moments couples and force couple systems by ahmad khan

ahmadi khan

1. Lecture #02
Moments, Couples, and Force
Couple Systems

2. Applied Mechanics:
Engineering Mechanics/ Mechanics:
:
It is the branch of engineering which
studies the effect of external forces
applied in any manner on a particle or
a body.Mom enm otsm , Cuptm lam oCfle ums
l enp n m Fteptm csCynm Feotm otsm
,stCfel m lam Cm ,lcA m Ftsum otsm
,lcA menm uysrnlysArsepswAyeAph
It is the branch of physical
science which deals with the
behavior of a body when the
body is at rest or in motion.

3. Body:
A body is defined as an object, which cannot retain its shape and size under the action
of a force system.
Rigid body: A rigid body is defined as a body, which can retain its shape and size even
if subjected to external forces.
In practice, there is small deformation of body under the action of a force system. Such
deformation is neglected and the body is treated as rigid body.
Particle: A particle is defined as a very small amount of matter, which may be
assumed to occupy a single point in space.
Practically, any object having very small dimensions as compared to its range of
motion can be called as a Particle.
Eg. Stars, planets, Rockets, Bullets etc.

4. Force:
The external agency, which tends to change the state of a body is known
as force.
A force is completely defined only when the following four characteristics are
specified:
Magnitude
Point of application
Line of action
Direction
A force (F) is a vector quantity which is represented graphically by a
straight line say ‘ab’ whose length is proportional to the magnitude of force
and the arrow shows the direction of force ‘ab’ as shown in Figure above.
Unit of force is Newton (N).

5. Force System:
When several forces of different magnitude
and direction act upon a body, they constitute
a system of forces.

6. Coplanar Force System:
Lines of action of all the forces lie in the same
plane in this system as shown in Fig. (A)
below.
Collinear Force System:
Lines of action of all the forces lie in the same
straight line in this system as shown in Fig.
(B) above.

:
Main types of force systems
are as follows

7. Concurrent Force System:
Lines of action of all the forces meet at a point in this system. The concurrent
forces may not be collinear or coplanar as shown in Fig. (C) above.
Parallel Force System:
Lines of action of all the forces are in parallel as shown in Fig. (D) above.

9. Non- Coplanar Force System:
Lines of action of all the forces does not lie in the same plane as shown in
Fig. (E) above.
Non- Concurrent Force System:
Lines of action of all the forces do not meet at a point in this system as
shown in Fig. (E & F) above.
Non-Parallel Force System:
Lines of action of all the forces are not in parallel as shown in Fig. (H)
above.

11. Others……..
Coplanar Concurrent Force System:
Lines of action of all the forces lie in the same plane and meet at a point shown
in Fig. (G) above.
Coplanar Non-Concurrent Force System:
Lines of action of all the forces lie in the same plane, but do not meet at a a
point as shown in Fig. (A) above. They may be in parallel.
Coplanar parallel Force System:
Lines of action of all the forces are in parallel in the same plane shown in Fig.
(D) above.
Coplanar, non-concurrent, non-parallel Force System:
The lines of action of all the forces are not in parallel, they do not meet at a
point but they are in the same plane as shown in Fig. (A) above.
Non- Coplanar, non-concurrent Force System:
The lines of action of all the forces do not lie in the same plane and do not
meet at a point as shown in Fig. (E) above.

12. Equivalent Forces
·We defined equivalent forces as being forces
with the same magnitude acting in the same
direction and acting along the same line of
action.

13. 4.1 Introduction to Moments
The tendency of a force to rotate a rigid body
about any defined axis is called the Moment
of the force about the axis
The turning effect caused by a force on the
body is called as a moment of force

14. MOMENT OF A FORCE - SCALAR FORMULATION (Section 4.1)
The moment, M, of a force about a point provides a measure of the tendency for
rotation (sometimes called a torque).
M
M = F * d

15. Moment caused by a Force
The Moment of Force (F) about an axis
through Point (A) or for short, the Moment of
F about A, is the product of the magnitude of
the force and the perpendicular distance
between Point (A) and the line of action of
Force (F)
M
A
= Fd

16. Units of a Moment
The units of a Moment are:
N·m in the SI system
ft·lbs or in·lbs in the US Customary system

17. APPLICATIONS
Beams are often used to bridge gaps in walls. We have to
know what the effect of the force on the beam will have on
the beam supports.
What do you think those impacts are at points A and B?

