civil & engg. engg. mechanics mechanical unit i · 2. two concurrent coplanar forces are acting...

Civil & Engg. Engg. Mechanics Mechanical Unit I

Prof. A.S. Rao RSREC

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UNIT 1: BASIC CONCEPTS 1. INTRODUCTION : Solids Rigid solids (Engg Mechanics deals with them) Deformable solids (Solid Mechanics deals with them) Matter (Substance) Fluids Liquids (Fluid Mechanics deals with them) Gasses (Thermodynamics deals with them) Engineering Mechanics is that branch of science which deals with the equilibrium (physical state) of rigid bodies which are at rest or in motion acted upon by forces. The part of E.M that deals with bodies at rest is known as “STATICS” and that part which deals with moving bodies is known as “DYNAMICS”. Therefore, Engg. Mechanics Statics is divided into Dynamics In the above statement, there are 3 new terms which are: rigid body, force and equilibrium; that are explained below. 1.1 Rigid body: The body that does not deform (no change in size and shape) or the distance between any two points of the body does not change under the action of applied forces.

Take a bar that is subjected to two loads/ forces at its ends. If it is a rigid bar, it will not bend and it remains horizontal as shown. No change in shape & size.

If the bar is not rigid, it will deform as shown in the fig after the forces or loads are applied. The distance b/w A & B changes. Shape also changes.

A

B

B

A



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1.2 Force: Force is an action which tends to change the state of rest or of uniform motion of a body. (An action which always tries to disturb the body). Force is a vector quantity. The S.I unit of force is ‘N’ (Newtons) To completely define a force, the following three are essential. 1. Magnitude of the force(P) 2. Direction of force (θ) 3.It’s point of application(O) ‘O’ P θ Q θ ‘O’ It is graphically represented by a single headed arrow. The direction of a force is the direction, along a straight line through its point of application in which the force tends to move a body to which it is applied. This line is called the line of action of the force. The arrow represents the line of action of the force and the head gives the direction. The point of application may be shown either at head or tail of the arrow. The length of the arrow represents the magnitude of the force to some scale. 1.3 Equilibrium: It is nothing but a state of balance between two opposing groups of forces. W W R The above diagram shows a ball resting on the ground: The weight of the ball ‘W’ is the action on the ground and ‘R’ is the Reaction from the ground. Here W & R are the two opposing groups, as the weight ‘W’ is trying to move the ball downwards whereas, the reaction ‘R’ is opposing or resisting the motion of the ball. But, from Newton’s third law, action and reaction are equal in magnitude and hence, the ball is in equilibrium.



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PART 1 - STATICS

2. FORCE SYSTEM or SYSTEM OF FORCES: A force system is a collection of forces acting at specified locations. It is a term used to describe a group of forces. (OR) When several forces of various magnitudes and directions act upon a body, they are said to constitute a system of forces. 3. CLASSIFICATION or TYPES OF FORCE SYSTEMS :

1. Coplanar forces: If the forces acting on a body are lying in a single plane, then they are said to be coplanar forces.

2. Concurrent forces: All the forces acting on a body, if either converge at or

diverge from a particular point, the forces are said to be concurrent forces. (OR) If the lines of action of all forces pass through a common point, then such forces are said to be concurrent.

Q4

Q3 Q2

Q1 P4

P3

P2

P1

P1, P2, P3 & P4 are coplanar. Q1, Q2, Q3 & Q4 are coplanar. P1, P2, Q2 & Q4 are not coplanar.

P

Q

S

T P, Q, S & T are concurrent forces.



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3. Collinear forces: If all the forces have lines of action lying on a particular line, then such forces are said to be collinear. (OR) If lines of action of all forces coincide, then these forces are said to be collinear.

In combination, we can find 1)coplanar, concurrent forces; 2)non-coplanar, concurrent forces; 3)coplanar, non-concurrent forces and 4)non-coplanar, non-concurrent forces.

