System-of-Coplanar-Forces

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Engineering Mechanics System of coplanar forces

The Automobile Society (India) Page 1

SYSTEM OF COPLANAR FORCES

INTRODUCTION

The Force is an external effort in the form of push or pull, which

(i) produces or tries to produce motion in a body at rest, or(ii) stops or tries to stop a moving body, or(iii) changes or tries to change the direction of motion of the body.

Force rate of change of momentum rate of change of (mass velocity)

Since mass do not change

Force mass rate of change of velocity

mass acceleration

F ma ...(1.1)= k m a

Where, F is the force, m is the mass and a is the acceleration and k is the constant of proportionality.In all the systems, unit of force is so selected that the constant of the proportionality becomes

unity. For example, in S.I. system, unit of force is Newton, which is defined as the force that is required to move

one kilogram (kg) mass at an acceleration of 1 m/sec2.

One newton = 1 kg mass 1 m/sec2Th us k = 1

F = ma .. .( 1. 2)However in MKS acceleration used is one gravitational acceleration (9.81 m/sec2on earth

surface) and unit of force is defined as kg-wt.

ThusF in kg wt = mg . . . (1 .3 )

Thus 1 kg-wt = 9 .81 newton (1.4)

It may be noted that in usage kg-wt is often called as kg only.

The following examples illustrate the above definition:

When we push a ball lying on the ground, it starts rolling. The force exertedhas thus produces motion in the ball. However, when we push a heavy stone,

it does not move. The effort made in this case has only tried to producemotion, but has not succeeded.

When a piece of stone tied to one end of a string is whirled in a circle, aconstant force has to be exerted by the hand along the string. This is because

the natural tendency of a body is to move along a straight line. Force is spent

in changing the direction of motion of the body from straight line path to the

circular path.


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Characteristics of a force:-

To define force we need four characteristics which are as under-

Magnitude of the force (ex- 10kg or 5N, 15 tonnes) Direction of line along which force is acts (ex- towards E,W,N,S) Nature of the force (ex- Push or Pull) Point at which the force acts.

SYSTEM OF FORCES

When two or more forces act on a body, they are called to for a system of forces.

Following systems of forces are important from the subject point of view:

1. Coplanar forces: The forces, whose lines of action lie on the same plane, are known ascoplanar forces.2. Collinear forces: The forces, whose lines of action lie on the same line, are known ascollinear forces.

3. Concurrent forces: The forces, which meet at one point, are known as concurrent forces.The concurrent forces may or may not b collinear.

4. Coplanar concurrent forces: The forces, which meet at one point and their line of actionalso lay on the same plane, are known as coplanar concurrent forces.

5. Coplanar non-concurrent forces: The forces, which do not meet at one point, but theirlines of action lie on the same, are known as coplanar non-concurrent forces.

6. Non-Coplanar concurrent forces: The forces, which meet at one point, but their lines ofaction do not lie on the same plane, are known as non-coplanar concurrent forces.

7. Non-Coplanar non-concurrent forces:The forces, which do not meet at one point andtheir lines of action do not lie on the same plane, are called non-coplanar non-concurrent

forces.

TYPE OF FORCE SYSTEM POSSIBLE RESULTANT

Concurrent Force

Coplanar, Non-Concurrent Force or a couple

Parallel, Non-Coplanar, Non-Concurrent Force or a couple

Non-parallel, Non-Coplanar, Non-Concurrent Force or a couple


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RESULTANT FORCE

If a number of forces, P,Q,R .etc, are acting simultaneously on a particle, then it ispossible to find out a single force which could replace them i.e. which would produce the same

effect as produces by all the given force. This single force is called resultant force and the given

forces P, Q, R etc are called component forces.

Methods For The Resultant Force

Though there are many methods for finding out the result force of a number of given

forces, yet the following are important from the subject point of view.

1. Analytical method 2. Methods of resolution

Analytical Method For Resultant Force

The resultant force of a given system of force may be found out analytically by the

following methods.

1. Parallelogram law of forces 2. Graphical method

Parallelogram Law of Forces

It state If two forces, acting simultaneously on a particle, be represented in magnitudeand direction by the two adjacent sides of a parallelogram; their resultant may be represented in

magnitude and direction by the diagonal of the parallelogram, which passes through their point

of intersection.

Mathematically, resultant force.

Where P and Q = Forces whose resultant is required to be found out

= Angle between the forces P and Q, and = Angle which the resultant force makes with one for the forces (say P)

Note: If the angle () which the resultant force makes with other force Q,Then


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Cor.

