moments couples and force couple systems by ahmad khan

Lecture #02Moments, Couples, and Force

Couple Systems

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.

It is the branch of physical science which deals with the behavior of a body when the body is at rest or in motion.

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.

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).

Force System:

When several forces of different magnitude and direction act upon a body, they constitute a system of forces.

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

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.

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.

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.

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.

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

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

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) MA = Fd

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

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?

APPLICATIONS

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

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

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.

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

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 MO is either clockwise or

counter-clockwise, depending on the tendency for rotation.

In the 2-D case, the magnitude of the moment is Mo = F d

READING QUIZ

1. What is the moment of the 10 N force about point A (MA)?

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

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.

Example #1

in. 12lb 100

in. 1260cosin.24

O

O

Md

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 OM

Example #1

in. 8.20in. lb 1200

in. 8.20in. lb 1200

in. 8.2060sinin. 24

F

FFdM

d

O

b) Horizontal force at A that produces the same moment,

lb 7.57F

Example #1

in. 42in. lb 1200

in. 42in. lb 1200

F

FFdMO

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 50F

Example #1

in. 5cos60

in. 5lb 402

in. lb 1200lb 240in. lb 1200

OB

d

dFdMO

d) To determine the point of application of a 240 lb force to produce the same moment,

in. 10OB

Example #1e) 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.

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 MA using a scalar analysis.

Given: A 20 lb force is applied to the hammer.

Find: The moment of the force at A.

Plan:

xy

GROUP PROBLEM SOLVING (cont.)

Solution:

+ Fy = 20 sin 30° lb

+ Fx = 20 cos 30° lb

xy

+ MA = {–(20 cos 30°)lb (18 in) – (20 sin 20°)lb (5 in)}

= – 351.77 lb·in = 352 lb·in (clockwise)

Moments in 3D4.5 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.

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)

Notation for Moments In simpler terms the Moment of a Force about

the y-axis (My) 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 (Mx, My, and Mz)

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

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 = Fdc

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)

Moment of a CoupleTwo 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.

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

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

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 (Mx, My, and Mz).

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.

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.

b-tech 3rd semester8/12/2016 48

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


(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.’

BC

AO

P2

P1

R

In the above figure, P1 and P2, 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 = P1

2 + P22 + 2P1P2Cos α

Inclination of the resultant w.r.t. the force P1 is given by

= tan-1 [( P2 Sin ) / ( P1 + P2 Cos )]

Derivation:From right angle triangle BCD

BD = Q sinθCD = Q cosθUsing Pythagorus theorem to the ∆OCDOC2 = CD2 + OD2

OC2 = CD2 + (OB + BD) 2

R2 = (P + Q cosθ)2 + (Q sinθ)2

R2 = P2 + Q2 cos2θ + 2PQ cosθ + Q2 sin2θ R2 = P2 + Q2 + 2PQ cosθ R = -----------------------------(1) Angle α of resultant R with force P is given

by, α = tan-1 ----------------------(2)

Particular cases:

When θ = 900 R= When θ = 00 R= P + Q (acting along

Same Direction) When θ = 1800 R= P – Q (acting in

Opposite Direction)

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

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.


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 moves a body from point A to point B.


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


Worked Example 3.2 Find the work done in raising 100 kg of water

through a vertical distance of 3m. Solution


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.


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


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.


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:


Worked example 3.3 What is the potential energy of a 10kg massa) 100m above the surface of the earthb) at the bottom of a vertical mine shaft 1000m

deep. Solution


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.


Formulae for kinetic energy


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.



Centroids