4. PRINCIPLES OF STATICS The general problem of statics consists of finding the conditions that such a system must satisfy in order to have equilibrium of the body. The various methods of solution of this problem are based on several axioms, called the “Principles of Statics”. They are:

1. Parallelogram Law / Triangle Law 2. Equilibrium Law 3. Law of Superposition 4. Law of Transmissibility 5. Law of Action and Reaction

1 A).Parallelogram Law: If two forces acting at a point are represented in magnitude and direction by the adjacent sides of a parallelogram, then the diagonal passing through their point of intersection represents the resultant in both magnitude and direction. 1 B). Triangle Law: If two forces acting at a point are represented by the two sides of a triangle taken in order, then their sum or resultant is represented by the third side taken in an opposite order. P Q R Q R Q P R

β θ θ α P

P



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The Resultant force is given by R = √P2+Q2+2PQ.cos θ

And also tanα = Q.sin θ / (P + Q.cos θ ) tanβ = P.sin θ / (Q + P.cos θ)

Where, ‘α’ is angle between P & R and ‘β’ is angle between Q & R, so that θ =α+β Note: For different values of ‘θ’, ‘R’ value can be found out.

2.Equilibrium Law: Two forces can be in equilibrium only if they are equal in magnitude, opposite in direction, and collinear in action.

From the principle of the parallelogram of forces, it follows that two forces applied at one point can always be replaced by their resultant which is equivalent to them. Thus, we conclude that two concurrent forces can be in equilibrium only if their resultant is zero. P P Q Q

3.Law of Superposition: The action of a given system of forces on a rigid body will in no way be changed if we add to or subtract from them another system of forces in equilibrium. T S P Q = P Q T S 4.Law of Transmissibility: The point of application of a force may be transmitted along its line of action without changing the effect of the force on any rigid body to which it may be applied.

P Q ≡ Q P ≡ Q P

The Forces T and T are removed and S & S are added which are in equilibrium.



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5.Law of Action and Reaction: Any pressure on a support causes an equal and opposite pressure from the support so that action and reaction are two equal and opposite forces. This last principle of statics is nothing but Newton’s third law.

5. Free-Body Diagram (FBD): The sketch in which the body is completely isolated from its supports and in which all forces (both actions and reactions) acting on it are shown by vectors is called a Free-body Diagram of that body. The given or applied forces are called as ‘active forces’ including weight of the body, and the resistance offered by supports are called as ‘reactive forces’. To have equilibrium of the body, it is necessary that the active forces and reactive forces together represent a system of forces in equilibrium. Consider a ball hanging from a wall as shown below.

The ball exerts a down ward pull on the string BC and also pushes to the left against the wall at A. These actions of the constrained ball against its supports induce reactions form the supports on the ball.

A

B Reaction from string

Reaction from wall onto ball

Action of ball on wall

Action of ball on string



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Problems (on Parallelogram law):

1. Two concurrent coplanar forces of magnitude 120N, 100N are acting at an angle 600 on a rigid body. Find the magnitude of their resultant and also its direction.

2. Two concurrent coplanar forces are acting on a rigid body. The bigger force is 200N. The angle b/w the two forces is 1200. Calculate the magnitude of the smaller force.

3. Two concurrent coplanar forces are acting on a rigid body. If the angle b/w the two forces is 900, the resultant R comes out to be √40 N. Else if the angle b/w these two forces is 600, the resultant R comes out to be √52 N. Calculate the magnitude of these two forces.

4. Two forces P & Q are acting at a point ‘O’. The resultant force ‘R’ is 400N and angles ‘α’ & ‘ β’ are 350 and 250 respectively. Find the forces P and Q.

6. COMPOSITION OF FORCES: The reduction of a given system of forces to the simplest system (Resultant) that will be its equivalent is called the problem of composition of forces. If a rigid body is acted upon by several (more than two) coplanar concurrent forces, the resultant of the given forces can be found out by constructing polygon of forces. Polygon law which is equivalent to the repeated application of parallelogram law can be applied to determine the resultant of a number of concurrent coplanar forces. 6.1 Law of Polygon of forces: “ If a number of coplanar forces are acting at a point such that they can be represented in magnitude and direction by the sides of a polygon taken in

A

B Reaction from string

Reaction from wall onto ball

Action of ball on wall

W W

F.B.D of ball F.B.D of wall A ball of weight ‘W’ leaning against a wall with the help of a string.

