Fundamental of Statics (Part 3)

Department of
Mechanical
Engineering.
Prof. Malay Badodariya
+91 9429 158833
FundamentalOfMachineDesign (01ME0504)
Unit 2: Fundamental of
Statics
(Part 3)

Topic will be cover:
1. Coplanar non concurrent forces:
a) Moment of a force
b) Couple
c) Characteristics of a couple
d) Equivalent Couple
e) Force Couple System
f) Varignon’s Principle of Moments (Law of
Moments)
g) Condition of equilibrium for coplanar non
concurrent forces
h) Example
Department of Mechanical
Engineering.
malay.badodariya@gmail.com

Topic will be cover:
1. Coplanar non concurrent forces:
a) Moment of a force
b) Couple
c) Characteristics of a couple
d) Equivalent Couple
e) Force Couple System
f) Varignon’s Principle of Moments (Law of
Moments)
g) Condition of equilibrium for coplanar non
concurrent forces
h) Example
Engineering.
malay.badodariya@marwadie
ducation.edu.in
Unit no.: 2
Fundamental Of
Statics
Subject Name: FMD
01ME0305

a)
Moment of a
Force
The moment of a force is equal to the product of the
force and the perpendicular distance of line of action of
force from the point about which moment is required.
M = F l
Where, M = Moment, F= Force,
l = Perpendicular distance between the
line of action of force and the point
about which moment is required
 Moment produce turning effect of the body.
 Unit of moment is N.m or KN.m.
 Example of moments are:
i. To tight the nut by spanner
ii. To open or close the door

Sign of Moments:
 Generally, Clockwise moment is taken as positive and
Anticlockwise moment is negative.

b)
Couple
 Two equal and opposite forces whose line of action are
different form a couple.
 Resultant or net force of couple is zero.
 Hence couple acting on body do not create any
translatory motion of the body.
 Couple produce only rotational motion of the body.

Types of Couple:
 There are two type of couple
i. Clockwise couple
ii. Anticlockwise couple
It is taken as positive It is taken as Negative

Arm of a Couple:
 The perpendicular distance between the line of action of
two forces forming couple is known as the Arm of
couple.
 In figure, X = arm of couple.

Moment of Couple:
 Moment of couple = Force XArm of couple
M = P X
 Unit of couple is N.m. or kN.m.
Example:
i. Steering Wheel of car
ii. Wheel valve of water supply pipe line.

c)
Characteristics
of a couple
A couple has the following characteristics.
i. The algebric sum of the forces, forming the couple is
zero.
ii. The algebric sum of the moment of the forces, forming
the couple, about any point is the same and equal to
the moment of the couple it self.
Moment of couple, M = P.a
Now, taking moment at ‘O’.
𝑴 𝒐 = P(a+X) – P.X
= Pa + PX – PX
𝑴 𝒐 = P.a

iii. A couple cannot be balanced by a single force, but can
be balanced only by a couple, but of opposite nature.
iv. If number of coplanar couples are acting on the body,
than they can be reduced to a single couple, whose
magnitude will be equal to the algebric sum of
moments of all the couples.

d)
Equivalent
couples
 Couple are said to be equivalent couples, if they produce
similar effects on a body.
 Consider three different systems acting on a rectangular
block as shown below.
 In the three cases above, all the couples produce moment
= 20,000 N.mm = 20 N.m (C.W.)
 Hence, these three couples are called equivalent couples.

e)
ForceCouple
system
 When a system consists of forces along with moments or
couple, it is known as force couple system.
 This concept is helpful in shifting a force away from its line
of action.
 A given force ‘F’ applied to a body at any point ‘A’ can
always be replaced by an equal force applied at another
point ‘B’ together with a couple which will be equivalent
to the original force.

i. Force F is acting at point A.We want to shift the force F
fromA to B.
ii. Apply two equal and opposite forces at ‘B’ parallel to F,
the magnitude of that force must be equal to the force
‘F’ acting at ‘A’.
iii. Let ‘X’ is the perpendicular distance between line of
action of forces.