1. If i.e., when the forces act along the same, thenR = P + Q (since )

2. If i.e., when the forces act at right angle, then (since )3. If i.e., when the forces act along the same straight line but in oppositedirection, then

(since )In this case, the resultant force will act in the direction of the greater force.

4. If the two force are equal i.e., when P = Q, then = ( )

Example.1 Two forces of 100 N and 150 N are acting simultaneously at a point. What is the

resultant of these two forces, if the angle between them is 45?

Solution. Given: P = 100N ; Q = 150 N and We know that resultant of the two forces,

()

N

N= 250 N (Ans).

Example 1.2. Two forces act at an angle of 120. The bigger force is of 40 N and the resultant is

perpendicular to the smaller one. Find the smaller force.


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Solution Given: From the geometry figure, we find that

Let Q = Smaller force in N

We know that

()

N(Ans).General Laws for the Resultant Force

The resultant force, of a given system of forces, may also be found out by the following general

laws:

1. Triangle law of force 2. Polygon law of forces.

Triangle Law of Forces

It states, If two forces acting simultaneously on a particle be represented in magnitude and

direction by the two sides of a triangle, taken in order; their resultant may be represented in

magnitude and direction by the third side of the triangle, taken in opposite order.


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Polygon Law of Forces

It is an extension of Triangle Law of Forces for more than two force which states, If a number

of forces acting simultaneously on a particle, be represented in magnitude and direction, by the

sides of a polygon taken in order; then the resultant of all these forces may be represented, in

magnitude and direction, by the closing side of the polygon, taken in opposite order.

Graphical (Vector) Method for the Resultant Force

It is another name for finding out the magnitude and direction of the resultant force by the

Polygon law of forces. It is done as discussed below:

1. Construction of space diagram (position diagram)- It means the construction of adiagram showing the various forces (or loads) along with their magnitude and lines of

action.

2. Use of Bows notations- All the forces in the space diagram are named by using theBows notations. It is a convenient method in which every force (or load) is named bytwo capital letters, placed on its either side in the space diagram.

3. Construction of vector diagram (force diagram)- It means the construction of a diagramstarting from a convenient point and then go on adding all the forces vectorially one by

one (keeping in view the directions of the forces) to some suitable scale.

Now the closing side of the polygon, taken in opposite order, will give the magnitude of the

resultant force (to the scale) and its direction.

Example-1.3. A particle is acted upon by three forces equal to 50 N, 100 N and 130 N, along the

three sides of an equilateral triangle, taken in order. Find graphically the magnitude and directionof the resultant forces.

Solution. The system of given forces is shown in below figure

First of all, name the forces according to Bows notation as shown in fir. The 50 N force is nameas AB, 100 N force as BC and 130 N force as CD, such that DA gives the resultant of these

three forces.


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Now draw the vector diagram for the given system of forces as shown in (b) and as discussed

below.

1. Select some suitable point a and draw ab equal to 50 N to some suitable scale and parallelto the 50 N force of the space diagram.

2. Through b, draw bc equal to 100 N to the scale and parallel to the 100 N force of thespace diagram.

3. Similarly through c, draw cd equal to 130 N to the scale and parallel to the 130 N force tothe space diagram.

4. Join ad, which gives the magnitude as well as direction of the resultant force.5. By measurement, we find the magnitude of the resultant force is equal to 70 N and acting

at an angle of 200 with ab. Ans.


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RESULTANT OF CONCURRENT FORCES

Resultantof a force system is a force or a couple that will have the same effect to the body, both

in translation and rotation, if all the forces are removed and replaced by the resultant.

The equations involving the resultant of force system are the following

1.The x-component of the resultant is equal to the summation of forces in the x-direction.

2.The y-component of the resultant is equal to the summation of forces in the y-direction.

3.The z-component of the resultant is equal to the summation of forces in the z-direction.

Note that according to the type of force system, one or two or three of the equations

above will be used in finding the resultant

CONCURRENT FORCE SYSTEM

Concurrent force system: The resultant of a concurrent forces system can be defined as the

simplest single force which can replace the original system without changing its external effect

on a rigid body

A concurrent force system contain force whose line-of-action meet at some one point.

Force may be Tensile(Pulling) as well as Compressive (Pushing)


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PARALLEL FORCE SYSTEM

Parallel forces can be in the same or in opposite directions. The sign of the direction can be

chosen arbitrarily, meaning, taking one direction as positive makes the opposite directionnegative. The complete definition of the resultant is according to its magnitude, direction, and

line of action.

