6 kN, 7 kN and 8 kN act at one of theangular points of a regular hexagon and pass through the other angularpoints as shown in figure 1.24. Find the magnitude and direction of theresultant of the force system.

8kN

7kN 6kN

5kN

4kN

30° 30°30°

30°

Figure 1.24

Solution:

Resolving all the forces along x and y directions.

∑Fx = 4 + 5 cos 30º + 6 cos 60º – 8 cos 60º

= 7.33 kN

∑Fy = 5 sin 30º + 6 sin 60º + 7 + 8 sin 60º

= 21.62 kN

Resultant, R 22

x yF F

2 27.33 21.62

= 22.23 kN

y

x

F 21.62tan 2.95

F 7.33

= 71.3º

10. A system of four forces act on a bolt as shown in figure 1.25. Determine theresultant of the forces on the bolt.

80N Y

X

150N

100N

110N

30°15°

20°

Figure 1.25

Solution:

Resolve all the forces along x and y directions.

∑Fx = 150 cos 30º + 100 cos 15º – 80 cos 70º

= 199.134 N

∑Fy = 150 sin 30º – 100 sin 15º + 80 sin 70º – 110

= 14.293 N

Resultant, R 22

x yF F

2 2199.134 14.293

= 199.64 N

y

x

F 14.293tan

F 199.134

= 4.10º

11. A force is represented by a vector F = (8i – 6j)N. Find the magnitude of the force and counter clockwise angle (ccw) it makes with + ve x - axis?

Solution:

F = (8i – 6j)N

Component of force along x - axis = Fx

= 8N

Component of force along y - axis = Fy

= –6 N

Magnitude of the force 2 2x yF F

2 28 6 10N

Direction of resultant,

6tan

8

= tan–1 (–6/8) = –36º 521

Counterclock wise angle (ccw) = 360 – 36º 521 = 323.48º

12. Three forces +20N, –10N and +30N are acting perpendicular to xz planeas shown in figure 1.26. The lines of action of all the forces are parallel to axis. The co-ordinate of the point of action of these forces along x and z directions are respectively (2, 3), (4, 2) and (7, 4). All distances beingrefered in metres find out, a) The magnitude of the resultant force, b) The location-of the resultant.

Solution:

Resolve all forces along y – axis.

Resultant R= ∑Fy = 20 – 10 + 30 = 40 N()

In order to determine the location of resultant from oz, equate the moment of resultant with moment of the algebraic sum of all the force with respect to plane.

3m2m2m

20N 10N

y

xO

30N

Figure 1.26

R × x = Mz

= Mo

40 × x = (20 × 2) – (10 × 4) + (30 × 7)

210

x 5.25m40

To determine the z co-ordinates of the resultant from ox, equate the moment of resultant with moment of the algebraic sum of all the forces with respect to yz plane.

x oR z M M

40 × z = –(30 × 4) – (20 × 3) + 10 × 2 = –160 N.M

160

z 4.0m40

O

y

z

30N 20N 10N

2m1m1m

However, the z co-ordinate must be such that a force of 40 N (acting up) will have negative moment of 100 N-M. Therefore, R must be to the left of O. In this case the z co-ordinate is positive when it is to the left of O.

z = + 4.0 m

The resultant of 40N () locating at (5.25, 4.0)m

13. A force ‘F’ acts at the origin of a co-ordinate system in a direction defined by the angles

x= 69.3º and

2 = 57.9º. If the components of the force

‘F’ along y - directions is –174 N. Determine

a) The angle y

b) Other components of the force ‘F’

c) The magnitude of the force ‘F’ .... (Nov. 2001)

Solution:

Given, x

= 69.3º and z

= 57.9º, Fy

= –174 N

We know,cos2

x + cos2

y + cos2

2= 1

cos2 (69.30º) + cos2 (57.9º) + cos2

y= 1

or cos2 y = 0.5928

or cos y = ± 0.7699

y

= 40º or y

= 140º

Magnitude of the force cannot be negative though its components may be negative, hence

y should be 140º.