Action of string on wall

A force (action) towards left is exerted by the ball on the wall and at the same time wall exerts an equal & opp. reaction on the ball towards right.



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an order, their resultant is represented in both magnitude and direction by the closing side of the polygon taken in the opposite order” . We may say that the resultant of any system of concurrent forces in a plane is obtained as the geometric sum of the given forces. Note: 1) The resultant R does not depend upon the order in which the forces are chosen to draw the polygon. 2) If the polygon turns to be a closed polygon, then the resultant is zero and the given system of forces is in equilibrium. (If the end of the last vector coincides with the beginning of the first, the resultant R is equal to zero and the given system of forces is in equilibrium.) Problem: 7. RESOLUTION OF A FORCE: “ The replacement of a single force by several components which will be equivalent in action to the given force is called the problem of resolution of a force” . 7.1 Principle of Resolution: “The algebraic sum of the resolved parts of a number of forces, in a given direction, is equal to the resolved part of their resultant in the same direction”.

By using the parallelogram law, we can resolve a given force R into any two components P and Q intersecting at a point on its line of action as shown below.

Problems:

1. Resolve the force R (=100N) into two directions which are given by the angles α=300 & β=450.

2. In the above problem if R=100N, P=73.25N & α=300, find the values of Q & β.

Case1: The directions of both components (α, β) are given; their magnitudes (P & Q) can be determined. Case2: Both the magnitude and direction of one component (P & α) are given; the magnitude and direction of the other (Q & β) can be determined. Note: Formulae in parallelogram law can be used.

α

Q

R

P

β



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7.2 Resolution of a force into rectangular components: Often it is required to resolve a given force into components which are perpendicular to each other. Such components are called rectangular components.

8. EQUILIBRIUM OF CONCURRENT COPLANAR FORCES : If a body known to be in equilibrium is acted upon by several concurrent, coplanar forces, then these forces, or rather their free vectors, when geometrically added must form a closed polygon. That is, when the resultant of all the forces acting on a body is zero, the body is said to be in equilibrium.

Problem: P) Five strings are tied at a point and are pulled in all directions, equally spaced (720 each) from one another. If the magnitude of pulls on 3 consecutive strings is 50N, 70N & 60N respectively, find the magnitude of pull on the remaining two strings. (Hint: Equations of Equilibrium OR Force Polygon (graphical method) may be used)

Let force F is to be resolved into two rectangular components along x and y axes. If ‘θ’ is the angle between the force F and the x-axis, then from trigonometry,

Fx = F.Cos θ Fy = F. Sin θ

F = √Fx

2+Fy2 and tan θ = Fy/ Fx

F

Fx

Fy

θ

y

x

For the resultant R to be zero, it’s each of the two rectangular components Rx and Ry must be separately equal to zero. If R = √Rx

2+Ry2 = 0

Then, Rx = 0 & Ry =0 Therefore, Rx = F1x + F2x + F3x + F4x = 0

Ry = F1y + F2y + F3y + F4y = 0 That is, ΣFx = 0 & ΣFy = 0 The above equations are called equations of equilibrium.

F1 F2

F4 F3



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9. EQUILIBRIUM OF THREE COPLANAR FORCES : 9.1 Theorem of Three Forces: “Three nonparallel forces can be in equilibrium only when they lie in one plane, intersecting in one point, and their free vectors build a closed triangle”. This statement is called theorem of three forces. 9.2 Lami’s Theorem: “If three coplanar concurrent forces are in equilibrium, then ratio of each force and the sine of the included angle between the other two forces are constant”. Lami’s theorem can be proved by considering a force system which consists of three concurrent coplanar forces F1, F2 & F3 as shown in the following figure. By constructing ‘triangle of forces’ and applying the ‘sine’ rule, the Lami’s theorem can be proved.

