UNIT – I
Introduction to Mechanics
Engineering Mechanics
Mechanics of solids
Mechanics of fluid
Ideal fluids
Rigid Bodies
Viscous fluids
Compressible fluids
Statics
Kinematics
Dynamics
Kinetics
Deformable bodies
Strength of Material
Theory of Elasticity
Theory of Plasticity
Units of Quantities (SI Unit)
Quantity
Unit
Symbol
Formula
-
m/s2
rad
Rad
Acceleration
Metre /(Second)2
Angle
Radian
Angular Acceleration
radian/(Second)2
-
Rad/s2
Angular Velocity
Radian/second
-
Rad/s
Area
Square meter
-
m2
Density
Kilogram/(meter)3
-
Kg/m3
Energy
Joule
J
Nm
Force
Newton
N
Kg m/s2
Frequency
Hertz
Hz
(l/s)
Length
Meter
M
M
Mars
Kilogram
Kg
Kg
Moment of force
Newton-metre
-
Nm
Power
Watt
W
J/s
Prevue
Pascal
Pa
N/m2
Stress
Pascal
Pa
N/m2
Time
Second
S
S
Velocity
Metre/second
-
m/s
GE6253-ENGINEERING MECHANICS
UNIT-I BASICS AND STATICS OF PARTICLES
Introduction
Engg. Mechanics is a branch of science which deals with the behavior of a
body when the body is at rest or motion.
Engg. Mechanics
Statics
(Study of a body when the
body is at rest)
Dynamics
(Study of a body when it is in
motion)
Kinematics
(Study of a body in motion
without considering the forces
that causes motion)
Kinetics
(Study of a body in motion
considering the forces that
causes motion)
Terms used in Engineering Mechanics
Vector quantity: A quantity which is completely specified by magnitude and
direction is known as vector quantity. (Eg.) Velocity, Acceleration, Force &
Momentum.
Scalar quantity: A quantity which is completely specified only by magnitude is
known as scalar quantity. (Eg) Mass, Time , Length etc.
Particle: A particle is a body of negligible dimensions and the mass of the particle is
considered to be concentrated at a point.
Rigid body: A body which dos not deform under the action of applied force.
Mass: The quantity of matter contained in a body is called as mass.
Weight: The force with which a body is attracted towards the centre of the earth.
W = mg
Unit of measurement: A physical quantity can be measured by comparing the sample
with a known standard amount. The known amount used as a reference in the
measurement of physical quantities called unit.
Types of units
1. Basic units
2. Derived units
Basic unit: Used in the measurement of basic (or) fundamental quantities (ie) mass,
length, time.
Derived unit: Used in the measurement of the physical quantities other than
fundamental quantities. (E.g) Area, Volume, Energy.
System of Units
1. FPS (Foot Pound System)
2. CGS (cm, gram, sec)
3. MKS (m, kg, sec)
4. SI (International System of units)
Laws of Mechanics
Newton’s first law of motion: A body remains in its state of rest or motion unless a a
external force acting on it.
Newton’s second law of motion: The accelaration of a particle is proportional to the
resultant force acting on it and is in the direction of the force applied.
F = ma
Newton’s Third law of motion: To every action there exists an equal and opposite
reaction.
Gravitation law of attraction
States that any two bodies in the universe attract each other with a force that is
directly forces proportional to the product of their masses and inversely proportional
to the square of the distance between them.
m1
m2
F
m1 m2
r2
m m 
F  G  12 2
 r 
r
G – Universal Gravitational constant
G =6.673 x 10-11 Nm2/kg2
G value – Henry Cavendish – After Newton’s death
Earth’s standard acceleration due to gravity g = 9.80665 m/s2 (32.1740 ft/s2)
An object falling near the earth’s surface increases its velocity by 9.80655 m/s for
each second of its descent.
Parallelogram law of forces
When two forces acting simultaneously at a point, can be expressed in both
magnitude and direction by the two close sides of parallelogram drawn on a point,
resultant is expressed completely, both in direction and magnitude by the diagonal of
the parallelogram going through the point.
C
B
Q
R
R
Q
O
 - Angle of resultant with x-axis
B
β
α
A
a
a
b
c


Sin 
Sin 
Sin  2
γ
b
P
α - Angle b/w two forces
Sine law:
C

𝛼
P
c
α
A
D
Corrine Law:
Direction of Resultant
a = b +c – 2bc cos α
2
2
2
C
b = a +c – 2ac cosβ
2
2
2
R
c = a +b – 2ab cos γ
2
2
2
Q

O
P
180−𝛼
α
A
CD tan  =
CD
Q Sin 

OD P  Q cos 
 Q Sin  

 P Q cos  
 = tan-1 
Magnitude of Resultant (R)
DAC = DOB = α, AC = Q
In triangle ACD
In triangle OCD
AD = AC cos α = Q cos α
OC2 = OD2 + OC2
CD = AC Sin α = Q sin α
OC = R, OD = OA + AD
= P + Q cos α

OC2
=
OD2 + OC2
R2
=
(OA + AD)2 + DC2
=
OA2 + AD2 + 2OA.AD + DC2
R2
=
P2 + Q2 cos2 α + 2 PQ cos α + Q2 sin2 α
R
=
P 2  Q 2  2 PQ cos 
( sin2 α + cos2 α = 1)
Case 1: If two forces P and Q acts at right angles, then
𝛼 = 90°
We know, magnitude of resultant.
R
=
P 2  Q 2  2 PQ cos  
R
=
P2  Q2
P 2  Q 2  2 PQ cos 90
[ cos 90 = 0]
We know, direction of resultant
 Q Sin  
  tan 1
 = tan 
 P  Q cos  
-1
 Q Sin 90  1 


0 
 P  Q cos 90 0 
Q

P
 = tan-1 
Case 2: The two forces P & Q are equal and are acting at an angle α between them. (P
= Q).
R
=
P 2  Q 2  2 PQ cos 
=
P 2  P 2  2 P 2 cos 
=
2 P 2 (1  cos  )
=
2P 2 2 cos 2
R
=
2 P cos
Q
=

2

2

2 
 1 cos   2 cos 
2


4P 2 cos 2 
2

2
Problem 1: The resultant of the two forces, when they act at an angle of 60° is 14N. If
the same forces are acting at right angles, their resultant is
136 N . Determine the
magnitude of the two forces.
Soln.:
Case 1
Case 2
R1 = 14N
R2 = 136 N
α1 = 60°
α2 = 90°
For case 1
R
=
14
=
196
=
√𝑃2 + 𝑄2 + 2𝑃𝑄 𝐶𝑜𝑠 𝛼, 14 = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝐶𝑜𝑠 60°1/2
P 2  Q 2  PQ
P2 + Q2 + PQ
 (1)
For case 2
R
=
136
=
P 2  Q 2 or
P2 + Q2
136 
P 2 Q2
 (2)
196-136
=
P 2  Qx  PQ  P 2  Q 2
60
=
PQ
(3) x 2 =>
120
=
2 PQ  (4)
(4) + (2)=>
136 + 120
=
P2 + Q2 + 2PQ
256
=
P2 + Q2 + 2PQ
(1) – (2)
=>
 (3)
(16)2 =
(P&Q)2
P+Q =
16
P
16 – Q
=
 (5)
Substitute (5) in (3)
60
=
(16-Q) Q
60
=
16Q – Q2
Q2 – 16Q + 60 = 0
This is a quadratic equation, so
Q
=
 b  b 2  4ac
2a
=
16  256  240
2
=
=
If
a=1
b = -16
c = 60
16 ± √16
16 ± 4
=
∴ 𝑄 = 10&6
2
2
16  16
16  4