F = –174 / cos 140º = 227.14 NF

x= F cos

x= 227.14 × cos (69.3º) = 80.28 N

Fz

= F cosz

= 227.14 × cos (57.9º) = 120.70 N..

14. The x, y, z components of a force are 30 kN, –24 kN and 18 kN respectively. Find the component of this force along the line joining A (1, 2,3) and B (–1, –2, 2).

Solution:

Given Fx= 30 kN, F

y= –24 kN and F

z= 18 kN.

Force vector, x y zF F i F j F k

30i 24j 18k

Also, the component of the force along AB

ABˆF · n

where, n̂ = Unit vector along AB

AB

AB��������������

B A B A B A

2 2 2

B A B A B A

x x i y y j z z K

x x y y z z

2 2 2

2i 4j 5k

2 4 5

ABn̂ 0.30i 0.60 j 0.75k

Component of force along AB

ABˆF n

30i 24j 18k 0.30i 0.60j 0.75k

= –9 + 14.4 + 13.5

= 18.9 N

y1

x

Ftan 50.10º

F

15. If five forces act on a particle as shown in figure 1.27, and if the algebraicsum of horizontal components of all forces is +250 N, calculate the magnitude of ‘F’ and the resultant of all the forces.

y

F

xO

30°45°

3

4

25 kN

10 kN

7 kN

Figure 1.27

Solution:

From figure,

Resolve all the forces along x - axis

∑Fx = F cos (36.87º) + 25 cos 45º – 10 cos 30º – 7

= F cos 36.87º + 4.59

But, ∑Fx = +250N (Given)

or + 250= F cos 36.87º + 4.59

F = 363.28 kN

Resolve all the forces along y - axis.

∑Fy = 363.28 sin 36.87º – 25 sin 45º + 10 sin 30º

= –299 kN

Resultant, R 22

x yF F

2 2250 299 389.8kN

16. Examine the equilibrium condition for the particle as shown in figure 1.28.y

xO

60°

60°

300 N

173 .2 N

400 N

200 N

Figure 1.28

Solution:

The necessary condition for equilibrium of particle are,

∑Fx = 0 and ∑Fy = 0

∑Fx = 300 – 200 cos 60º – 400 cos 60º = 0∑Fy = –173.2 – 200 sin 60º + 400 sin 60º = 0

Hence the particle will be in equilibrium.

17. Two cables are tied together at ‘N’ and loaded as shown in figure 1.29.Determine the tension in cables MN and ON,

20°

60°

N

O

M500N

Figure 1.29

Solution:

Let, T1 and T

2 be the tension in cable MN and ON respectively.

The particle N will be in equilibrium if the resultant of all the forces acting on it is zero.

yy

x x

W = 500N W = 500N

20°70°

60°140°

150°

T 1 T 1

T 2T 2

Apply lami’s theorem.

1 2500 T T

sin140º sin150º sin 70º

1

500 sin150ºT 388.93 N

sin140º

and

2

500 sin 70ºT 730.96 N

sin140º

18. Determine the tension in cable PQ and PR required to hold 50 kg weightshown in figure 1.30.

20°

450N

R

30°Q

P

40 kg

Figure 1.30

Solution:y

x 450N

W = 40 kg = 40 × 9 .81 = 392 .4 N

P30°50°

T 2

T 1

Let T1 and T

2 be the tension in cable PQ and PR.

Resolve all the forces along x - axis.

∑Fx = –T1 cos 30º – T

2 cos 50º + 450

Resolve all the forces along y - axis

∑Fy = T1 sin 30º + T

2 sin 50º – 392.4

For equilibrium condition,

∑Fx = 0 and ∑Fy=. .0

–T1 cos 30º – T

2 cos 50º + 150 = 0

T1 sin 30º + T

2 sin 50º – 392.4 = 0

or – 0.87 T1 – 0.642 T

2 + 450 = 0 ....(1)

0.5 T1 + 0.767 T

2 – 392.4 = 0 ....(2)

Solving (1) and (2)

Add

1 2

1 2

2

T 0.738 T 517.24 0

T 1.534 T 784.8 0

0.796 T 267.56 0

2

267.56T 336.05 N

0.796

21

T sin 50º 392.4 336.05 sin 50º 392.4T

sin 30º sin 30º

= 269.9 N

19. A sphere weighing 300 N is tied to a smooth wall by a string as shown in figure 1.31. Find the tension in the string and reaction ‘R’ of the wall.