Q

P

60N 70N

50N

720



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Problems:

1. A traffic signal of mass 50kg is hung with the help of two strings as shown in fig. Find the tension developed in both the strings.

F1

F2

F3

β

γ

α

Therefore, as per Lami’s Theorem, F1 = F2 = F3 Sin α sin β sin γ

Triangle of forces

Given system of forces

γ

α

β

For the given system of forces, triangle of forces can be constructed as shown below.

F3

F2

F1

From property of triangle using Sine rule, F1/sin(180- α) = F2/sin(180- β) = F3/sin(180- γ) From which, F1/sin(α) = F2/sin(β) = F3/sin(γ)

300

A B

C 450

Road



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2. In the four bar mechanism ABCD shown in figure, determine the force ‘P’ required for equilibrium.

10. MOMENT OF A FORCE : The effectiveness or importance of a force, as regards its tendency to produce rotation of a body about a fixed point, is called the moment of the force with respect to that point. And this moment can be measured by the product of the magnitude of the force and the (perpendicular) distance from the point to the line of action of the force.

The point O is called the moment centre and the distance ‘d’ is called the arm of the force. Moment about point ‘O’ is given by Mo = F * d The unit of moment is ‘N-m’ (Newton-meter)

The force F1 has a tendency to produce an anti-clockwise moment about the moment centre ‘O’ and the force F2 has a tendency to produce a clockwise moment. Total moment about ‘O’ is, Mo = F2*d2 - F1*d1

Sign convention: Clock wise moments are considered as +ve.

O

F d

F1 F2

O

d1 d2

B C

P 380N

D A

500 450 450 750



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Problems:

1. Find the moment of the 100N force about hinge of the sluice gate shown in following figure.

2. Find the moment of 50N force acting at B about point A as shown in following

figure.

10.1 Theorem of Varignon: The moment of the resultant of two concurrent forces with respect to a centre in their plane is equal to the algebraic sum of the moments of the components with respect to the same center. Proof: Consider a force F acting at a point ‘A’ and having components F1 & F2 in any two directions as shown below.

Q

P

d2d1

A

MA = - (P.d1 + Q.d2)

d

Hinge

2m

600

100N

A

600

50N

0.3m

0.5m

B



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we get, F.d = F1.d1 + F2.d2 …. (2.13)

Hence, the moment of a force about an axis is equal to the sum of the moments of its components about the same axis.



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Problems: P) Find the magnitude, direction and position of the Resultant force of the following system.

11. PARALLEL FORCES IN A PLANE : 11.1 Definition: A set of forces whose lines of action are parallel to each other are called parallel forces. As parallel forces are not concurrent, parallelogram law cannot be applied. 11.2 Types of parallel forces: 1)Like Parallel forces: When the two parallel forces act in the same direction, they are called as like parallel forces. These forces can be equal or unequal in magnitude.

Resultant force, R = F1 + F2

2)Unlike Unequal Parallel forces: when the two parallel forces act in the opposite directions and are unequal in magnitude.

Resultant force, R = F1 - F2

3)Unlike Equal Parallel forces: When the two parallel forces act in opposite directions and are equal in magnitude. Resultant force, R = F1 – F1 = 0

6kN

9kN

5kN

12kN

8kN 7kN

1m

2m

2m



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12. COUPLE: Definition : A system of two equal parallel forces acting in opposite directions (Unlike

Equal Parallel forces) cannot be replaced by a single force. In such a case, the two forces

form a ‘couple’ which has a tendency to rotate the body. The distance (d) between the

lines of action of these two forces is termed as ‘arm’ of the couple.

Moment of a couple: The rotational tendency of a couple is measured by its moment. The moment of a couple is the product of the either one of the forces forming the couple and arm of the couple. From the above fig., the moment of the couple, M = F * d Sense of a Couple Clockwise couple (+ve) Anti-clockwise couple (-ve) Note: Two couples acting in a plane can be in equilibrium if their moments are equal in magnitude and opposite in direction.