2
2
 Q  10  6
Q = 10, P = 6
Q = 6, P = 10
Two forces are 10 N & 6 N
Vectorial Representation of force & moments
The quantities which possess magnitude as well as direction are called as
vector quantities.
Symbol of vector ‘P’ is represented with an arrow such as 𝑃⃗
Magnitude of vector is represented by P or P.
Free vector can be moved anywhere in space provided it maintains the same direction
and magnitude.
Sliding vector may be applied at any point along its line of action.
Bound vector It will remain at the same point of application.
Negative vector The negative of a vector P is the vector-P which has same magnitude
& inclination but in opposite direction.
Vector operations
Vector addition
Triangle law – if two forces acting at a point are represented by the two sides of
a triangle taken in order, then their resultant is represented by the third side taken in an
opposite order.
⃗
𝑄
P
𝜃
Q
𝑃⃗
⃗
𝑅⃗ = 𝑃⃗ + 𝑄
P Q  Q  P
Commutative
P  Q  S   P  Q   S  P  Q  S    P  S   Q
(𝑚 + 𝑛)𝑃⃗ = 𝑚𝑃⃗ + 𝑛𝑃⃗ Distributive
Associative
Vector subtraction
It is the addition of corresponding negative vectors.
Problem 4: Find the vector (𝐴̅ + 2𝐵̅ + 3𝐶̅ ) in terms of 𝑖̅, 𝑗̅, 𝑘̅ and also find it’s
̅ + 4𝑘̅
magnitude where 𝐴̅ = 4𝑖̅ − 𝑗̅ − 2𝑘̅ , 𝐵̅ = 5𝑖̅ + 2𝑗̅ − 3𝑘̅ , 𝐶̅ = 2𝑖̅ − 6𝑗
Soln:
Multiplication of vectors by scalars
The product of scalar and vector gives vector quantity. The vector 𝑝̅ is
multiplied by a scalar gives a vector 𝑚𝑝̅
(𝑚 + 𝑛)𝑝̅ = 𝑚𝑝̅ + 𝑛𝑝̅
Dot or scalar product of vectors
The dot or scalar products of two vectors 𝐴̅ &𝐵̅ written as 𝐴̅ + 𝐵̅ is a scalar and
is defined as the product of the magnitude of the two vectors and the coline of their
included angel 𝜃
𝐴̅. 𝐵̅ = 𝐵̅. 𝐴̅
𝑖̅. 𝑗̅ = 𝑗̅. 𝑘̅ = 𝑘̅ . 𝑖̅ = (1)(1) cos 90°
𝐴̅. (𝐵̅ + 𝐶̅ ) = 𝐴̅. 𝐵̅ + 𝐴̅. 𝐶̅
=0
𝐴̅ = 𝐴𝑥 𝑖̅ + 𝐴𝑦 𝑗̅ + 𝐴𝑧 𝑘̅
𝑖̅. 𝑖̅ = 𝑗̅. 𝑗̅ = 𝑘̅ . 𝑘̅ = (1)(1) cos 0
𝐵̅ = 𝐵𝑥 𝑖̅ + 𝐵𝑦 𝑗̅ + 𝐵𝑧 𝑘̅
=1
𝐴̅. 𝐵̅ = 𝐴𝑥 . 𝐵𝑥 + 𝐴𝑦 . 𝐵𝑦 + 𝐴𝑧 . 𝐵𝑧
𝑎̅. 𝑏̅ = |𝑎̅|. |𝑏̅| cos 𝜃
𝐴̅. 𝐴̅ = 𝐴2𝑥 + 𝐴2𝑦 + 𝐴2𝑧
Cross (or) Vector product of vectors
Y
Y
𝑃̅ = 𝐵̅ × 𝐴̅
𝑃̅ = 𝐴̅ × 𝐵̅
X
X
𝜃
Z
𝜃
Z
𝐴̅
𝐴̅
𝐵̅
𝐵̅
̅× B
̅
̅ ≠B
̅ × A
A
𝑎̅ × ̅𝑏 = |𝑎̅||𝑏̅|𝑆𝑖𝑛𝜃 𝑛̂
̅× B
̅)
̅ = −(B
̅ × A
A
𝑖̅, 𝑗̅, 𝑘̅ 𝑖̅. 𝑖̅ = 𝑗̅. 𝑗̅ = 𝑘̅ . 𝑘̅ = 0
𝑖̅ × 𝑗̅ = 𝑘̅ , 𝑗̅ × 𝑘̅ = 𝑖̅, 𝑘̅ × 𝑖̅ = 𝑗̅
𝐴̅ = 𝐴𝑥 𝑖̅ + 𝐴𝑦 𝑗̅ + 𝐴𝑧 𝑘̅
𝐵̅ = 𝐵𝑥 𝑖̅ + 𝐵𝑦 𝑗̅ + 𝐵𝑧 𝑘̅
𝑖 𝑗 𝑘
𝐴
𝐴 = | 𝑥 𝐵𝑦 𝐶𝑧 |
𝐵𝑥 𝐵𝑦 𝐵𝑧
Portion vector
𝑟̅ = 𝑥 𝑖̅ + 𝑦 𝑗̅ + 2 𝑘̅
Y
𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2
(x,y,z)
𝑟̅
𝐹̅
Unit vector= 𝑛̂ = |𝐹̅|
z
X
y
Z
x
Problem 6 : Determine the unit vector along the line which originates at the point
(4, 1, -2) and passes through the pint (2, 2, 6).
Solution:
𝑂(0,0,0)
𝐴̅(4,1, −2)
𝐵̅(2,2,6)
𝑛̂ =
̅̅̅̅
𝐴𝐵
−2𝑖̅ + 𝑗̅ + 8𝑘̅
=
|̅̅̅̅
8.3
𝐴𝐵 |
Victorial Representation of Forces, Moments & couples:Force: It is an action executed on a body which changes or tends to change the state
of rest or uniform motion of the body. Force is a vector quantity i.e, it has both
magnitude and direction. Some characteristics of force are
1. Magnitude
2. Line of Action
3. Direction
Magnitude: The magnitude of force is denoted by certain member of units. In SI units
magnitude of force denoted by Newton.
Line of Action: Line of action of the force is denoted by a line through which it acts.
Direction: It represents the direction of force. The direction of the force is represented
by arrow heads.
Moment: Moment 𝑚
̅ refers to turning effort produced about point ‘O’ by the force ‘F’
𝑚
⃗⃗ = 𝑟̅ × 𝐹̅
Where 𝑟̅ is the portion vector of any point on the line of action of the force.
Moment of force about a point:
r
F
O
Product of the force and perpendicular distance of the line of action of the force from
a point is known as line of action of the force
M = r × F
If the tendency of this moment 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 the body is to rotate the body in anticlockwise direction then it is
called as anticlockwise moment.
CCW or ACW moment considered as positive.
CW moment is considered as negative.
Problem 7: Find the moment of force about A, If the forces acting are 300 KN, 400
KN & 500 KN as shown in fig.
300 KN
𝑚 = 𝑟 × 𝐹
5𝑚
2𝑚
= (−300 × 2) + (400 × 3) + (500 × 5)
3𝑚
500 KN
400 KN
= − 600 + 1200 + 2500
= 3100 KN
Problem 8: The force F̅ = (6𝑖̅ – 3𝑗̅– 4𝑘̅ )N is acting at a point P whose position
vector from the origin ‘O’ is (8𝑖̅ + 3𝑗̅– 4𝑘̅ )𝑚𝑚 find the moment of the force about
the origin.
F̅ = (6𝑖̅ – 3𝑗̅– 4𝑘̅ )N
r̅ = (8𝑖̅ + 3𝑗̅– 4𝑘̅ )𝑚𝑚
𝑚
̅ = 𝑟̅ × 𝐹̅
𝑖̅ 𝑗̅ ̅𝑘
=|8 6 −4|
6 −3−4
[
(
)
(
)]
[
(
𝑖̅ 6 × −4 − −3 × −4 − 𝑗̅ 8 × −4) − (6 × −4)] + 𝑘̅ [(8 × −3) − (6 × 6)]
𝑖̅[−24 − 12] − 𝑗̅[−32 + 24] + 𝑘̅ [−24 − 36]
m
̅ = (−36𝑖̅ + 8𝑗̅– 60𝑘̅ )Nm
Problem 9: A force vector of magnitude 900N starts at the point (−4, 2,1) and
terminates at the point (1, −3,6). Find the moment of the force about X, Y, Z axes.
Solution:
F = 900 N; A (−4, 2, 1) , B (1, −3, 6)
̅̅̅̅ = 𝑂𝐵
̅̅̅̅ – 𝑂𝐴
̅̅̅̅ = (𝑖̅ − 3𝑗̅ + 6𝑘̅ ) – (−4𝑖̅ + 2𝑗̅ + 𝑘̅ )
𝐴𝐵
𝐹̅ = 𝐹. 𝑛̂ 𝐴𝐵
̅̅̅̅
𝐴𝐵 = 5𝑖̅ – 5𝑗̅ + 5𝑘̅
|̅̅̅̅
𝐴𝐵 |=
𝑛̂ AB =
5 2  (5) 2  (5) 2