2 0 °

W = 30 0 N

Figure 1.31

Solution:

F.B.D of sphere

160°

11 0°

R

W = 300N

T

70°

T

R

W = 300 N

Let R be the reaction of the wall and T be the tension in the string.

Apply lami’s theorem.300 T R

sin110º sin 90º sin160º

300 sin 90ºT 319.25 N

sin110º

300 sin160ºR 109.19 N

sin110º

20. A chord supported at A and B carries a load of W at ‘C’ as shown in

figure 1.32. Find the value of W so that CD remains horizontal.

60° 30°B

C

W 20 kN

D

A

Figure 1.32

Solution:

Draw the F.B.D of the system.

T D C

150°

120°

W = 20 kN

T D B

T D C

30°

W = 20 kN

T D B

D

Let TDC

and TDB

be the tension in the chord DC and DB respectively.

Apply lami’s theorem.

DCDB T20 T

sin150º sin 90º sin120º

DB

20 sin 90ºT 40 kN

sin150º

DC

20 sin120ºT 34.64 kN

sin150º

150°

120°

W

T C A

T C D = 34 .64 kN

60°

W

T C A

T C A D C = T = 34 .64 kN

C

Let TCA

be the tension is the chord CA.

Apply lami’s theorem.

CD CAT T W

sin150º sin 90º sin120º

CDCA

T sin 90º 34.64 sin 90ºT 69.28kN

sin150º sin150º

and 34.64 sin120º

W 60 kNsin150º

21. Determine the tension in various segments of the connected flexible cables shown in figure 1.33.

300N

400N

30°

Q

60°

R

S

P

T

45°

Figure 1.33

Solution:

Consider the free body diagram of joint ‘S’T S Q T S T

30° 45°S

400N

T S Q T S T

400N

105°

135°120°

Let TSQ

and TST

be the tensions in the cable SQ and ST respectively.

Apply lami’s theorem.

SQSTTT400

sin105º sin120º sin135º

ST

400 sin120ºT 358.7 N

sin105º

and,SQ

400 sin135ºT 292.82 N

sin105º

Consider F.B.D of joint Q.

Q

60°

30°T Q P

T Q R

T Q S = T = 292 .82NS Q300N

Resolve all forces along x - axis.

∑Fx = 0

∑Fx = TQR

cos 60º + 292.82 cos 30º – TQP

= 0

TQR

cos 60º + 292.82 cos 30º – TQP

= 0

Resolve all forces along y - axis.

∑Fy = 0

TQR

sin 60º – 292.82 sin 30º – 300 = 0

QR

292.82 sin 30º 300T 515.47 N

sin 60º

and TQP

= TQR

cos 60º + 292.82 cos 30º

= 514.47 cos 60º + 292.82 cos 30º

= 511.32 N

22. A wire fixed at two points A and D as shown in figure 1.34. Two weights of 10 kN and 30 kN are supported at B and C respectively. When equilibrium is reached it is found that the inclination of AB is 20º and thatof CD is 50º to the verticle, determine the tension in the segment AB, BCand CD of the wire and also the inclination of BC to the verticle.

30 kN

30 kN

20°

A

B

C

50°

D

Figure 1.34

Consider the F.B.D of joint B and ‘C’.

30 kN

T C D

40°C

T C B

T B C

T B A

30 kN

70°

B

For joint B

Resolve all the forces along x and y axis.