1)Like parallel forces 2)Unlike unequal parallel 3)Unlike equal parallel

F

F

d

P

Q Q P

The forces ‘P’ are forming a clock wise couple of magnitude ‘P*d’ & the forces ‘Q’ are forming an anti- clock wise couple of magnitude ‘Q*b’. These two couples will be in equilibrium if P.d = Q.b

d

b

P

Q

P

Q



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Characteristics of a Couple:

1. The algebraic sum of the forces, constituting the couple is zero. 2. The algebraic sum of the moments of the forces, constituting the couple, about any

point is the same, and equal to the moment of the couple itself. 3. A couple cannot be balanced by a single force. But it can be balanced by a couple

of opposite sense of same magnitude. 4. Any no. of coplanar couples can be reduced to a single couple, whose magnitude

will be equal to the algebraic sum of the moments of all the couples. 5. The translatory effect of a couple on the body is zero.

Problems:

1. P) A square ABCD has sides equal to 200mm. Forces of 150N each act along AB & CD and 250N each act along CB & AD. Find the moment of the couple that keeps the system in equilibrium.

RESOLUTION OF A FORCE INTO A FORCE AND A COUPLE : (Replacement of a Force by an Equivalent ‘Force –Couple’ system) It will be advantageous to resolve a force acting at a point on a body into a force acting at some other suitable point on the body and a couple as shown in the following figure.

P

P

P P

P

M=P.d

d

Let a force ‘P’ is acting on a bar as shown.

Apply two equal and opposite forces ‘P’ at a point (whose distance is ‘d’ from the previous point) to where the force is to be shifted as shown.

Now the unlike equal parallel forces ‘P’ form an anti-clockwise couple ‘M’ whose moment value is P*d.

Thus a force ‘P’ can be shifted from one point to the other by the same force ‘P’ and a couple ‘M’ (=P*d).



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Moment Vs. Couple: A moment is developed when a force is acting at a distance from the specified point which is nothing but the rotating capacity of that force about the specified point, where as a couple is formed due to two unlike equal parallel forces acting on a rigid body. Moment of a force = (force)X( perpendicular distance of that force from the specified point) Moment of a couple = (force forming couple)X(arm of the couple) EQUILIBRIUM OF A COPLANAR FORCE SYSTEM Equilibrium of a system of Concurrent Coplanar forces: The resultant R of a system of concurrent forces is zero only when the following two conditions are satisfied. ∑Fx = 0 (i.e., Rx = 0) ------- (1) & ∑ Fy = 0 (i.e., Ry = 0) -------(2) The algebraic sum of the components of the forces acting in each of the two mutually perpendicular (x & y) directions, is zero. Equilibrium of a system of non-concurrent coplanar forces: A body is said to be in equilibrium when it does not have any translator or rotary motion in any direction. This means when the body is in equilibrium under the action of coplanar forces, the following simultaneous conditions are to be satisfied.

1. The algebraic sum of the components of forces along each of the two mutually perpendicular (x & y) directions, is zero.

∑Fx = 0 (i.e., Rx = 0) --------(1) & ∑ Fy = 0 (i.e., Ry = 0) --------(2)

2. The algebraic sum of the moments of all forces acting on the body about the third perpendicular (z) direction is zero.

∑Mz = 0 --------(3)



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SUPPORTS: Supports offer resistance against movement or rotation or both. Types of Supports:

1) Frictionless support: The reaction acts normal to the surface at the point of contact as shown in Fig.

2) Roller and Knife Edge (Simple) Supports: They are common type of constraints and always exert their reaction normal to the surface on which they rest.

3) Hinged Support: A Hinge provides resistance against any type of movement

(ie., in both the directions) and hence offers two reactions; one in horizontal direction and the other in vertical direction (both are of two way type). These two reactions can be combined into a single reaction. It allows rotation freely.

a)Sphere resting on a horizontal plane b)Rod resting inside a sphere

Beam supported on a roller and a knife edge support



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4) Built-in or Fixed Support: This support provides resistance against both movement (horizontal & vertical) and rotation. Hence, it offers total three reactions namely one horizontal reaction, one vertical reaction and a reactive couple (moment).