5i  5 j  5k
75
= 75
𝑛̂𝐴𝐵 =
̅̅̅̅
𝐴𝐵
̅̅̅̅|
|𝐴𝐵
= (0.5773 𝑖̅ – 0.5773𝑗̅ + 0.5773𝑘̅ )
𝐹̅ = 𝐹. 𝑛̂ 𝐴𝐵
= 900 (0.5773𝑖̅ – 0.5773 𝑗̅ + 0.5773 𝑘̅ )
𝐹̅ = 519.6 𝑖̅– 519.6 𝑗̅ + 519.6𝑘̅
Moment of the force about the origin
𝑚
̅ = 𝑟̅𝑂𝐴 × 𝐹̅𝐴𝐵
A (−4,2,1) B (1, −3,6)
𝑚
̅ = 𝑖̅ [(2 × 519.6) – (−519.6 × 1)] – 𝑗̅ [(−4 × 519.6) – (519.6 × 1)] −
𝑘̅ [(−4 × −519.6) − (519.6 × 2)]
𝑚
̅ = 1558.8𝑖̅ + 2598𝑗̅ + 1039.2 𝑘̅
Moment about
𝑥 𝑎𝑥𝑖𝑠 = 1558.8𝑁𝑚
𝑦 𝑎𝑥𝑖𝑠 = 2598 𝑁𝑚
𝑧 𝑎𝑥𝑖𝑠 = 1039.2 𝑁𝑚
Effects of a force :
A force may produce the following effects in a on which it acts
i.
ii.
iii.
It make a body at rest to move
It may retard the motion of a body
It produces internal stresses in a body
System of Forces :When several number of forces act on a body simultaneously then
they are called to form a system of forces.
Coplanar forces: Forces whose line of action lie on the same plane are known as
coplanar forces.
Collinear forces: The forces whose line of action lie on the same line are known
collinear forces.
Concurrent forces: All forces whose line of action meet at one point are known
concurrent forces.
Non-coplanar forces : Line of action of all the forces do not lie in the same plane.
Coplanar – concurrent forces: Line of action of all forces pass through a single point
and forces lie in the same plane.
Coplanar – noncurrent forces: Line of action of forces do not pass through a single
point the lie in the same plane.
Non – coplanar, concurrent forces: All forces do not lie in the same plane but their
line of action pass through a single point.
Non – coplanar, Non concurrent forces: Their line of action of all the forces do not
pass through a single point and they do not lie in the same plane.
Coplanar parallel forces: The line of action of all the forces lie in the same plane and
are parallel to each other.
Like collinear forces: Acting in same line and in the same direction.
Unlike collinear forces: Acting in the same line but in opposite direction.
𝐹4
𝐹5
𝐹1
𝐹1
𝐹3
𝐹2
𝐹1
𝐹2
𝐹3
𝐹2
Collinear forces
𝐹4
Coplanar Forces
Concurrent forces
𝐹4
𝐹1
𝐹1
𝐹1
𝐹2
𝐹3
𝐹3
𝐹2
𝐹3
𝐹2
Coplanar Concurrent Forces
Non-Coplanar Concurrent
Forces
𝐹1
𝐹1
Coplanar parallel
forces
𝐹1
𝐹2
𝐹2
𝐹3
Coplanar Non-concurrent
forces
𝐹2
Non-Coplanar
NonConcurrent Forces
Like collinear forces
𝐹1
𝐹2
Unlike collinear
forces
Like parallel forces: The line of action of forces are parallel to each other and acting
in the same direction.
𝐹1
𝐹2
Unlike parallel forces: The line of action of forces are parallel to each other and acting
in the opposite direction.
𝐹1
𝐹2
Resultant: When a body is acted upon by a number of forces simultaneously, it is
possible to find out a single force which would produce the same effort as produced
by all the given forces acting on the body. This single force is known as resultant
force.
Equilibrium : If a body remains in a state of rest under the action of two or more
forces then the forces are said to be in equilibrium.
Equilibrant: When a body is not in equilibrium under the action of several forces, a
single force can be applied to keep the body in equilibrium. This single force is known
as equilibrant.
Resultant of collinear forces:
(i)
Like collinear forces
(ii) Unlike collinear forces
𝐴
𝐹1
𝐹1
𝑅 = 𝐹1 + 𝐹2 + 𝐹3
𝐹2
𝐹2
𝐹3
𝑅 = 𝐹1 − 𝐹2 + 𝐹3
=
The direction of the resultant
force depends on the sign of
the resultant force obtained
𝐹3
𝐵
Resultant in Same direction
Resultant of concurrent forces:
(i)
Resultant of two concurrent forces
(ii)
Resultant of more than two concurrent forces
Resultant of two concurrent forces:
It can be determined in the following two ways
a) Analytical method – Parallelogram law of forces
b) Graphical method
𝑄
𝑄
𝑃
𝑃
𝑅 =𝑃+𝑄
Resultant force of several forces:
The process of determination of resultant of the given several forces is known
as composition of forces.
The resultant of more than two concurrent forces can be determined by the
following three methods.
(i) Analytical method (by wing method of resolution)
(ii) Lami’s theorem used for three concurrent forces
(iii) Polygon law (Graphical method)
Determination of resultant force using method of resolution:
Resolution of forces: The process of splitting up a given force into two components
along two reference axis, without changing its effect on a particle is called as
resolution of forces.
𝑌
𝐶
𝑌
𝐴
=
𝐹
𝐶
𝐹V
𝜃
O
𝐵
𝑋
O
𝐹𝐻
𝐵
𝑋
Magnitude of components:
If θ is the inclination of the force F with respect to OX axis, then
Horizontal component of the force along OY axis
FH = F cos θ
Vertical component of the force along OY axis
FV = F sin θ
Problem 11: Determine the magnitude and direction of the resultant force for the
system of concurrent forces shown in fig.
𝑌
150 𝑁
20∘
𝑋
′
400 𝑁
30∘
50∘
𝑋
300 𝑁
𝑌′
Solution
Horizontal
Component ∑𝐹𝐻 = 150 𝑐𝑜𝑠 20∘ + 300 𝑐𝑜𝑠 50∘ – 400 𝑐𝑜𝑠 30∘
∑ 𝐹𝐻 = − 12.62 𝑁
Vertical
Component ∑𝐹𝑉 = 150 𝑠𝑖𝑛 20∘ − 300 𝑠𝑖𝑛 50∘ – 400 𝑠𝑖𝑛 30∘
∑ 𝐹𝑉 = − 378.51𝑁
Magnitude of the resultant R
R=
 F    Fv
2
2