= 0T

BC cos – T

BA cos 70º = 0

or

BABA BC

T cosT 2.92 T cos

cos 70º

Similarly, = 0

–TBC

sin – 10 + TBA

sin 70º = 0

–TBC

sin + 2.92 TBC

cos sin 70º = 10

–TBC

sin + 2.74 TBC

cos = 10

TBC

(2.74 cos – sin) = 10 ....(1)

For joint C

∑Fx = 0

TCB

cos 40º – TCB

cos = 0

or TCD

= 1.30 TCB

cos

Similarly, = 0

TCD

sin 40º + TCB

sin – 30º = 0

1.30 TCB

cos sin 40º + TCB

sin = 30

0.839 TCB

cos + TCB

sin = 30

TCB

(0.839 cos + sin)= 30 ....(2)

Divide (1) / (2)

BCT CB

2.74 cos sin

T

BC CB

10T T

300.839 cos sin

(0.839 cos + sin) 10 = 30 (2.74 cos – sin)

8.39 cos + 10 sin= 82.2 cos – 30 sin

40 sin = 73.81 cos

sin 73.81

cos 40

tan = 1.8452

= tan–1 (1.8452) = 61.54º

Put = 61.64º in Equation (1)

BC

10T

2.74 cos 61.54º sin 61.54º

= 23.44 N

TBA

= 2.92 TBC

cos = 2.92 × 23.44 cos (61.45º)

= 32.61 kN

and TCD

= 1.30 TCD

cos

= 1.30 × 23.44 cos (61.54º)

= 14.52 kN.

23. A fine light string ABCDE whose extremity ‘A’ is fixed, has weights W

1 and W

2 attached to it at B and C. It passes round a small smooth pulley

at D carrying a weight of 40 kg at the free end E as shown in figure 1.35.

150° 120°

D

ECB

W 1 W 2

40 kg

Figure 1.35

If in the position of equilibrium, BC is horizontal and AB and CD makes150º and 120º with BC, find i) Tensions in the portion AB, BC and DC ofthe string, ii) Magnitudes of W

1 and W

2.

Solution:

C150°

120°

T C B

T = 40× 9.81 = 392 .4N

C D

W 2

B

150°

120° T B C

T B A

W 1

F.B.D of joint B F.B.B of joint C

For joint C

Applying lami’s equation at joint C.

CB 2T W 392.4

sin150º sin120º sin 90º

CB

sin150º 392.4T 196.2 N

sin 90º

and,2

sin120º 392.4W 339.82 N

sin 90º

For joint B

Applying lami’s equation at ‘B’

BA 1T W 196.2

sin 90º sin150º sin120º

BA

196.2 sin 90ºT 226.6 N

sin120º

and, 1

196.2 sin150ºW 113.27 N

sin120º

24. Determine the horizontal force F to be applied to a block of weight 2 kN tohold it in position on a smooth inclined plane AB which makes an angle of30º with horizontal axis, shown in figure 1.36.

30°30°

R

B

W

A

FW =2kN

Figure 1.36

Solution:

150°

120°30°

F

W = 2000N

R

F.B.D

Apply lami’s equationF 2000 R

sin150º sin120º sin 90º

2000 sin150ºF 1154.70 N

sin120º

2000 sin 90ºR 2309.4 N

sin120º

25. A ball weighing 1000 N is at rest in a right angled trough as shown infigure 1.37. Determine the forces exerted on the sides of the through at D and E. Assume are the surfaces to be smooth.

DE

1000N

50°40°

Figure 1.37

Solution:

50° 40°R D R E

1000N R DR E

1000N

90°

140°130°

F.B.D of Ball

Apply lami’s theorem.

D E1000 R R

sin 90º sin130º sin140º

D

1000 sin130ºR 766.04 N

sin 90º

and,E

1000 sin140ºR 642.78 N

sin 90º

26. Two identical rollers each weighing 200 N are placed in a trough as shownin figure 1.38. Assuming all surfaces of contact are smooth, find thereactions developed at the point of contact surfaces.

40°

AB

D

C

Figure 1.38

Solution:

R D

O 1

O 2

R B

R A

R C200N

200N

40°

40°

50°

50°

50°

40°

O 2 R D

R B

R C

200N

90°

140°

50° 40°130°

R A R B

200N

F.B.D of Balls

For Ball O1

Apply lami’s theorem,

B A200 R R

sin 90º sin140º sin130º

B

200 sin140ºR 128.6 N

sin 90º

A

200 sin130ºR 153.20 N

sin 90º

For Ball O2

Resolve all forces along x and y axis.