18. APPLICATIONS
Carpenters often use a hammer in this way to pull a stubborn nail.
Through what sort of action does the force F
H
at the handle pull the nail?
How can you mathematically model the effect of force F
H
at point O?

19. Properties of a Moment
Moments not only have
a magnitude, they also
have a sense to them.
The sense of a moment
is clockwise or counter-
clockwise depending
on which way it will
tend to make the object
rotate

20. Properties of a Moment
The sense of a Moment is defined by the
direction it is acting on the Axis and can be
found using Right Hand Rule.

21. Varignon’s Theorem
The moment of a Force about any axis is
equal to the sum of the moments of its
components about that axis
This means that resolving or replacing forces
with their resultant force will not affect the
moment on the object being analyzed

22. MOMENT OF A FORCE - SCALAR FORMULATION (continued)
As shown, d is the perpendicular distance from point O to the line of action of
the force.
In 2-D, the direction of M
O
is either clockwise or
counter-clockwise, depending on the tendency for rotation.
In the 2-D case, the magnitude of the moment is M
o
= F d

23. READING QUIZ
1. What is the moment of the 10 N force about point A
(M
A
)?
A) 3 N·m B) 36 N·m C) 12 N·m
D) (12/3) N·m E) 7 N·m
• A
d = 3 m
F = 12 N

24. Example #1
A 100-lb vertical force is applied to
the end of a lever which is attached
to a shaft at O.
Determine:
a)Moment about O,
b)Horizontal force at A which
creates the same moment,
c)Smallest force at A which
produces the same moment,
d)Location for a 240-lb vertical
force to produce the same
moment,
e)Whether any of the forces from
b, c, and d is equivalent to the
original force.

25. Example #1
( )
( )( )in. 12lb 100
in. 1260cosin.24
=
=°=
=
O
O
M
d
FdM
a) Moment about O is equal to the product of the
force and the perpendicular distance between the
line of action of the force and O. Since the force
tends to rotate the lever clockwise, the moment
vector is into the plane of the paper.
in lb 1200×=
O
M

26. Example #1
( )
( )
in. 8.20
in. lb 1200
in. 8.20in. lb 1200
in. 8.2060sinin. 24
×
=
=×
=
=°=
F
F
FdM
d
O
b) Horizontal force at A that produces the same
moment,
lb 7.57=F

27. Example #1
( )
in. 42
in. lb 1200
in. 42in. lb 1200
×
=
=×
=
F
F
FdM
O
c) The smallest force at A to produce the same
moment occurs when the perpendicular distance is
a maximum or when F is perpendicular to OA.
lb 50=F

28. Example #1
( )
in. 5cos60
in. 5
lb 402
in. lb 1200
lb 240in. lb 1200
=°
=
×
=
=×
=
OB
d
d
FdM
O
d) To determine the point of application of a 240 lb
force to produce the same moment,
in. 10=OB

29. Example #1
e)Although each of the forces in parts b), c), and d)
produces the same moment as the 100 lb force, none
are of the same magnitude and sense, or on the same
line of action. None of the forces is equivalent to the
100 lb force.

33. GROUP PROBLEM SOLVING
Since this is a 2-D problem:
1) Resolve the 20 lb force along the
handle’s x and y axes.
2) Determine M
A
using a scalar
analysis.
Given: A 20 lb force is applied to the
hammer.
Find: The moment of the force at A.
Plan:
x
y

34. GROUP PROBLEM SOLVING (cont.)
Solution:
+ ­ F
y
= 20 sin 30° lb
+ ® F
x
= 20 cos 30° lb
x
y
+ M
A
= {–(20 cos 30°)lb (18 in) – (20 sin 20°)lb (5 in)}
= – 351.77 lb·in = 352 lb·in (clockwise)

37. Moments in 3D
4.5 Moment of a Force about a
Specific Axis
In 2D bodies the moment is due to a force
contained in the plane of action perpendicular
to the axis it is acting around. This makes the
analysis very easy.
In 3D situations, this is very seldom found to
be the case.