BEAMS: Definition of a beam: Beam is a structural member which generally carries transverse loads. Transverse loads are those which act perpendicular to the axis of the member. Generally beam is shown/represented by a single line (its axis).

Types of loading on the beams:

1) Point or Concentrated Loads: Loads which are assumed to ‘act’ or ‘be concentrated’ at a point. They are distributed over an infinitesimally small area and hence treated as point or concentrated loads.

2) Distributed load: The load which acts over a length or area or volume of a

body.

Beam simply bends without any rotation at fixed support.

Axis

Beam with loads



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A) Uniformly distributed load (udl): If the intensity of the distributed load is constant through out the length (or area or volume), it is called as ‘uniformly distributed load’. it may be called as rectangular loading. Total load = intensity of load * length of load The total load is assumed to act at it’s mid length.

B) Uniformly varying load (uvl): If the intensity of the distributed load is not constant, but uniformly varies along the length, it is called as ‘uniformly varying load’. It may be called as Triangular loading. Total load = ½ *intensity of load * length of load. The total load is assumed to ‘act’ or ‘be concentrated’ at 2/3rd of the length of load from zero intensity end.

The units of ‘intensity of loading’ if it is distributed over a length: ‘N/m’ The units of ‘intensity of loading’ if it is distributed over an area : ‘N/m2’ It is also called as ‘pressure’. Ex: Pressure exerted by water or soil etc. The units of ‘intensity of loading’ if it is distributed over a volume: ‘N/m3’

It is nothing but the weight of a body which is distributed over entire volume of the body.

L L

2L/3 L/2

Beam with a uvl Beam with a udl

w N/m

Intensity of loading

Length of loading (m)



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Types of beams:

1) Simply Supported (S.S) beam: A beam supported over two simple supports like knife-edge or roller or hinge supports. Generally a S.S beam is supported by a roller at one end and a hinge at the other end.

2) Fixed beam: A beam with two ends fixed is called as a fixed beam.

3) Cantilever: A beam with one end fixed and the other end free (with out any

support) is called as a cantilever.

4) Propped cantilever: A beam with one end fixed and the other end simply supported i.e., a cantilever with a prop.

5) Continuous beam: A beam with ‘more than two supports’ or ‘more than one span’.

Propped Cantilever Continuous beam

Fixed beam Cantilever

lengthspan



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Span of the beam: The centre to centre (c/c) distance between the two adjacent supports of the beam is known as ‘span’. However, in case of cantilevers, length (instead of span) is specified. In the above example, length of the cantilever = l.

QUIZ QUESTIONS 1. The two essential properties of force are { c }

a)magnitude and sense b)sense and direction c)magnitude and direction d)none of these 2. Collinear forces are { b } a)concurrent b)coplanar c)concurrent and coplanar d)none 3. According to law of triangle of forces { c } a)three concurrent forces will be in equilibrium b)three concurrent forces can be represented by a triangle , each side being

proportional to the force c)if three forces acting upon a particle are represented in magnitude and

direction by the side of a triangle taken in order, they will be in equilibrium. d)if three concurrent forces are in equilibrium, each force is proportional to

the sine of the angle between the other two. 4. According to the principle of transmissibility of forces, the effect of a force upon a body is { b } a)maximum when it acts at the centre of gravity of a body b)same at every point in its line of action c)minimum when it acts at the centre of gravity of a body d)different at different at different points in its line of action 5. The simplest possible resultant of a coplanar force system is { c } a)a single moment b)a single force c)a single force or a single moment d)none of these