iv. The force at ‘A’ and the force in opposite direction at ‘B’
forms a couple of magnitude F.X as shown in fig. c.
v. Thus, a single force ‘F’ acting at ‘A’ is replaced by a force
couple system acting at ‘B’.

f)
Varignon’s
Principle of
moments.
 “The algebric sum of moments of all forces about a point
is equal to the moment of resultant about the same
point.”
 Now consider non concurrent coplanar forces for finding
resultant force.
Example:
F1 F2 F3 F4
X1
X2
X3
A
R
X
Now as per definition,
⅀𝑀𝐴 = 𝑅. 𝑋
 Consider all clock wise moment is
+ve. & counter Clock wise moment is
–ve.
 Now find momentWRT point ‘A’:
i. Force F1 = F1.0 = 0 (perpendicular
distance is zero)

Example:
F1 F2 F3 F4
X1
X2
X3
A
R
X
 Consider all clock wise moment is
+ve. & counter Clock wise moment is
–ve.
 Now find momentWRT point ‘A’:
i. Force F1 = F1.0 = 0 (perpendicular
distance is zero)
ii. Force F2 = - F2.X1 (CCW Moment for
point A)
iii. Force F3 = F3.X2 (CW moment for
point A)
iv. Force F4 = - F4.X3 (CCW Moment for
point A)
v. Resultant Force R = R.X (CW
moment for point A)

Example:
F1 F2 F3 F4
X1
X2
X3
A
R
X
⅀𝑀𝐴 = 0 + (- F2.X1) + (F3.X2) + (- F4.X3)
⅀𝑀𝐴 = Algebraic sum of the moment of
all the forces.

g)
Condition of
Equilibrium for
Coplanar Non
Concurrent
Forces
 If a body is acted upon by a number of co-planar non-
concurrent forces, it may have one of the following states:
i. The body may move in any one direction.
ii. The body may rotate about itself without moving.
iii. The body may move in any one direction, and at the
same time it may also rotate about itself.
iv. The body maybe completely at rest.
 Now we will discuss the above four states.

 If the body moves in any direction, it means that there is
a resultant force acting on it.
 A little consideration will show that if the body is to be at
rest or in equilibrium, the resultant force causing
movement must be zero.
 i.e.,
⅀H = 0 & ⅀V = 0
i.The body may move in any one direction.

 If the body rotates about itself without moving it means
that there is a single resultant couple acting on it.
rest or in equilibrium, the moment of the couple causing
rotation must be zero. i.e.
⅀M = 0
ii.The body may rotate about itself without moving.

 If the body moves in any direction and at the same time
it rotates about itself.
 It means that there is a resultant force and also a
resultant couple acting on it.
rest or in equilibrium the resultant force causing
movement and the resultant moment of the couple
causing rotation must be zero. i.e.
⅀H = 0 & ⅀V = 0
⅀M = 0
iii.The body may move in any one direction, and at the
same time it may also rotate about itself.

 If the body is completely at rest it means that there is
neither a resultant force nor a couple acting on it. In this
case following conditions are satisfied.
⅀H = 0 & ⅀V = 0
⅀M = 0
iv. The body maybe completely at rest.

Example on Moment Of Forces
Example 1: ABCD is a square of 5 m. side forces of 100 N, 250 N, 150 N and 300 N act along BA, DA, DC, BC
respectively. Find the algebric sum of the moments of all the forces about the centre point of the square. (Write
ABCD on square in the clockwise sense)
Solution:
Given Data,
5m
5 m
O
A B
D C
300 N
100 N
250 N
150 N
Moment about centre of square:
For 100 N Force = - 100 X 2.5 = - 250 Nm (CCW)
For 250 N Force = 250 X 2.5 = 625 Nm (CW)
For 150 N Force = - 150 X 2.5 = - 375 Nm (CCW)
For 300 N Force = 300 X 2.5 = 750 Nm (CW)
𝑀 𝑂 = −250 + 625 + −375 + 750 = 750 𝑁𝑚
HereValue of 𝑀 𝑂 is positive so moment will Clockwise
(CW) Direction.