4
12.622   378.512
 378.72 N
Direction of the resultant
𝛼 = tan−1 (
∑ 𝐹𝑉
)
∑ 𝐹𝐻
  378.51 
= tan -1   12.62 
 = 88° 5′ 25′′
Both ∑ Fv & ∑ FH are negative so the resultant lies in III quadrant
𝑌
𝑋′
𝑋
80∘ 5′ 55′′
𝑅 = 378.72 𝑁
𝑌′
Problem 15: The resultant of the force system shown in fig. is 520N along the
negative direction of y axis. Determine P and θ
𝑌
360 𝑁
200 𝑁
3
3
2
4
12
𝑋
𝜃
5
260 𝑁
𝑃
Solution:
Resultant is 520 N acting along the negative (↓) direction of y axis.
As resultant force is a truly vertical downward force, ∑ H = 0 and
∑ 𝑉 = 𝑅 = −520𝑁.
Let,
F1 = 200 N,
3
θ1 = tan−1 ( ) = 36.87°
4
F2 = P, angle is θ
12
F3 = 260 N, θ3 = tan−1 ( ) = 67.38°
5
3
F4 = 360 N, θ4 = tan−1 ( ) = 56.31°
2
Algebraic sum of horizontal forces,
∑ H = 200 cos 36.87 + P cosθ – 260 cos 67.38 – 360 cos 56.31
0 = 160 + P cosθ – 100 − 199.69
0 = P cosθ − 139.69
 P cosθ = 139.69N ⟶ (1)
Algebraic sum of vertical forces,
Problem:16: A car is pulled by means of two cars as shown in figure. If the resultant
of the two forces acting on the car A is 40KN being directed along the positive
direction of X axis, determine the angle Q of the cable attached to the car at B, such
that the force in cable AB is minimum. What is the magnitude of force in each cable
when this occurs?
Car B
20°
Car A
Axis
𝜃
Car C
To find 𝜃 for 𝐹𝐴𝐵 minimum:
Here 𝐹𝐴𝐵 is minimum. For this condition the angle between 𝐹𝐴𝐵 and 𝐹𝐴𝐶 should be
equal to 900 (In the triangle length of the side will be minimum only when the sides
are perpendicular to each other (  ABC, 𝑎𝑏 ⊥𝑟 𝑡𝑜 𝑏𝑐)
Triangle law
R = 40KN
𝑎
700
200
𝑐
& FAC ⊥𝑟 to each other]
𝐹𝐴𝐶
200
𝐹𝐴𝐵
[FAB will be least minimum when FAB
𝜃 = 700
𝑏
Forces in the cables
𝐹𝐴𝐵
40
FAB
FAC
=
=
0
0
sin 90
sin 20
sin 700
FAB
40
=
=> FAB = 13.68 KN
0
sin 20
sin 900
A
20°
70°
FAC
40
=
=> FAC = 37.587 KN
0
sin 70
sin 900
𝑅 = 40𝐾𝑁
𝐹𝐴𝐶
Method of resolution (Alternate method)
Resolving forces horizontally
∑ H = FAC cos 200 + FAB cos700 → (1)
∑ H = 0.939 FAC + 0.342 FAB
Resolving forces vertically
∑ V = FAC sin200 − FAB cos700
= 0.342 FAC – 0.939 FAB → (2)
Since the resultant force acting in positive direction of x axis, (given in problem)
R = ∑ M = 40 KN & ∑ V = 0
(1), (2) 
40 = 0.939 FAC + 0.342 FAB
0 = 0.342 FAC – 0.939 FAB
By solving we get
FAB = 13.68 KN and FAC = 37.587 KN
Equilibrium of a particle
𝑌
𝐹2
𝜃2
𝜃1
𝑌
𝑌
𝐹1
𝑅
𝑅
𝑋
𝛼
𝜃
𝛼
𝑋
𝛼
𝑋
𝐹3
𝐸
(a)
(b)
Resultant (R)
(c)
Resultant (R) and
Equilibrant (E)
Equilibrant (E) is equal to the Resultant force (R) in magnitude and direction,
collinear but opposite in direction.
Conditions for Equilibrium
For equilibrium of the force system, the resultant is zero.
i.e, R = 0
R=
 H   V 
2
2
Hence both ∑ H and ∑ V are to be zero for equilibrium condition.
 H = 0 indicates that   
(ie, sum of the left hand forces = sum of the right hand forces)
V  0indieates that   
(ie, sum of the vertical upward forces = sum of the vertical downward forces)
Principle of Equilibrium
Equilibrium principles are developed from the force law of equilibrium
(i.e, ∑ F = 0).
1. Two force principle:
𝐹2
“If the body is subject to two forces, then the body
𝐹1
will be in equilibrium if the two forces are collinear,
equal & opposite”.
2. Three force Principle:
𝐹1
𝐹3
𝑅
𝐹2
“If a body is subjected to three forces, then the body will be in equilibrium, if the
resultant of any two forces is equal, opposite and collinear with the third force”
R is the resultant of 𝐹1 and 𝐹2 .
And R = 𝐹3
Solution of equilibrium of coplanar force:
Lami’s theorem:
“If three coplanar forces acting at a point be in equilibrium, then each force is
proportional to the sine of the angle between the other two”.
𝑃
𝑄
Mathematically
𝑃
𝑄
𝑆
=
=
𝑆𝑖𝑛 𝛼
𝑆𝑖𝑛 𝛽
𝑆𝑖𝑛 𝛾
𝛾
𝛽
𝛼
𝑆
Problem 18: The forces shown in fig are acting on a particle and keep the particle in
equilibrium. The magnitude of force F1 is 250 N. Find the magnitude of forces F2 and
F3.
Method I : Lami’s theorem
𝐹1
𝐹2
𝐹3
=
=
0
0
sin 120
sin 150
sin 900
F1 = 250N
𝐹2
600
300
F2 = 144.33 N, F3 = 288.67N
𝐹3
Method II: (By equations of equilibrium)
Since the particle is in equilibrium
 H  0; V
0
Resolving forces horizontally
∑ H = −F1 cos 600 + F2 cos 300
0 = F1 cos600 + F2 cos300
F2 =
250 cos 60