Span1 span2 span



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6. The moment of a force about a point is { a } a)a vector quantity b)a scalar quantity c)zero d)none of these 7. The moment of a couple is { a } a)a free vector b)a bound vector c)zero d)none 8. Reactive components of a ‘roller’ supported on a horizontal plane { a } a)only vertical force b)only horizontal force c)both vertical and horizontal forces d)none 9. Reactive components of a ‘hinge’ supported on a horizontal plane { c } a)only vertical force b)only horizontal force c)both vertical and horizontal forces d)none 10. Reactive components of a fixed support { c } a)only vertical force and fixity moment b) only horizontal force and fixity moment c)both vertical and horizontal forces and a fixity moment d)none of the above 11. A transversely loaded beam will be unstable, if it is supported over { d } a)one fixed & other hinged b)one fixed & other roller c)one roller and one hinge d)both rollers 12. For two unlike equal parallel forces, there exists { b } a)a resultant force b)a resultant moment c)a resultant force and a moment d)none 13. When trying to turn a key into a lock, the following is applied { d } a)coplanar force b)non-coplanar force c)lever d)couple 14. Self weight of a block resting on an inclined plane acts { a } a)vertically downwards b)horizontally c)along the inclined plane d)normal to incline Note: Weight (Gravitational force) of a body always acts vertically downwards. 15. The algebraic sum of the resolved parts of a number of forces in a given direction Is equal to the resolved component of the resultant in that direction.



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16. The resultant of two equal forces of magnitude ‘P/4’ which are acting at right

angles = P/√8

17. “Every particle (body) remains at rest or continues to move in a straight line with uniform velocity, if there is no unbalanced force acting on it” is known as Newton’s First Law (of motion). 18. Bodies which do not deform under the action of applied forces are known as Rigid bodies. 19. If two forces P & Q act at a point and the angle between the two forces be ‘α’ ,

then the resultant is given by R = √P2+Q2+2PQ.cosα

And the angle made by the resultant with the direction of force P is given by tanθ = Q.sinα / (P + Q.cosα ) P R θ

Q

20. The moment of a force about a point = (Force) * (perpendicular distance of the line of action of the force from that point). O

P Area gives moment 21. A force causes linear displacement (movement), while moment causes angular

displacement (rotation).

22. A body will be in equilibrium if, 1)resultant force in any direction is zero and 2)the net (sum) moment of the forces about any point is zero.

23. Gravitational law of attraction is given by F = G. m1 .m2 / r2. 24. Coplanar forces mean the forces that are acting in one plane. 25. Concurrent forces mean the forces are intersecting at a common point. 26. Collinear forces means the forces are having same line of action.



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27. The resultant R of three collinear forces F1, F2 & F3 acting in the same direction is given by R = F1 + F2 + F3. If the force F3 is acting in 0pposite direction then, their resultant R will be R = F1 + F2 – F3.

28. The resultant of three or more forces acting at a point is given by R = √∑Fx2 + ∑Fy2, where, Fx = algebraic sum of horizontal components of all forces. The angle made by resultant with horizontal is given by tanθ = ∑Fy/∑Fx = Ry/Rx

29. The resultant of several forces acting at a point is found graphically by using

polygon law of forces. 30. The resultant of two like parallel forces is the sum of the two forces and acts at a

point between these two in such a way that the resultant divides the distance in the ratio inversely proportional to the magnitudes of the forces.

31. When two equal and opposite parallel forces act on a body at some distance apart,

the two forces form a couple which has a tendency to rotate the body. The moment of this couple is the product of either one of the forces and the distance between these two forces.

32. A system of couple acting in one plane is in equilibrium if the algebraic sum of

their moments is equal to zero. 33. A given force P applied to a body at any point A can always be replaced by an

equal force applied at another point B in the same direction together with a couple.

34. If the resultant of a number of parallel forces is not zero, the system can be reduced to a single force, whose magnitude is equal to the algebraic sum of all forces. The point of application of this single force is obtained by equating the moment of this single force about any point to the algebraic sum of moments of all forces acting on the system about the same point.

35. Free Body Diagram (FBD) of a body is a diagram in which the body is completely

isolated from its supports and the supports are replaced by the reactions which these supports exert on the body. OR It is the sketch of a body in which all



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actions and reactions (in place of supports) which are required for equilibrium are shown.