Example 2: Find the moment of force of 30 kN about point ‘O’ as shown in figure.
Solution:
Given Data,
Split 30 kN force into two component, Horizontal
&Vertical.

Solution:
Given Data, Perpendicular Distance betweenVertical Force ( F sin 30) to ‘O’
O 60
A B
C
Cos 45 =
𝐴𝐵
𝐴𝐶
=
𝐴𝐵
2
𝑆𝑜, 𝐴𝐵 = 2 𝐶𝑜𝑠 45 =1.414 m
Sin 45 =
𝐵𝐶
𝐴𝐶
=
𝐵𝐶
2
𝑆𝑜, 𝐵𝐶 = 2 𝑆𝑖𝑛 45 =1.414 m
D E
F
Cos 60 =
𝐷𝐸
𝐷𝐹
=
𝐷𝐸
2
𝑆𝑜, 𝐷𝐸 = 2 𝐶𝑜𝑠 60 =1 m
Sin 60 =
𝐸𝐹
𝐷𝐹
=
𝐸𝐹
2
𝑆𝑜, 𝐸𝐹 = 2 𝑆𝑖𝑛 60 =1.73 m

Solution:
Given Data,
O
Moment of all forces about ‘O’:
For F Cos 30 N Force = ( 30 Cos 30 )(1.73+1.41) = 81.58 kNm
For F Sin 30 N Force = (30 Sin 30)(1+3+1.41) = 81.15 kNm
𝑀 𝑂 = 81.58 + 81.15 = 162.73 𝑘𝑁𝑚
HereValue of 𝑀 𝑂 is positive so moment will Clockwise
(CW) Direction.

Example 3: Find the moment of force of 125 N about point ‘O’ as shown in figure.
Solution:
Given Data,
M0 = -175.24 N.m

Example on Eqivalent Force Couple System
Example 4: An angle is subjected to 30 kN force as shown in figure. Replace this force as an equivalent force couple
system at point A.
Solution:
Given Data,
First Find angle θ,
Tan θ =
3
4
= 0.75
Θ = 𝑇𝑎𝑛−1
0.75 = 36.86.
F Cos θ = 30 Cos 36.86 = 24
F Sin θ = 30 Sin 36.86 = 18
Moment of all forces about ‘A’:
For F Cos θ N Force = (30 Cos 36.86 )(300)
For F Sin θ N Force = (30 Sin 36.86)(200)

Example 4: An angle is subjected to 30 kN force as shown in figure. Replace this force as an equivalent force couple
system at point A.
Solution:
Given Data,
Moment of all forces about ‘A’:
For F Cos θ N Force = (30 Cos 36.86 )(300) = - 7200.93 kNm
For F Sin θ N Force = (30 Sin 36.86)(200) = 3599.17 kNm
𝑀𝐴 = −7200.93 + 3599.17 = −3601.7 𝑘𝑁𝑚
HereValue of 𝑀 𝑂 is Negative so moment will Counter
Clockwise (CCW) Direction.

Example 5: Replace 200 N force by an equivalent force couple system about C for the system shown in figure.
Solution:
Given Data, Moment of all forces about ‘C’:
For F Cos θ N Force = (200 Cos 60 ) = 100 N
For F Sin θ N Force = (200 Sin 60 ) = 173.2 N
𝑀𝐴 = 100 (40) + 173.2 (20) = 7464 𝑁𝑐𝑚
HereValue of 𝑀 𝑂 is Positive so moment willClockwise (CW)
Direction.