= 144.33 N
cos 30
Resolving forces vertically
0 = F1 sin600 + F2 sin300 – F3
F3 = 250 sin600 + 144.33 sin300
F3 = 288.67 N.
Free body diagram:
 In equilibrium analysis of structures / machines, it is necessary to consider all
the forces acting on the body and exclude all the forces which are not directly
applied on it.
 The problem becomes much simple if each body is considered in isolation i.e,
separate from surronding body or bodies.
 Such a body which has been separated or isolated from the surronding bodies
is called free body.
 The sketch of showing all forces and moments acting on the body is called as
the free body diagram.
𝐴
𝑇
𝑂
𝑂
𝑊
𝐷𝐹
𝐴
𝐵
𝐶
𝑊𝐵
𝑊𝐶
Freebody diagram at B:
The forces acting at B are
i.
ii.
iii.
Weight of the body attached at B, acting downwards, let it be 𝑊𝐵
Tension on string AB, acting at B towards A, let it be 𝑇𝐵𝐴
Tension on string BC, acting at B towards C, let it be 𝑇𝐵𝐶
Freebody diagram at C:
i.
ii.
iii.
Weight of the body attached at C, acting downwards, let it be 𝑊𝐶
Tension on string CB, acting at C and towards B, let it be 𝑇𝐶𝐵
Tension on string CD, acting at C and towards D, let it be 𝑇𝐶𝐷
𝑇𝐵𝐴
𝐹𝐵𝐷 𝑎𝑡 𝐵
𝑇𝐵𝐶
𝑊𝐵
Now all the forces acting on the string ABCD.
𝐹𝐵𝐷 𝑎𝑡 𝐶
𝑇𝐶𝐷
𝑇𝐶𝐵
𝑊𝐶
𝐷𝐹
𝐴
𝐵
𝐶
𝑊𝐵
Equilibrium
𝑇𝐴𝐵 = 𝑇𝐵𝐴
𝑇𝐵𝐶 = 𝑇𝐶𝐵
𝑇𝐶𝐷 = 𝑇𝐷𝐶
𝑊𝐶
Action and Reaction:
𝑊
W(Action)
𝐴
R A (Reaction)
Consider a ball is place on a horizontal surface. The self-weight of the ball (w) is
acting vertically downwards, through its centre of gravity. This force is called as
“Action”.
Now, the ball can move horizontally. But its vertical downward motion is resisted due
to resisting force developed at support, acting vertically upwards. This force is called
Reaction.
Free Body Diagram
Bodies under Equilibrium
Free Body Diagram
𝑇𝐵𝐴
𝐴
𝐶
𝑅𝐶
𝐵
𝑊𝐵
𝑇𝐵𝐴
𝐴
𝐵
𝐶
𝑅𝐶
𝑊𝐵
𝑇𝐵𝐴
A
𝑅𝐶
C
B
𝑊𝐵
𝑅𝐷
D
𝐶
𝑅𝐵
𝑅𝐴
C
𝑊𝐶
A
B
D
TBA(F)
B
A
F
RD
RC
A
𝑇𝐸
𝑇𝐴𝐵(𝐹)
C
E
𝑊𝐴
𝑇𝐵𝐴 (𝐹) = 𝑇𝐴𝐵 (𝐹)
𝑊𝐶
D
.B
TBA(E)
F
𝑅𝐹
B
A
RC
E
.A
C
𝑇𝐴𝐵(𝐸)
𝑊𝐴
𝑇𝐵𝐴 (𝐸) = 𝑇𝐴𝐵 (𝐸)
𝑊𝐵
RD
D
Problem 19: An electric light fixture weighting 150 N hangs from a point C, by two
strings AC and BC as shown in fig. Determine the forces in the strings AC and BC.
𝑇𝐶𝐵
450
𝑇𝐶𝐴
300
𝐴
𝐵
60
450
0
450
300
600
𝐶
150 N
150 N
𝐹𝐵𝐷
From the Free body diagram,
Angle between 𝑇𝐶𝐵 & 150 𝑁 = 900 + 300 = 1200
Angle between 𝑇𝐶𝐴 & 150 𝑁 = 450 + 900 = 1350
Angle between 𝑇𝐶𝐵 & 𝑇𝐶𝐴 = 3600 – (1200 + 1350 ) = 1050
Now applying lami’s theorem at C,
𝑇𝐶𝐵
𝑇𝐶𝐴
150
=
=
sin 135° sin 120° sin 105°
∴
𝑇𝐶𝐴
150
=
sin 120° sin 105°
𝑇𝐶𝐴 = 134.49 𝑁
𝑇𝐶𝐵
150
=
°
sin 135
sin 105°
TCB = 109.81 N
Note: This problem can also be solved by applying equations of equilibrium at C.
Problem 21: A string ABCD attached to two fixed points A and D has two equal
weights of 1000N attached to it at B and C. The weights rest with the portions AB and
CD inclined at 300 and 600 respectively to the vertical as shown in figure. Find the
tensions in the portions AB, BC and CD of the string if the inclination of the portion
BC with the vertical is 1200.
A
D
60°
30°
120°
C
B
1000N
1000N
Solution:
Free Body Diagram
TDC
TAB
A
TBA T
BC
30°
D
60°
TCD
TCB
B
C
𝑇𝐴𝐵 = 𝑇𝐵𝐴
𝑇𝐵𝐶 = 𝑇𝐶𝐵
1000N
𝑇𝐶𝐷 = 𝑇𝐷𝐶
1000N
Applying Lami’s theorem at B,
𝑇𝐵𝐴
𝑇𝐵𝑐
1000
=
=
𝑆𝑖𝑛 60° 𝑆𝑖𝑛 150° 𝑆𝑖𝑛 150°
TBA
30
30°
°
∴ 𝑇𝐵𝐴 =
120°
60°
B
𝑇𝐵𝐴 =
60°
1000 sin 60°
= 1732𝑁
𝑆𝑖𝑛 150°
1000 sin 150°
= 1000𝑁
𝑆𝑖𝑛 150°
TBC
1000N
Applying Lami’s theorem at C,
TCB
TCD
60°
60°
30
°
60°
60°
30°
𝑇𝐶𝐵
𝑇𝐶𝐷
1000
=
=
𝑆𝑖𝑛 120° 𝑆𝑖𝑛 120° 𝑆𝑖𝑛 120°
𝑇𝐶𝐵 =
B
𝑇𝐶𝐷
1000N
1000 sin 120°
𝑆𝑖𝑛 120°
= 1000𝑁 = 𝑇𝐵𝐶
1000 sin 120°
=
= 1000𝑁
𝑆𝑖𝑛 120°
∴Tension in the string AB = 1732 N
Tension in the string BC = 1000 N
Tension in the string CD = 1000 N
Problem 22 : Three smooth pipes each weighting 20 KN and of diameter 60 cm are to
be placed in a rectangular channel with horizontal base as shown in fig. Calculate the
reactions at the points of contact between the pipes and between the channel and the
pipes. Take width of channel as 160cm.
C
20KN
B
A
𝐷
20KN
20KN
𝐸
𝐺
𝐹
Solution:
2
In triangle ABC side
AB = 160 - AD-BG
C
FCA
1
RD
3
20KN
FCB
FBC
FAC
A
B
20KN
20KN
160cm
RE
=160 – 30 – 30
RG
=100 cm
AC = BC = 2 x Radius
= 60 cm
RF
Draw a vertical line through C, ⊥𝑟 to AB, to intersect at m
In  Bmc
Bm =
AB 100