Non-Concurrent force system: The resultant will not necessarily be a single force but a force

system comprising a force or a couple or a force and a couple. The type of force system as

classified below along with their possible

The resultant of non-concurrent force system is defined according to magnitude, inclination, and

position.The magnitude of the resultant can be found as follows

The inclination from the horizontal is defined by

The position of the resultant can be determined according to the principle of moments.


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Where,

Fx = component of forces in the x-direction

Fy = component of forces in the y-direction

Rx = component of the resultant in x-direction

Ry = component of the resultant in y-direction

R = magnitude of the resultant

x = angle made by a force from the x-axis

MO = moment of forces about any point O

d = moment arm

MR = moment at a point due to resultant force

ix = x-intercept of the resultant R

iy = y-intercept of the resultant R

NON-CONCURRENT NON-PARALLEL

The principles of equilibrium are also used to determine the resultant of non-concurrent, non-

parallel systems of forces. Simply put, all the lines of action of the forces in this system do not

meet at one point. The parallel force system was a special case of this type. Since all of these

forces are not entirely parallel, the position of the resultant can be established using the

graphical or algebraic methods of resolving co-planar forces.

There are a number of ways in which one could resolve the force system that is shown. One

graphical method would be to resolve a pair of forces using the parallelogram or triangle methodinto a resultant. The resultant would then be combined with one of the remaining forces and a

new resultant determined, and so on until all of the forces had been accounted for. This couldprove to be very cumbersome if there is a great number of a force. The algebraic solution to this

system would potentially be simpler if the forces that are applied to the system are easy to break

into components. The algebraic resolution of this force system is illustrated in the example

problem.


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Moment of a Force

Above, we discussed about the effects of forces, acting on a body, through their lines of action or

at the point of their intersection. But inMoment of Force , the effect of these forces, at some

other point, away from the point intersection on their lines of action.

It is the turning effect produced by a force, on the body, on which it acts. The moment of a force

is equal to the product of the force and the perpendicular distance of the point, about which themoment is required and the line of action of the force.

Mathematically, moment

Where Force acting on the body, and

L = Perpendicular distance between the point, about which the moment is

required and the line of action of the force.

Moment of Force about any point

Consider a force P represented, in magnitude and direction, by the

line AB. Let O be a point, about which the moment of this force isrequired to be found out, as shown in fig. From O, draw OC

perpendicular to AB. Join OA and OB.

Now moment of the force P about O

But is equal to twice the area of triangle, ABO.Thus the moment of a force, about any point, is equal to twice the area of the triangle, whosebase is the line representing the force and whose vertex is the point about which the moment is

taken.


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VARIGNONS THEOREM

It states that the moment of a force about any point is equal to the sum of the moments of its

components about the same point. This principle is also known as principle of moments.

Varignons theorem need not be restricted to the case of only two components but appliesequally well to a system of forces and its resultant. For this it can be slightly modified as, the

algebraic sum of the moments of a given system of forces about a point is equal to the momentof their resultant about the same point This principle of moment may be extended to any force

system.

Proof: Referring to the fig Let R be the resultant of forces F1 and F2 and B the moment centre.

Let d, d1 and d2 be the moment arms of the forces, R, F1 and F2 respectively from the moment

centre B. Then in this case, we have to prove that

Rd = F1d1+F2d2

Join AB and consider it as y axis and draw x axis at right angles to it at A (fig-below). Denoting

by the angle that R makes with x axis and nothing that the same angle is formed byperpendicular to R at B with Ab1. We can write;

Rd = R AB = AB (R )= AB Rx

Where Rx denotes the component of R in x direction.

Similarly, if F1x and F2x are the components of F1 and F2 in x direction, respective then and

From eqns (b) and (c)

F1d1 = AB F1 (b)

And

F2d2 = AB F2x (c)

From Eqns (b) and (c)


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F1d2, F2d2 = AB (F1x+F2x)

= AB Rx . (d)

From, equation (a) and (d) we get

Rd = F1d1+F2d2

If a system of forces consists of more than two forces, the above result can be extended as given

below:

Let F1, F2, F3 and F4 be four concurrent forces and R be their resultant. Let d1, d2, d3, d4 and a

be the distance of line of action of force F1, F2, F3 and F4 and R1 respectively from the moment

centre O. (ref fig - 2.7)

If R1 is the resultant of R1 of F1, F2, and its distance O is a1, then applying Varignons theorem.

F1a1 = F1d1+F2d2

If R2 is the resultant of R1and F1, (and hence of F1, F2, F3) and its distance from O is a2, then

applying Varignons theorem.