Example 6: Replace theforces acting on the rod by an equivalent single resultant force and couple system acting at
point A.
Solution:
Given Data,
⅀V = - 700 – 500 Sin 45 = -1053.55 N
⅀H = -200 – 500 Cos 45 = -553.55 N
Here ⅀V and ⅀H both are negative so R will be in third Quadrant
R = ⅀H2 + ⅀V2 = 1190.12 N
Tan θ =
⅀V
⅀H
=
−1053.55
−553.55
= 1.903
θ = 𝑇𝑎𝑛−1
1.903 = 62.28

Example 6: Replace theforces acting on the rod by an equivalent single resultant force and couple system acting at
point A.
Solution:
Given Data,
Moment about A,
𝑀𝐴 = 700 𝑋 0.5 + 500 𝐶𝑜𝑠 45 0.4 + 500 𝑆𝑖𝑛 45 0.5 + 0.5
= 844.97 Nm
Here Value of 𝑀𝐴 is positive so its CW.

Example on Co-Planar Non-Concurrent Forces
Example 7: Find magnitude and direction of resultant WRT ‘O’ for a system shown in figure.
Solution:
Given Data,
Find Magnitude:
⅀H = - 5 - (20 Cos 45) + 10 = - 9.14 kN
⅀V = (20 Sin 45) + 20 = 34.14 kN
Here value of ⅀H is negative and ⅀V is positive so resultant force will be in
2nd Quadrant.
Magnitude
R = ⅀H 2 + ⅀V 2 = −9.14 2 + 34.14 2 = 35.34 kN
Direction
Tan θ =
⅀V
⅀H
=
34.14
− 9.1 4
= -3.735, So θ = 75°
20 kN
10 kN5 kN
20 kN
O
35.34 kN
2 m
2m

Example 7: Find magnitude and direction of resultant WRT ‘O’ for a system shown in figure.
Solution:
Given Data,
Find position of R WRT ‘O’,
Let ‘R’ be at perpendicular distance ‘X’ from ‘O’,
Taking moment @ O,
- (R.X) – (20 Cos 45) (2) - (20) (2) = 0
- (R.X) = 28.28 + 40 = 68.28 kN.m
- (35.34)(X) = 68.28
X = 1.93 m From ‘O’
20 kN
10 kN5 kN
20 kN
O
35.34 kN
2 m
2m

Example 8: A square ABCD of side 1 m is acted upon by the force system as shown in figure. Find resultant of the
system, magnitude and direction and position WRT ‘A’. ‘E’ is centroid of the square.
Solution:
Given Data,
20 N
10 N
40 N
30 N
A
1 m
1m
B
CD
40 N.m
E
Find Magnitude:
⅀H = 10 – 30 = - 20 N
⅀V = 20 – 40 = - 20 N
Here value of ⅀H & ⅀V are negative so resultant force will be in 3rd
Quadrant.
Magnitude
R = ⅀H 2 + ⅀V 2 = −20 2 + −20 2 = 28.28 N
Direction
Tan θ =
⅀V
⅀H
= 1, So θ = 45°
28.28 N

Example 8: A square ABCD of side 1 m is acted upon by the force system as shown in figure. Find resultant of the
system, magnitude and direction and position WRT ‘A’. ‘E’ is centroid of the square.
Solution:
Given Data,
20 kN
10 kN
40 N
30 N
A
1 m
1m
B
CD
40 N.m
E
28.28 N
Find position of R WRT ‘A’,
Let ‘R’ be at perpendicular distance ‘X’ from ‘A’,
Taking moment @ A,
- (R.X) – (30)(1) - (20) (1) + 40= 0
- (R.X) = 50 - 40 = 10 N.m
- (28.28)(X) = 10
X = 0.354 m From ‘A’

Example 9: Determine resultant of force system shown in figure. The side of each small square is 1 m. The overall
size of body is 6 m X 4 m.
Solution:
Given Data,
First find the angle, θ1, θ2, θ3.
Value of 𝜃1,
Tan 𝜃1 =
2
1
= 2, 𝜃1 = 𝑇𝑎𝑛−1 2 = 63.43°
Value of 𝜃2,
Tan 𝜃2 =
1
1
= 1, 𝜃2 = 𝑇𝑎𝑛−1 1 = 45°
Value of 𝜃3,
Tan 𝜃3 =
1
3
= 0.333, 𝜃3 = 𝑇𝑎𝑛−1
0.333 = 18.43°
𝜃1 = 63.43°
𝜃2 = 45°
𝜃3 = 18.43°