38. Moments in 3D
The moment about an axis is still calculated
the same way (by a force in the plane
perpendicular to the axis) but most forces are
acting in abstract angles.
By resolving the abstract force into its
rectangular components (or into its
components perpendicular to the axis of
concern) the moment about the axis can then
be found the same way it was found in 2D –
M = Fd (where d is the distance between the
force and the axis of concern)

39. Notation for Moments
In simpler terms the Moment of a Force about
the y-axis (M
y
) can be found by using the
projection of the Force on the x-z Plane
The Notation used to denote Moments about
the Cartesian Axes are (M
x
, M
y
, and M
z
)

41. Force Couples
A Couple is defined as two Forces having the
same magnitude, parallel lines of action, and
opposite sense
In this situation, the sum of the forces in each
direction is zero, so a couple does not affect
the sum of forces equations
A force couple will however tend to rotate the
body it is acting on

42. Moment Due to a Force Couple
By multiplying the magnitude of one Force by the
distance between the Forces in the Couple, the
moment due to the couple can be calculated.
M = Fd
c
The couple will create a moment around an axis
perpendicular to the plane that the couple falls in.
Pay attention to the sense of the Moment (Right
Hand Rule)

43. Moment of a Couple
Two couples will have equal moments if
•
2211
dFdF=
•the two couples lie in parallel planes, and
•the two couples have the same sense or
the tendency to cause rotation in the same
direction.

44. Why do we use Force Couples?
The reason we use Force Couples to analyze
Moments is that the location of the axis the
Moment is calculated about does not matter
The Moment of a Couple is constant over the
entire body it is acting on

45. Couples are Free Vectors
The point of action of a Couple does not
matter
The plane that the Couple is acting in does
not matter
All that matters is the orientation of the plane
the Couple is acting in
Therefore, a Force Couple is said to be a
Free Vector and can be applied at any point
on the body it is acting

46. Resolution of Vectors
The Moment due to the Force Couple is
normally placed at the Cartesian Coordinate
Origin and resolved into its x, y, and z
components (M
x
, M
y
, and M
z
).

47. Vector Addition of Couples
By applying Varignon’s Theorem to the Forces in
the Couple, it can be proven that couples can be
added and resolved as Vectors.

48. rkRzBbT,zk:LDBb,fhu,
()B‘R,b’b(BMb FORCES, COUPLES AND
THEIR SYSTEMS
Forces
A force is defined as an action of one body
on another. It is a vector quantity because
its effect depends on the direction as well as
on the magnitude of the action. Because
force is a vector quantity, forces may be
combined according to the parallelogram law
of vector addition. Consider the figure below,
the force vector is F, and magnitude F.
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49. Therefore, the complete specification of the
action of a force must indicate its
magnitude, direction, and point of application
and it must be treated as a fixed vector.
Force is a vector quantity, it is important for
forces to obey the parallelogram law of
combination
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50. (1)Parallelogram law of forces : It is stated as
follows : ‘If two forces acting at a point are
represented in magnitude and direction by the
two adjacent sides of a parallelogram, then
the resultant of these two forces is
represented in magnitude and direction by the
diagonal of the parallelogram passing through
the same point.’

51. B
C
A
O
P
2
P
1
R
a
q
In the above figure, P
1
and P
2
, represented by the sides OA and OB have R as their
resultant represented by the diagonal OC of the parallelogram OACB.
It can be shown that the magnitude of the resultant is given by:
R = ÖP
1
2
+ P
2
2
+ 2P
1
P
2
Cos α
Inclination of the resultant w.r.t. the force
P
1
is given by
q = tan
-1
[( P
2
Sin a) / ( P
1
+ P
2
Cos a )]

52. Derivation:
From right angle triangle BCD
BD = Q sinθ
CD = Q cosθ
Using Pythagorus theorem to the ∆OCD
OC
2
= CD
2
+ OD
2