∑Fx = 0

RD – R

B cos 40º – R

C cos 50º = 0

RD = R

B cos 40º + R

C cos 50º .... (1)

∑Fy = 0

–200 – RB sin 40º + R

C sin 50º = 0

B

C

R sin 40º 200 128.6sin 40º 200R

sin 50º sin 50º

= 368.98 N

Substituting RD in equation (1)

RD = 128.6 cos 40º + 368.98 cos 50º

= 335.69 N

27. Two cylinders M and N rest in a channel as shown in figure 1.39. Thecylinder M has diameter of 100 mm and weighs 200 N, whereas the cylinder N has diameter of 180 mm and weighs 500 N. If the bottom width of the box is 180 mm with one side vertical and the other inclined at 60º, determine the pressures at all the four points of contact.

+

+

N

M

G

F

H

DC

A B180m mE

60°

Figure 1.39

Solution:

R H

R G

R G

R C

56°9

56°9 I

200N

30°J

(R – 500)E

N

I

MJ R H

200N

5 0 0N

9 0 m m

3 0 ° 6 0 °R F

R G

R E

A E L B

K

F.B.D of cylinders

From the geometry of the figure,

IAE 60º

AE = IE cot 60º = (180/2) × 0.577 ( cot = 1/tan)

= 52 mm

LB = 100/2 = 50 mm

EL = IK = 180 – 52 – 50 = 78 mm

IJ = 90 + 50 = 140 mm

IK 78

cos 0.5571IJ 140

= 50º 9’

Considering the equilibrium of cylinder M.

Apply lami’s equation at J.

GH RR 200

sin 90º 56º 9 ' sin 90º sin 180º 58º 9 '

H

200 sin 90º 56º9 'R

sin 180º 56º9 '

= 134.2 N

and G

200 sin 90ºR

sin 180º 56º 9 '

= 240.8 N

Considering the equilibrium of cylinder N.

Apply lami’s theorem.

E FR 500 240.8 R

sin 180º 30º 50º9 ' sin 60º sin 90º 50º9 '

F

240.8 sin 90º 56º9 'R

sin 60º

= 154.9 N

E

240.8 sin 180º 30º 56º9 'R 500

sin 60º

= 122.5 N

RE

= 122.5 + 500 = 622.5 N

1.4 Introduction to Equilibrium of Rigid bodiesIn the previous unit, we have analyzed the system of forces on a particle in which we

assumed each body as a particle. This assumption is valid, if the body has negligible size. But in many cases, this assumption may not be possible. When a body is not subjected to concurrent force system then the body is said to be a rigid body. A rigid body is a combination of particles in large numbers. For the equilibrium of rigid body, the size of the body, forces acting on different point of applications has to be taken into consideration.

The basic and important conditions to be considered for the equilibrium of a rigid body are the summation of forces and moments in each direction is taken as zero. By taking the above conditions, the forces and moments of forces are balanced in each direction. Another important consideration is that the system of external forces will not impart translational or rotational motion on the body taken. When a stationary body is

acted upon by some external forces, the body may start to rotate (or) may start to move about any point or an axis. If the body does not move or rotate about any point or an axis, then the body is said to be in equilibrium.

1.5 Moment of a ForceA force acting on a body tends either to displace in some direction or to rotate the

body about a point. The tendency of a force to turn a body about a point or an axis is called the moment of that force. Therefore, moment is defined as the product of the magnitude of the force and the perpendicular distance of the point from the line of action of the force.

F

A

d 1

d 2

o 2

o 1

Figure 1.40

Consider a force ‘F’ applied to a rigid body (Figure 1.40) at A. The force tends to rotate the body about O. The point ‘O’ is called the moment centre. The perpendicular distance between the line of action of force and the point about which moment is calculated is called moment arm.