 50 cm
2
2
Cos θ =
Bm 50

Bc
60
𝜃 = 32.550
(1),( 2) - more number of unknown forces
(3) – Two forces
Free body diagram of Pipe (3)
Now applying equation of equitation at C,
𝐶
33.550
33.550
∑𝐻 = 0
FAC cos 33.550 – FBC cos 33.550 = 0
𝐹𝐴𝐶
𝐹𝐵𝐶
FAC = FBC
20KN
V 0
FAC sin 33.550 + FBC sin 33.550 − 20 = 0
2FBC sin 33.550 = 20
FBC = 18.09 KN
FAC = 18.09 KN
Free body diagram at pipe (1)
FCA = FAC = 18.09 KN
33.550
RD
Applying equations of equilibrium
∑𝐻 = 0
R D – FAC cos 33.550 = 0
20KN
RE
R D = 15.07KN
∑𝑉 = 0
– FAC sin 33.550 − 20 + R E = 0
R E = 30 KN
P cos300 − R cos 53.130 = 0
P= R
cos 53.13 0
cos 30 0
P = 0.693 R
∑𝑉 = 0
P sin300 + R sin 53.130 − 10 = 0
Sub P = 0.693 R
0.693 R sin 300 + R sin 53.130 = 10
0.3464R + 0.8R = 10
R = 8.72 KN
P = 0.693 R
P = 0.693 × 8.72
P = 6.04 KN
∴The amount of force required to pull the roller to cross the obstacle is 6.04 KN
Rectangular components of a force:
Already we have seen the resolution of force into two components which one
perpendicular to each other.
𝐹̅
𝑌
Fy j
𝐹
Fy
𝑂
(𝑎)
FX
𝑗
X
𝑖
(𝑏)
𝑂
(𝑐)
Fx j
In fig (a) the force has been resolved into two component 𝐹𝑥 and 𝐹𝑦 along x and y
axes, respectively called rectangular components of the force F.
Where, 𝐹𝑥 = 𝐹 𝑐𝑜𝑠 𝜃 and 𝐹𝑦 = 𝐹 𝑠𝑖𝑛 𝜃
Now, in vector approach, introduce unit vectors i and j along the positive x and y axes.
(A unit vector is a vector of unit magnitude i.e., one unit in length)
To get the vector quantities of the given force along x and y axes, multiply the scalar
quantities Fx and Fy with respective unit vectors i and j.
Vector quantity along x-axis= 𝑠𝑐𝑎𝑟 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑎𝑙𝑜𝑛𝑔 𝑥 − 𝑎𝑥𝑖𝑠 × 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑎𝑙𝑜𝑛𝑔 𝑥 − 𝑎𝑥𝑖𝑠
⃗⃗⃗
𝐹𝑥 = 𝐹𝑥 . 𝑖
i.e.
Similarly
⃗⃗⃗𝑦 = 𝐹𝑦 . 𝑗
𝐹
By vector addition of forces, the force vector 𝐹 is equal to the sum of its components
of vector quantities in x and y directions.
⃗⃗⃗𝑦 (or) 𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗
𝐹 = ⃗⃗⃗
𝐹𝑥 + 𝐹
Magnitude of the force, F = √(𝐹𝑥 )2 + (𝐹𝑦 )2
Also, we know Fx = F cos  and Fy = F sin
⃗⃗ can be written as below,
∴ F
𝐹 = (𝐹 𝑐𝑜𝑠 )𝑖 + 𝐹 𝑠𝑖𝑛𝑗
Force vector in terms of x and y
For the force 𝐹
⃗F = F cosx i + F siny j
To express F interms of only cosine
(cos) ratios
sinx = cosy
⃗F = F cosx i + F cosy j
𝐹
Y
𝜃𝑦
𝜃𝑥
O
X
𝜃𝑥 and 𝜃𝑦 angle of force 𝐹 with x and
y axes respectively
Force vector in terms of co-ordinates
B
B (x2,y2)
Y
d
𝑑𝑦
A
A (x1,y1)
O
X
𝑑𝑥
Consider a force of magnitude ‘F’ directed from A(x1, y1) to B (x2, y2) as shown in fig.
Let d = distance between A and B
Resolving d in x and y directions
𝑑𝑥 = 𝑥2 – 𝑥1 𝑎𝑛𝑑 𝑑𝑦 = 𝑦2 – 𝑦1
∴ 𝑑̅ = 𝑑𝑥 𝑖 + 𝑑𝑦𝑗
= (𝑥2 – 𝑥1 ) 𝑖 + (𝑦2 – 𝑦1 ) 𝑗
x2  x1 2   y2  y1  2
Distance, d =
Unit vector along AB =𝜆𝐴𝐵 =
𝑑̅
𝑑
=
𝑑𝑥𝑖+𝑑𝑦𝑗
𝑑
⃗ = 𝜆𝐴𝐵 . 𝐹 = 𝐹 [𝑑𝑥𝑖+𝑑𝑦𝑗]
Force VectorF
𝑑
⃗ = 𝐹 [𝑑𝑥𝑖 + 𝑑𝑦𝑗] → (1)
F
𝑑
𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗 → (2)
Equating 1 & 2
𝐹𝑥 =
Fdx
Fdy
𝑎𝑛𝑑 𝐹𝑦 =
d
d
F cosθx =
F.dx
d
, F cosθy =
F.dy
d
cosθx =
θx = cos −1 (
dx
dy
, cosθy =
d
d
𝑑𝑥
𝑑𝑦
) , θy = cos −1 ( )
𝑑
𝑑
Problem 26: A force vector of magnitude 40 N, is directed from A (1, 4) to B (6, 7).
Determine
(i)
(ii)
(iii)
The components of the force along x, y axes
Angles with x and y axes
Specify the force vector
Solution:
d
𝑑𝑥 = 𝑥2 – 𝑥1 = 6 − 1 = 5
B (6,7)
𝑥2 , 𝑦2
𝑑𝑦 = 𝑦2 – 𝑦1 = 7 – 4 = 3
∴ 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐷 = √𝑑𝑥 2 + 𝑑𝑦 2
√52 + 32 = 5.83
A (1,4)
𝑥1 , 𝑦1
Components of the force (𝐹𝑥 & 𝐹𝑦 )
(i)
𝐹𝑥 = 𝐹
𝐹𝑦 = 𝐹
(ii)
𝑑𝑥
40 × 5
=
= 34.30 𝑁
𝑑
5.83
𝑑𝑦
40 × 3
=
= 20.58 𝑁
𝑑
5.83
Angle with co-ordinate axes
𝐹𝑥
𝑑𝑥
𝜃𝑥 = cos −1 ( ) (𝑂𝑟) ( )
𝐹
𝑑
 34.3 
 = 30.960
 40 
𝜃𝑥 = cos −1 
𝐹𝑦
𝑑𝑦
𝜃𝑦 = cos −1 ( ) (𝑂𝑟) ( )
𝐹
𝑑
 20.58 
 = 59.030
40


= cos −1 
(iii)
Force Vector
⃗⃗⃗
𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗
𝐹 = 34.30 𝑖 + 20.58 𝑗
Resultant force of coplanar concurrent force system
Let R be the resultant force
R in Rectangular co-ordinates
Magnitude of resultant force R =
Direction of the resultant force,
Rx 2  Ry 2
𝜃𝑥 = cos −1 (
𝑅𝑦
𝑅𝑥
) 𝑎𝑛𝑑 cos −1 ( )
𝑅
𝑅
If a number of concurrent forces acts on a particle, then the resultant force is
determined by summing their rectangular components.