Example 9: Determine resultant of force system shown in figure. The side of each small square is 1 m. The overall
size of body is 6 m X 4 m.
Solution:
Given Data,
𝜃1 = 63.43°
𝜃2 = 45°
𝜃3 = 18.43°
Find Magnitude:
⅀H = 10 + 40 Cos 63.43° - 30 Cos 45° + 50 Cos 18.43° = 54.11 N
⅀V = 20 + 40 Sin 63.43° - 30 Sin 45° - 50 Sin 18.43° = 18.76 N
Here value of ⅀H & ⅀V are Positive so resultant force will be in 1st
Quadrant.
Magnitude
R = ⅀H 2 + ⅀V 2 = 54.11 2 + 18.76 2 = 57.27 N
Direction
Tan θ =
⅀V
⅀H
=
18.76
54.11
, So θ = 19.08°
19.08°
57.27 N

Example 10: Determine the magnitude, Direction and position of resultant force, of the force system given in figure
with reference to point A.
Solution:
Given Data,
Find Magnitude:
⅀H = 25 – 20 Cos 45° + 45 Cos 45° = 42.68 kN
⅀V = 20 + 20 Sin 45° + 45 Sin 45° = 65.96 kN
Here value of ⅀H & ⅀V are Positive so resultant force will be in 1st
Quadrant.
Magnitude
R = ⅀H 2 + ⅀V 2 = 42.68 2 + 65.96 2 = 78.56 kN
Direction
Tan θ =
⅀V
⅀H
=
65.96
42 .68
, So θ = 57.08°

Example 10: Determine the magnitude, Direction and position of resultant force, of the force system given in figure
with reference to point A.
Solution:
Given Data,
Find position of R WRT ‘A’,
Let ‘R’ be at perpendicular distance ‘X’ from ‘A’,
Taking moment @ A,
- (R.X) – (20)(6) + (25) (8) – (20 Sin 45)(2) - (20 Cos 45 )(4) = 0
- (R.X) - 120 + 200 - 28.28 - 56.57 = 0
- (R.X) = -200 +120 + 28.28 +56.57 = 4.85
- (78.56)(X) = 4.85
X = 0.062 m From ‘A’
“45 kN force is acting at 45° and its line of action
pass through ‘A’ so Moment will be zero.”

Example 11: A Lamina is subjected to a system of forces and a couple as shown in figure. Determine the resultant.
Also locate the points of intersection of the resultant with the arm AB, BC and CD of the lamina.
Solution:
Given Data,
Find Magnitude:
⅀H = 50 + 75 Cos 60 = 87.5 N
⅀V = -100 + 75 Sin 60 = -35.05 N
Here value of ⅀H are positive & ⅀V are negative so resultant force will be in
4th Quadrant.
Magnitude
R = ⅀H 2 + ⅀V 2 = 87.5 2 + −35.05 2 = 94.25 N
Direction
Tan θ =
⅀V
⅀H
=
35.05
87.5
, So θ = 21.83°19.08°
94.25 N

Solution:
Given Data, Find position of R WRT ‘B’,
Let ‘R’ be at perpendicular distance ‘d’ from ‘B’,
Taking moment @ B,
(R.d) + (100)(2) - (50) (1) – 75 = 0
(94.25)(d) = -75
d = 0.796 m From ‘B’

Solution:
Given Data, Now Find value of Y and X,
Consider Δ BEC,
Sin α =
𝑑
𝑋
Sin 21.83° =
0.796
𝑋
, So X =
0.796
Sin 21.83°
= 2.14 m
Consider Δ ABC,
Tan α =
𝑌
𝑋
Tan 21.83° =
𝑌
2.14
, SoY = 2.14Tan 21.83° = 0.857 m

Engineering.
malay.badodariya@marwadie
ducation.edu.in

Fundamental of Statics (Part 3)

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