R2a2 = R1a1+F3d3

=F1d1+F2d2+F3d3

Now considering R2 and F4, we can write

Ra = R2a2 + F4d4

Since R is the resultant of R2 and F4 (i.e. F1, F2, F3 and F4)

Ra = F1d1+F2d1+F3d3+F4d4 . (2.5)Thus, the moment of the resultant of a number of forces about a moment centre is equal to the

sum of the moment of its component forces about the same moment centre.


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Example Find the moment of 100 N force acting at B about point A as shown in fig 2.8

Solution 100 N force may be resolved into its horizontal components as 100 andvertical component 100 . From Varignons theorem moment of 100 N force about thepoint A is equal to sum of the moment of its components about A .

Taking clockwise moment as positive,

Anticlockwise

Example What will be the y intercept of the 5000N force if its moment about A is 8000 N-

m in Fig.2.9

Solution: 5000 N force is shifted to a point B along its line of action (law of transmissibility) and

it is resolved into its x and y components (Fx and Fy as shown in above fig)

And


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By varignons theorem, moment of 5000 N force about A is equal to moment of its component

force about the same point.

Couples

A pair of two equal and unlike parallel forces (i.e. forces equal in magnitude, with lines of action

parallel to each other and acting in opposite directions) is known as a couple.

Arm of a Couple

The Perpendicular distance (a), between the lines of action of the two equal and opposite parallel

forces, is known as arm of the couple asshown in fig.

Moment of a Couple

The moment of a couple is the product of the force (i.e. one of the forces of the two equal andopposite parallel forces) and the arm of the couple.

Mathematically Moment of a couple = P Where P = Magnitude of the force, and

Arm of the coupleClassification of Couples

The couples may be, broadly, classified into the following two categories depending upon their

direction, in which the couple tends to rotate the body on which it acts.


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1. Clockwise couple, 2. Anticlockwise couple

It is the moment of a force, whose effect is to turn or rotate the body, in the same direction in

which hands of a clock move as shown in fig.

2. It is the moment of a force, whose effect is to turn or rotate the body, in the opposite

direction in which the hands of a clock move as shown in the fig.

Note:- The general convention is to take clockwise moment as positive and anticlockwise

moment as negative.

INTRODUCTION TO CENTROID AND CENTRE OF GRAVITY

It has been established, since long that every particle of a body is attracted by the earth towardsits centre. The force of attraction, which is proportional to the mass of the particle, acts vertically

downwards and is known as weight of the body. As the distance between the different particles

of a body and the centre of the earth is the same, therefore these forces may be taken to act along

parallel lines.

Center of Gravity: -It can be defined as the point through which the whole weight of a body

may be assumed to act.

Centroid: - The plane figures (like triangle, quadrilateral, circle etc.) have only areas, but no

mass. The centre of area of such figures is known as centeroid. The method of finding out the

centeroid of a figure is the same as that of finding out the centre of gravity of a body.

In other words the point where the whole area of the figure is assumed to be concentrated.

Moment of Inertia

The moment of inertia (I) is the capacity of a cross-section to resist bending. It is alwaysconsidered with respect to a reference axis and how that cross-sectional area is distributed about

the reference axis, usually a centroidal axis. Also known as the second moment of the area, the

moment of inertia is expressed mathematically as whereA is the area of theplane of the object andy is the distance between the centroid of the object and the x-axis.

Engineers use the moment of inertia to determine the state of stress in a section, calculate the

resistance to buckling, and determine the amount of deflection in a beam.


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Units of Moment of Inertia

The unit of moment of inertia of a plane area depends upon the units of the area and the length.

e.g.

1. If the area is in m2 and the length is also in m, the moment of inertia expressed in m4.2. If area is in cm2 and the length is also in cm, then moment of inertia is expressed in cm4.3. If the area in mm2 and the length is also in mm, then moment of inertia is expressed in

mm4.

Theorem of Moment of Inertia

1. Perpendicular axis2. Parallel axis

Theorem of Perpendicular axis

It states, If Ixx and Iyy be the moment of inertia of a plane section about two perpendicular axismeeting at O, the moment of Inertia Izz about the axis z-z, perpendicular to the plane and passing

through the intersection of X-X & Y-Y is given by the relation:-

Izz = Ixx+Iyy

Theorem of Parallel axis

It states, If the moment of Inertia of a plane are about an axis through its centre of gravity be

denoted by Ig, the moment of Inertia of the area about any other axis AB, parallel to the first, and

at a distance h, from the centre of gravity is given by:-

IAB = Ig+a-h2

where, IAB = M.I of the area about an axis A.B

IG = M.I of the area about its C.G

A = Area of section

H = distance b/w C.G of the section & axis AB.

System-of-Coplanar-Forces

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