Fx and

Fy are the algebraic sum of x and y components.
Vector Approach for a 3 – Dimensional Problem
Rectangular components of a force in space
Consider a force 𝐹 acting in a space, at the point of origin O. This Force 𝐹 can be
resolved into three components along x, y and z axes and we may write as
Where Fx = Vector components of the force vector 𝐹 along x axis
llly Fy and Fz.
In fig OM, ON and OP are the scalar components of force vector 𝐹 .
Let 𝜃𝑥 , 𝜃𝑦 and 𝜃𝑧 are the angles made by 𝐹 with x, y and z axes respectively.
The scalar quantities of force 𝐹 along x, y and z axes
Where F = magnitude of 𝐹
Now to get vector quantities multiply the scalar quantities by the respective unit
vectors.
 The force vector F in space can be written as
Magnitude of the force F is given by,
⃗⃗ | = F = √Fx 2 + Fy 2 + Fz 2
|F
Direction cosines
The cosines of θx, θy and θz are known as the direction cosines of the force ⃗⃗⃗
𝐹.
Let
Cos θx = l; cos θy = m; cos θz = n
There three angles are related by
cos 2 θx + cos 2 θ y + cos 2 θ z = 1
i.e, l2 + m2 + n2 =1
Problem 28 : A force of 400 N forms angle of 600, 450 and 1200 respectively with x,
y and z axes. Find the components along three coordinate axes and express the force
in Cartesian (i.e Rectangular) coordinates.
Solution: Also find angle of the force.
Magnitude of force = F = 400 N
𝜃𝑥 = 600 ; 𝜃𝑦 = 450 ; 𝜃𝑧 = 1200
Component of force along X axis
𝐹𝑥 = 𝐹 𝑐𝑜𝑠𝜃𝑥
= 400 𝑐𝑜𝑠 600 = 200𝑁
Lll y
𝐹𝑦 = 𝐹 𝑐𝑜𝑠 𝜃𝑦 = 400 𝑐𝑜𝑠 450 = 282.84
𝐹𝑧 = 𝐹 𝑐𝑜𝑠 𝜃𝑧 = 400 𝑐𝑜𝑠 1200 = − 200 𝑁
And expressing the force in Cartesian coordinates we get
⃗⃗⃗ = Fxi + Fyj + Fzk
𝐹
⃗⃗⃗
𝐹 = 200i+ 282 – 84j– 200k
To find the angle, we need F
Magnitude, F =
Fx  Fy  F2
2
2
2
= 150 2   2002  300 2 = 390.51N
We know,
𝐹𝑥 = 𝐹 𝑐𝑜𝑠 𝜃𝑥 Or 𝑐𝑜𝑠𝜃𝑥 =
𝐹𝑥
𝐹
𝐹𝑥
150
) = 67.410
𝜃𝑥 = cos −1 ( ) = cos −1 (
𝐹
390.51
𝐹𝑦
−200
) = 120.80
𝜃𝑦 = cos −1 ( ) = cos −1 (
𝐹
390.51
𝐹𝑧
300
) = 39.800
𝜃𝑧 = cos −1 ( ) = cos −1 (
𝐹
390.51
To check
Force vector in co-ordinates
Y
𝐵(𝑥2 , 𝑦2 , 𝑧2 )
𝐹̅
N
p
𝐴(𝑥1 , 𝑦1 , 𝑧1 )
m
X
Z
Consider the vector ⃗⃗⃗⃗⃗
𝐴𝐵 , joining A and B of co-ordinates (x1, y1, z1) and (x2, y2, z2)
respectively.
Let d = Distance between A and B
Now this distance can be resolved into three components along the three co-ordinate
axes are dx, dy and dz (which are equal to Am, AN, and Ap)
Distances interms of co-ordinates
𝑑𝑥 = 𝑥2 − 𝑥1 ; 𝑑𝑦 = 𝑦2 − 𝑦1 ; 𝑑𝑧 = 𝑧2 − 𝑧1

 vector AB  ( x2  x1 ) i  ( y2  y1 ) j  ( z2  z1 )k

because AB  dxi  dyj  dzk
The distance d interms of co-ordinates
d  (dx) 2  (dy ) 2  (dz ) 2
d  ( x2  x1 ) 2  ( y2  y1 ) 2  ( z 2  z1 ) 2
Unit vector along
⃗⃗⃗⃗⃗
𝐴𝐵
 AB 
AB
AB

dxi  dyj  dzk
d
𝐹 = 𝐹. 𝛼𝐴𝐵
= 𝐹. (
𝑑𝑥𝑖 + 𝑑𝑦𝑗 + 𝑑𝑧𝑘
)
𝑑
Scalar components
𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗 + 𝐹𝑧 𝑘
𝐹𝑑𝑥 𝐹
= (𝑥2 − 𝑥1 )
𝑑
𝑑
𝐹𝑥 =
IIIly
𝐹𝑦 =
𝐹𝑑𝑦
𝑑
𝐹
𝐹𝑑𝑧
𝑑
𝑑
= (𝑦2 − 𝑦1 );𝐹𝑧 =
𝐹
= (𝑧2 − 𝑧1 )
𝑑
Angle of inclination with co-ordinate axes
We know Fx  F cos  x and Fx 
F cos  x 
Fdx
d
Fd x
d
 dx 
1  x  x 
  cos  2 1 
d 
 d 
 x  cos 1 
IIIly
 dy 
1  y  y1 
  cos  2

d 
 d 
 y  cos 1 
 dz 
1  z  z 
  cos  2 1 
d 
 d 
 z  cos 1 
Problem 29: A force vector of magnitude 100N, is represented by a line AB of coordinates A (1,2,3) and B (5,8,12). Determine
(i) The components of the force along x, y and z axes
(ii) Angles with x, y and z axes
(iii) Specify the force vector.
Solution:
Let the distance of AB=d
The components of distance d, along the co-ordinate axes
d x  x2  x1  5  1  4
d y  y2  y1  8  2  6
d z  z 2  z1  12  3  9
 The distance of AB  d  dx 2  dy 2  dz 2
 42  62  92 11.53
(i)
Components of the force along the axes
Using 𝐹𝑥 =
111y
𝐹𝑧 =
𝐹𝑦 =
𝐹𝑑𝑥
𝑑
𝐹𝑑𝑦
𝑑
=
=
100×4
11.53
100×6
11.53
= 34.69𝑁
= 52.03𝑁
𝐹𝑑𝑧 100 × 9
=
= 78.05𝑁
𝑑
11.53
(ii)
Angles with x, y and z axes
4
Using 𝜃𝑥 = cos −1 (𝑑𝑥
) = cos−1 (
) = 69.70 )
𝑑
11.53
𝑑𝑦
6
𝜃𝑦 = cos −1 ( ) = cos−1 (
) = 58.640 )
𝑑
11.53
𝑑𝑧
9
𝜃𝑧 = 𝑐𝑜𝑠 −1 ( ) = 𝑐𝑜𝑠 −1 (
) = 38.680 )
𝑑
11.53
To check
cos2 𝜃𝑥 + cos2 𝜃𝑦 + cos2 𝜃𝑧 = 1
cos2 69.7 + cos2 58.64 + 𝑐𝑜𝑠 2 38.68 = 1
0.120 + 0.270 + 0.609 = 1
0.991 ≈ 1
(iii)
Force vector
The force vector 𝐹 is given by
𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗 + 𝐹𝑧 𝑘
 𝐹 = 34.69𝑖 + 52.03𝑗 + 78.05 𝑘
Resultant force for a force system in space
Let 𝑅⃗ be the resultant force, writing 𝑅⃗ in Cartesian co-ordinates, we get
⃗⃗⃗ = 𝑅𝑥 𝑖 + 𝑅𝑦 𝑗 + 𝑅𝑧 𝑘
𝑅
Magnitude of resultant force R =
Rx 2  R y  2  Rz 2
Where Rx, Ry and Rz are the scalar components of Resultant force along x, y and z
axes respectively.
Direction of resultant force
Let θx, θy and θz be the angle of resultant force with x, y and z axes respectively
111y
𝑅
𝜃𝑥 = cos −1 ( 𝑥)
𝑅
𝑅𝑦
𝜃𝑌 = cos −1 ( )
𝑅
𝑅
𝜃𝑍 = cos −1 ( 𝑧)
𝑅
Resultant force of concurrent spatial forces
The resultant of concurrent spatial forces ⃗⃗⃗⃗
F1 , ⃗⃗⃗⃗
F2 etc., acting on a particle is given by
𝑅⃗ = 𝐹1 + 𝐹2 + ⋯
= (𝐹1𝑥 𝑖 + 𝐹1𝑥 𝑗 + 𝐹1𝑧 𝑘) + (𝐹2𝑥 𝑖 + 𝐹2𝑦 𝑗 + 𝐹3𝑧 𝑘) + ⋯
𝑅⃗ = (∑ 𝐹𝑥 )𝑖 + (∑ 𝐹𝑦 )𝑗 + (∑ 𝐹𝑧 )𝑘
𝑅⃗ = 𝑅𝑥 𝑖 + 𝑅𝑦 𝑗 + 𝑅𝑧 𝑘
 𝑅𝑥 = ∑ 𝐹𝑥 ; 𝑅𝑦 = ∑ 𝐹𝑦 ; 𝑅𝑧 = ∑ 𝐹𝑧
Equilibrium of particles in space
When a particle is subjected to concurrent force in space, for equilibrium condition
resultant force𝑅⃗ = 0
𝑅𝑥 𝑖 + 𝑅𝑦 𝑗 + 𝑅𝑧 𝑘 = 0
(∑ 𝐹𝑥 )𝑖 + (∑ 𝐹𝑦 )𝑗 + (∑ 𝐹𝑧 )𝑘 = 0
∴ ∑ 𝐹𝑥 = 0; ∑ 𝐹𝑦 = 0; ∑ 𝐹𝑧 = 0
Problem 30:
The lines of action of three forces are concurrent at the origin ‘O’, passes through
point A, B and c having coordinates (3,0,-3), (2,-2,4) and (-1,2,4) respectively. If the
magnitude of the forces are 10N,30N and 40N, find the magnitude and direction of
their resultant.
B(2,-2,4)
Solution:
A(3,0,-3)
FA = 10N, FB = 30N & Fc = 40N
30KN
Consider the force passing through A
O
10KN
Position vector
𝑟𝑂𝐴 = (𝑥2 − 𝑥1 )𝑖 + (𝑦2 − 𝑦1 )𝑗 + (𝑧2 − 𝑧1 )𝑘
⃗⃗⃗⃗⃗⃗
40KN
𝑟𝑂𝐴 = 3𝑖 − 3𝑘
⃗⃗⃗⃗⃗⃗
C(-1,2,4)
Unit vector along OA,𝜆𝑂𝐴 =
⃗⃗⃗⃗⃗⃗⃗⃗
𝑟𝑂𝐴
𝑟
3𝑖 − 3𝑘
𝜆𝐴𝐵 =
√32 + 32
=
3𝑖 − 3𝑘
4.24
3𝑖−3𝑘
⃗⃗⃗⃗
] = 7.07𝑖 − 7.07𝑘
𝐹𝐴 = 𝐹𝐴 . 𝜆𝑂𝐴 = 10 [
4−2𝑘
Consider the force passing through B
Position vector ̅̅̅̅
𝑟𝑂𝐵 = (𝑥2 − 𝑥1 )𝑖 + (𝑦2 − 𝑦1 )𝑗 + (𝑧2 − 𝑧1 )𝑘
= 2i – 2 j +4k
Unit vector along OB, 𝜆𝑂𝐵 =
̅̅̅̅̅̅
𝑟𝑂𝐵
2i  2 j  4k
𝑟
2 2 4
2
2
2