36. Rigid body is a body whose shape and size cannot be changed due to action of

forces. 37. The resultant of two equal forces acting at a point is equal to either of them. Then

the angle between the two forces will be 1200. P

R = √P2 + P2 + 2.P.P.cos120 = P 120 P 38. If two concurrent coplanar forces P & Q are acting on a rigid body at an angle

1200, find the value of force ‘Q’ if P=40N and the angle (β) between ‘Q’ and resultant ‘R’ is 900. Ans: Q=20N

39. Find the magnitude of the two concurrent coplanar forces (P&Q) acting on a rigid body, if the resultant ‘R’ comes out to be 3.16N if θ = 900 and 3.6N if θ = 600.

Ans: P= 3N & Q=1N



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EXERCISE PROBLEMS :

1. Define the following a) Rigid body b) Principle of Transmissibility c) Triangle law of forces d) Deformable body

2. a) State and prove Lami’s theorem b) State and prove Theorem of Varignon.

3. a) Show how force on a body can be replaced by an equivalent force couple system.

b) Define moment and couple and differentiate them. 4. a) Write brief note on F.B.D. and explain with suitable examples. b) What is free vector? Give example of free vector.

θ

W

150

100N

450

Find Tension developed in the string and Compression developed in the bar.

Find Tension developed in the string and reaction developed at the ground.

30cm 150

24cm

3cm

Find the Pull exerted on the nail by nail puller.

Length of string=20cm Radius of cylinder=10cm

Find the tension developed in the string.

Wt. of bar, W = 100N

W



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29

100N

2m

x=? 0.5m

200N

W

300 450

Find the distance ‘x’ so that bar remains horizontal.

α

θ

a

P

Find the reaction developed at wall

P

Q

θ

Find the angle ‘θ’ which keeps the bar of length ‘l’ in equilibrium, if weight of the bar is ‘W’.

α Find the angle ‘α’ required for equilibrium and also the reaction exerted by the ground. Take pulleys as perfectly smooth.

l

A

B Hint: the reactions RA, RB and weight W should be concurrent and coplanar.

100N

50N

20N-m

A B

1.5m 2.5m

1m

1m

Find the reaction developed at the hinge ‘A’ of a Bell Crank loaded as shown in fig.



30

30

P) A spherical ball of weight ‘W’ rests in a triangular groove whose sides are inclined at angles α & β to the horizontal. Find the reactions at the surfaces of contact. If a similar ball is now placed, so as to rest above the first ball and one side of which is inclined at ‘α’. Also find the reaction on the lower ball from the surface which is inclined at ‘β’. P) A ladle is lifted by means of 3 chains each 2m in length. The upper ends of the chains are attached to a ring, while the lower ends are attached to three hooks, fixed to the ladle forming an equilateral triangle of 1.2m side as shown in fig. If the weight of the ladle is 5kN, find the tension developed in each chain.

P) A compound lever shown is required to lift a load of 10kN with an effort ‘P’. Find the magnitude of the effort ‘P’.

1.2m

2m

5kN

8cm 30cm

5cm 45cm 10kN

P



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31

P) Two identical cylinders of weight 100N each and 100mm in radius are supporting an another cylinder of weight 200N and radius 300mm as shown in fig. Find the tension developed in the string that is connecting the lower cylinders.

P) A hollow cylinder of radius 10cm is opened at both ends and rests on a smooth horizontal surface as shown in fig. Inside the cylinder there are two spheres having weights W1 and W2 and radii r1 and r2 respectively. The lower sphere also rests on the horizontal surface. Neglecting friction find the minimum weight ‘Q’ of the cylinder in order that will not tip over. Take W1=100N, W2=150N, r1=5cm and r2=7cm. Note: When tipping occurs, there will be no contact b/w cylinder and horizontal surface at ‘B’. Hence, reaction at B i.e., RB becomes zero.

700mm

W1

W2

BA

20cm

civil & engg. engg. mechanics mechanical unit i · 2. two concurrent coplanar forces are acting...

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