Mathematically,

M = F × d

where, F = Force applied

d = Moment arm

If point O1 is taken as the moment centre,

M1= F × d

1 CCW (Counter clockwise sense)

If point O2 is taken as the moment centre,

M2= F × d

2 CW (Clockwise sense)

The tendency of the moment is to rotate the body in the clockwise direction about O. Hence the moment is called clockwise moment. If the tendency of a moment is to rotate the body in anticlockwise direction, then that moment is known as anticlockwise moment.

1.6 Varignon’s Theorem (Principle of Moments)

It states that the sum of moments of individual forces of a system of concurrent forces about any point is equal to the moment of their resultant about the same point.

o

A

d 2

d 1

d

F 1

F 2

F

Figure 1.41

Refering to Figure 2.41, F1 and F

2 are a system of coplaner forces. F is their

resultant.

According to Varignon’s theorem, moment about O is given by,

F1d

1 + F

2d

2= Fd

1.7 Couple

A couple is defined as two parallel forces that have the same magnitude, opposite directions and are separated by a perpendicular distance ‘d’ as shown in figure 2.42. The perpendicular distance between the parallel forces is called arm of the couple. Couple has a tendency to rotate the body.

– F

F

Figure 1.42

Moment of a couple is the product of any one of the parallel forces and perpendicular distance between the forces.

Moment of couple = Force × arm of the couple

1.8 Equivalent Forces and Couples

The two forces having the same magnitude, direction and line of action but acting at different points producing the same external static effect on the rigid body are said to be equivalent forces.

If two couples produce the same moment on the rigid body, they are called equivalent couples.

1.9 Equivalent Force Couple System

A system of forces acting on a rigid body can be replaced by a single force acting at a given point and a couple. This force and couple is known as equivalent force couple system.

1.10 Moment and Couple

The couple is a pure turning effect which may be moved anywhere in its own plane or into a parallel plane without change of its effect on the body. But moment of a force must include a description of the reference axis about which the moment is taken.

1.11 Resolution of a Force into a Force and a Couple

Let us consider two points M and N acting on a rigid body as shown infigure 1.43. A force ‘F’ is acting at point M. The perpendicular distance between line of action of ‘F’ and the point N is d.

N

M = F d

(c )

F

N

M

F

F

F

(b )

==

N

M

F

(a )

d d

Figure 1.43

Apply equal and opposite forces at N (Figure 1.43(b)) so that the equilibrium of the system is not disturbed. The system of forces shown in figure 1.43 (b) is same as that shown in figure 1.43 (a). Now the force at M and the downward force at N form a couple of magnitude M = F ×d (Clockwise). Hence at N, we have a couple M and a force ‘F’ as

shown in figure 1.43(c). The system shown in figure 1.43(c) is same as that shown in figure 1.43(b).

1.12 Resultant of Coplanar Non-Concurrent Forces in Plane

o

F 3

F 1

F 2(a )

F 1

F 2

F 3

o

(b )

R

o

(c )

R

R

R

A

d

o

(d )

y R

xo

(e )

Figure 1.44

In many situations, it may be required to reduce a system of forces into a single equivalent force or into a couple of force.

Figure 1.44 shows three non concurrent coplanar forces F1, F

2 and F

3 acting on a

rigid body. Each of these forces can be replaced by a force of same magnitude and direction, acting at ‘O’ and a couple. The sum of couples due to F

1, F

2 and F

3 as M

O, as

shown in figure 1.44(b). The forces become concurrent, when all acting at O.

In figure 1.44(c), the resultant of the concurrent forces, R, is shown along with MO.

Thus a system of coplanar noncurrent forces has been reduced to a couple and a single force.

Now, introduce equal and opposite force ‘R’ at point ‘A’ located at a distance ‘d’ from O. Such that the couple due to downward Force ‘R’ at A and R at O should nullify the effect of M

O. The couple due to R is counterclockwise while M

O is clockwise.

  R × d = MO

Finally, we are left with a single Force R acting A. Thus a single force resultant for a system of coplanar nonconcurrent forces has been obtained.

The following equations are used to solve the above problem analytically.

Resultant, 2 2R Fx Fy

Fy

tanFx

, where = angle made by R with respect to x-axis

Distance, OM

dR

where, Fx = Algebraic sum of x components of all forces.