53. OC
2
= CD
2
+ (OB + BD)
2
R
2
= (P + Q cosθ)
2
+ (Q sinθ)
2
R
2
= P
2
+ Q
2
cos
2
θ + 2PQ cosθ + Q
2
sin
2
θ
R
2
= P
2
+ Q
2
+ 2PQ cosθ
R = -----------------------------(1)
Angle α of resultant R with force P is given
by,
 α = tan
-1
----------------------(2)

54. Particular cases:
When θ = 90
0
R=
When θ = 0
0
R= P + Q (acting along
Same Direction)
When θ = 180
0
R= P – Q (acting in
Opposite Direction)

56. In the case of this radio tower, if you know the forces in the three
cables, how would you determine the resultant force acting at D, the
top of the tower?
Assignment

58. Couple
When two equal unlike parallel forces are
brought together no resultant can be formed
instead further pairs of equal unlike parallel
forces will be produced, each pair being
equivalent to the original set. Such pairs of
forces therefore have no resultant. A
force set of this kind is termed a couple.
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59. Momentos,n Cnunposlausandomp, Work done by a constant force

When the point at which a force acts moves, the force is
said to have done work.
When the force is constant, the work done is defined as
the product of the force and distance moved.

Consider the example in Figure, a force F
acting at the angle q moves a body from point
A to point B.
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60. Worked Example 3.1
How much work is done when a force of 5 kN
moves its point of application 600mm in the
direction of the force.
Solution
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61. Worked Example 3.2
Find the work done in raising 100 kg of water
through a vertical distance of 3m.
Solution
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62. Energy
A body which has the capacity to do work is
said to possess energy.
For example , water in a reservoir is said to
possesses energy as it could be used to drive
a turbine lower down the valley. There are
many forms of energy e.g. electrical,
chemical heat, nuclear, mechanical etc.
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63. The SI units are the same as those for work,
Joules J.
In this module only purely mechanical energy
will be considered. This may be of two
kinds, potential and kinetic
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64. Potential Energy
There are different forms of potential energy two
examples are:
 i) a pile driver raised ready to fall on to its target
possesses gravitational potential energy while (ii) a coiled
spring which is compressed possesses an internal
potential energy.Only gravitational potential energy will
be considered here. It may be described as energy due to
position relative to a standard position (normally chosen
to be he earth's surface.)
The potential energy of a body may be defined as the
amount of work it would do if it were to move from the
its current position to the standard position.
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65. Formulae for gravitational
potential energy
A body is at rest on the earth's surface. It is then
raised a vertical distance h above the surface. The
work required to do this is the force required times
the distance h.
Since the force required is it's weight, and weight, W
= mg, then the work required is mgh.
The body now possesses this amount of energy -
stored as potential energy - it has the capacity to do
this amount of work, and would do so if allowed to
fall to earth.
Potential energy is thus given by:
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66. Worked example 3.3
What is the potential energy of a 10kg mass
a)100m above the surface of the earth
b)at the bottom of a vertical mine shaft 1000m
deep.
Solution
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67. Kinetic energy
Kinetic energy may be described as energy due to
motion.
The kinetic energy of a body may be defined as the
amount of work it can do before being brought to
rest.
For example when a hammer is used to knock in a
nail, work is done on the nail by the hammer and
hence the hammer must have possessed energy.
Only linear motion will be considered here.


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68. Formulae for kinetic energy
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69. MadFracSyASbrifwFhzSMadFryw.cSid.S
FgawrSivvjwMiFwydc CENTRES OF GRAVITY, CENTROIDS AND
THEIR APPLICATIONS
Centre of Gravity
We have assumed all along that the attraction
exerted by the earth on a rigid body could be
represented by a single force W. This force is
called the force of gravity. In actual sense, the
earth exerts a force on each particles forming
the body. The action of the earth on a rigid body
can thus be represented by a large number of
small forces distributed over the entire body. All
these small forces can be replaced by a single
equivalent force, W, that is the resultant.
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71. MtC xouIP Centroids
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