2i  2 j  4k
4.899
̅̅̅
𝐹
𝐵 = 𝐹𝐵 𝜆𝑂𝐵
 2i  2 j  4k 
= 12.247i – 12.247j+24.49k
4.899 
= 30 

Consider the force through C
Position vector𝑟̅̅̅̅
𝑂𝐶 = (𝑥2 − 𝑥1 )𝑖 + (𝑦2 − 𝑦1 )𝑗 + (𝑧2 − 𝑧1 ) 𝑘
= -i+2j+4k
Unit vector along OC,
𝜆𝑜𝑐 =
̅̅̅̅̅
𝑟𝑜𝑐
𝑟

 i  2 j  4k
1 2 4
2
2
2
=
 i  2 j  4k
4.582
̅̅̅
∴𝐹
𝐶 = 𝐹𝐶 . 𝜆𝑂𝐶
  i  2 j  4k 
= -8.729i + 17.459j + 34.919k
4.582 
= 40 

Resultant force
The resultant force vector is obtained by adding force vector 𝐹𝐴 , 𝐹𝐵 𝑎𝑛𝑑 𝐹𝐶
̅̅̅
̅̅̅ ̅̅̅
∴ 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒 𝑅̅ = 𝐹
𝐴 + 𝐹𝐵 + 𝐹𝐶
i.e𝑅̅ = (7.07𝑖 − 7.07𝑘) + (12.247𝑖 − 12.247𝑗 + 24.49𝑘) + (−8729𝑖 + 17.459𝑗 + 34.919𝑘)
𝑅̅ = 10.59𝑖 + 5.212𝑗 + 52.343𝑘
Magnitude of resultant force
R = 10.592  5.2122  52.3432 = 53.657N
Direction of Resultant force
Let the angler of resultant force with x,y and z axes are θx, θy, θz respectively
𝑅
10.59
𝑅
53.657
Using 𝜃𝑥 = cos −1 ( 𝑥) = cos −1 (
𝑅𝑦
5.212
𝑅
53.657
lllly 𝜃𝑦 = cos −1 ( ) = cos −1 (
𝜃𝑧 = cos −1 (
) = 78.610
) = 84.420
𝑅𝑧
52.343
) = cos −1 (
) = 12.70
𝑅
53.657
To check
𝑐𝑜𝑠 2 𝜃𝑥 + 𝑐𝑜𝑠 2 𝜃𝑦 + 𝑐𝑜𝑠 2 𝜃𝑧 = 1
cos 2 78.61 + cos2 84.42 + cos 2 12.7  1 𝑜. 𝑘
Problem 31: The tension in cables AB and AC are 100N and 120N respectively in fig.
Determine the magnitude of the resultant force acting at A.
Solution:
Y
Considering the tension in cable AB
A
The force is directed from A to B.
A (0, 4, 0),
x1, y1, z1,
4m
B (0, 0, 4)
X
x2, y2, z2
C
B
A, B – Coordinates
Z
2m
Position vector ̅̅̅̅
rAB = (x2 − x1 ) i + (y2 − y1 ) j + (z2 − z1 )k
= −4𝑗 + 4𝑘
4m
̅̅̅̅ =
 Unit vector along 𝐴𝐵
̅̅̅̅̅̅
𝑟𝐴𝐵
=
𝑟
−4𝑗+4𝑘
√42 +42
=
−4𝑗+4𝑘
5.656
Tension in cable
̅̅̅̅
𝑇𝐴𝐵 = 𝑇𝐴𝐵 . 𝜆𝐴𝐵
−4𝑗+4𝑘
= 100 [
5.656
]
̅̅̅̅̅
T
AB = − 70.72j + 70.72k
Considering tension in cable AC
Now the force is dissected from A to C
A (0, 4, 0)
B (2, 0, 4)
x1, y1, z1
x2,y2,z2
A, B – Coordinates
 Position vector ̅̅̅̅
rAC = (x2 − x1 ) i + (y2 − y1 ) j + (z2 − z1 )k
= 2𝑖 – 4 𝑗 + 4 𝑘
𝑟𝐴𝐶
2𝑖−4𝑗+4𝑘
̅̅̅̅ , 𝜆𝐴𝐶 = ̅̅̅̅̅
Unit vector along 𝐴𝐶
= √22 2 2
𝑟
=
+4 +4
2𝑖−4𝑗+4𝑘
6
 Tension in cable ̅̅̅̅
𝐴𝐶 , ̅̅̅̅
𝑇𝐴𝐶 = 𝑇𝐴𝐶 . 𝜆𝐴𝐶
= 120 [
2𝑖−4𝑗+4𝑘
6
]
̅̅̅̅
𝑇𝐴𝐶 = 40𝑖 – 80𝑗 + 80𝑘
Resultant force
𝑅̅ = ̅̅̅̅
𝑇𝐴𝐵 + ̅̅̅̅
𝑇𝐴𝐶
= [−70.72j + 70.72k] + [40i + 80j + 80k]
Now i, j and k components
𝑅̅ = 40𝑖 – 150.72𝑗 + 150.72𝑘
 Magnitude of Resultant force R =
40 2  (150.72) 2  (150.  72) 2
R = 216.87N
Principle of Transmissibility
B
B
F
=
A
F
A
The condition of equilibrium or motion of a rigid body remain, unchanged if a force
acting at a given point of the rigid body is replaced by a force of same magnitude and
direction, but acting at a different point provided that the two forces have the same
line of action.