Fy = Algebraic sum of y components of all forces.

=Algebraic sum of moments of all forces about any arbitrary point O.

d = Perpendicular distance between the point O and the line ofaction of the single force resultant ‘R’.

1.13 Equilibrium Conditions for Coplanar Non-Concurrent Forces in PlaneA body is said to be in equilibrium when it does not have any rotatory motion in any

direction. This means that when the body is in equilibrium, the following two simultaneous conditions are to be satisfied.

i) The algebraic sum of the components of the forces along each of the two mutually perpendicular directions, is zero.

ii) The algebraic sum of the components of the moments acting on the body about each of the two mutually perpendicular axes, is zero.

Mathematically,

Fx = 0 : Algebraic sum of the components of the forces along x axis is zero.

Fy = 0 : Algebraic sum of the components of the forces along y axis is zero.

= 0 : Algebraic sum of the components of the moments along x and y axis is zero.

Review Questions1. Define Engineering mechanics.

2. State the branches of Engineering Mechanics.

3. Define Statics and Dynamics.

4. Distinguish between Particle and Rigid body.

5. State Newton’s law of motion

6. Explain the following:

i) Parallelogram law of forces ii) Triangle law of forces

iii)Principle of transmissibility iv)Lami’s theorem

7. Define force. State its characteristics.

8. Describe the classification of force system.

9. What is resultant force? Discuss the methods of determining the resultant oftwo concurrent forces.

10. What is meant by resolution of a force?

11. Distinguish between resultant and equilibriant.

12. State the conditions of equilibrium of a particle in plane.

13. What is a free body diagram?

14. Write a force in space in cartesian co-ordinates.

15. Explain the method of finding the resultant of concurrent forces in space.

16. Briefly explain equilibrium of a particle.

17. State the necessary and sufficient conditions for equilibrium of a particle in two dimensions.

18. Explain the equilibrium of a particle in space.

19. State the necessary and sufficient conditions for equilibrium of a particle in space.

20. What are equivalent forces?

21. Find the magnitude and direction of resultant from the following cases.

i) 200 N inclined at 30º towards North of East

ii) 250 N towards North

iii)300 N towards North West

iv)350 N inclined at 40º towards South of West

(Ans: Magnitude, R = 456 N and = 47.7º)

22. Two cylinders of radius 300mm and 150mm are lying in a channel with their axes horizontal. The larger one, of weight 1800N, lies below the smaller one, ofweight, 600N as shown in figure. Take the width of the channel to be 0.8m. Determine the reactions at all points of contacts assuming smooth surface for all contacts.

0 .8 m m

0 1

0 2

A

D

C

F

E

B

(Ans: RC

= 743.3 N, RBA

= RAB

= 955.5 N, RE = 2400 N, R

F = 743.3 N)

23. A 100 N force is acting away from the origin at 25º to the horizontal. Another force

of 150 N is acting towards the origin at 115º to the horizontal. Find the total verticle

and horizontal components of these two forces. Also find the magnitude and direction of resultant.

(Ans: H = 154 N,V = –93.68 N, R = 180.25 and = 31.31º

24. A lamp weighing 10N is suspended from the ceiling by a chain. It is pulled aside by a horizontal cord until the chain makes an angle of 60º with the ceiling as shown in figure. Find the tension in the chain and the cord by applying lami’s theorem.

A

6 0 °

C ha in

C ord

B

2 0 N

O

(Ans: TOB

= 5.78 N, TOA

= 11.6 N)

25. Determine the magnitude and direction of the resultant of forces acting at ‘O’as shown in figure below. (Ans: R = 21.70 N, = 79.37º)

2 5 N

1 25

2 6 N2 0 N

1 0 N

3 0 º

26. A force acts at the origin in a direction defined by the angles y = 65 and z = 40. Knowing that the x component of force is -90kN, determine (a) the other components and magnitude of force. (b) the value of x.

27. The x,y,z components of a force are 36kN, -24kN and 24kN respectively. Find the component of this force along the line joining A(1,2,-3) and B(-1,-2,2).