Chapter 2

Chapter 2

2 - 1

Fundamental of Statics

Introduction

• The objective for the current chapter is to investigate the effects of forces

on particles:

- replacing multiple forces acting on a particle with a single

equivalent or resultant force,

- relations between forces acting on a particle that is in a

state of equilibrium.

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Resultant of Two Forces

• force: action of one body on another;

characterized by its point of application,

magnitude, line of action, and sense.

• Experimental evidence shows that the

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• Experimental evidence shows that the

combined effect of two forces may be

represented by a single resultant force.

• The resultant is equivalent to the diagonal of

a parallelogram which contains the two

forces in adjacent legs.

• Force is a vector quantity.

Vectors• Vector: parameters possessing magnitude and direction

which add according to the parallelogram law. Examples:

displacements, velocities, accelerations.

• Vector classifications:

- Free vectors may be freely moved in space without

• Scalar: parameters possessing magnitude but not

direction. Examples: mass, volume, temperature

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changing their effect on an analysis.

- Sliding vectors may be applied anywhere along their

line of action without affecting an analysis.

• Equal vectors have the same magnitude and direction.

• Negative vector of a given vector has the same magnitude

and the opposite direction.

External and Internal Forces

• Forces acting on rigid bodies are

divided into two groups:

- External forces

- Internal forces

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• External forces are shown in a

free-body diagram.

• If unopposed, each external force can impart a motion of

translation or rotation, or both.

Principle of Transmissibility: Equivalent Forces

• Principle of Transmissibility -

Conditions of equilibrium or motion are

not affected by transmitting a force

along its line of action.

NOTE: F and F’ are equivalent forces.

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• Moving the point of application of

the force F to the rear bumper

does not affect the motion or the

other forces acting on the truck.

Classification of A Force System

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Classification of a Force System

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Resultant of Several Forces

• When a number of coplanar forces are acting on a rigid body

then these forces can be replaced by a single force which has

the same effect on the rigid body as that of all the forces acting

together, then this single force is known as the resultant of

several forces.several forces.

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Resultant of Several Forces

• Definition:

• A single force which can replace a number of forces acting one a rigid body,

without causing any change in the external effects on the body is known as

the resultant force.

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Resultant of Coplanar Forces

• The resultant of coplanar forces may be determined by

following two method:

1. Analytical method

2. Graphical method

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Resultant of Collinear Coplanar Forces

• Analytical Method

• The resultant is obtained by adding all

the forces if they are acting in the same

direction.

• If any one of the forces is acting in the• If any one of the forces is acting in the

opposite direction, then resulting is

obtained by subtracting that forces.

R= F1 + F2 - F3

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Resultant of Collinear Coplanar Forces

• Graphical Method

• Some suitable scale is chosen and vectors are drawn to the

chosen scale.

• These vectors are added/or subtracted to find the resultant.

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F1 F2 F3

R= F1 + F2 + F3

a b c d

Resultant of Concurrent Coplanar Forces

• Concurrent coplanar forces are those forces which act in the

same plane and they intersect or meet at a common point.

• Following two cases are consider:

i. When two forces act at a point.i. When two forces act at a point.

ii. When more than two forces act at a point.

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• When two forces act at a point.

• (a) Analytical Method:

– When two forces act at a point,

their resultant is found by the law

of parallelogram of forces.

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of parallelogram of forces.

• The magnitude of Resultant force R

• The direction of Resultant force R with the force P

+=

−

α

αθ

cos

sintan 1

QP

Q


• (b) Graphical Method

i. Choose a convenient scale to represent the forces P and

Q.

ii.From point O, draw a vector OA= P.

iii.Now from point O, draw another vector OB= Q and at

an angle of α as shown in fig.an angle of α as shown in fig.

iv.Complete the parallelogram by drawing lines AC║ to

OB and BC ║ to OA.

v. Measure the length OC.

vi.Resultant R = OC x Chosen Scale

• The Direction of resultant is given by angle θ.• Measure the angle θ.

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• When more than two forces act at a point.

• Analytical Method

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22 )()( ∑∑ += VHR∑∑

=H

Vθtan


• (b) Graphical Method

• The resultant of several forces acting at a point is found graphically with the

help of Polygon law of forces.

• Polygon law 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

the same order, then their resultant is represented in magnitude and

direction by the closing side of the polygon taken in the opposite order.”

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o

F1

F2

F4

F3

F1

F2

F3

F4

R a

b

c

d

e


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Vector Addition: The Order Does NOT Matter

Coplanar Parallel Forces

• A parallel coplanar force system consists of two

or more forces whose lines of action are parallel

to each other.

• Two parallel forces will not intersect at a point.• Two parallel forces will not intersect at a point.

• The line of action of forces are parallel so that

for finding the resultant of two parallel forces,

the parallelogram cannot be drawn.

• The resultant of such forces can be determined

by applying the principle of moments.

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Coplanar Parallel Forces

• Moment of Forces

• The tendency of a force to producerotation of body about some axis or pointis called the moment of a force.

• moment of a force about a point

• moment of a force about an axis• moment of a force about an axis

• moment due to a couple

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Unit: force x distance =F*L = N-m, kN-m (SI unit)

The moment (m) of the force F about O is given by,

M=F x d

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Moment of Forces

• The tendency of moment is to rotate about the body

in the clockwise direction about O is called

clockwise moment.

• If the tendency of a moment is to rotate the body in

anti clockwise direction, then that moment isanti clockwise direction, then that moment is

known as anti clock wise moment.

• Sign conv. : clockwise (-ve),

counterclockwise (+ve)

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Resultant Moment of Forces

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• The resultant moment of F1, F2, and F3 about O

= -F1 x r1 – F2 x r2 + F3 x r3

Resultant Moment of Forces

• For Example:

• Find the resultant moment about point A.

• Soln:

• Forces at point A and B passes through Point

A.

• So perpendicular distances from A on the line

CD20N

30N

2m

• So perpendicular distances from A on the line

of action of these forces will be zero.

• Hence their moments about point A will be

zero.

• Moment of force at C about point A:

20 x 2 =40N(CCW)

• Moment of force at D about point A :

30 x 2= 60N(CCW)

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A B

40N

10N2m

So resultant moment at point A = 40 + 60 = 100N(CCW)

Principle of Moments

• The Principle of Moments, also known as Varignon's Theorem, states that

the moment of any force about any point is equal to the algebraic sum of the

moments of the its components of that force about that point.

• As with the summation of force combining to get resultant force

= + + +ur uur uur uur

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• Similar resultant comes from the addition of moments

1 2 nR F F F= + + +ur uur uur uur

K

0 R 1 1 2 2 n nM R d F d F d F d= = + + +uuur ur uur uur uur

K

Principle of Moments

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Types of Parallel Forces

• Two important types of parallel forces

1. Like parallel forces

2. Unlike parallel forces

• Like Parallel forces• Like Parallel forces

• Two parallel forces which are acting in the same direction

are known as like parallel forces.

• The magnitude of a forces may be equal or unequal.

• Unlike Parallel forces

• Two parallel forces which are acting in the opposite

direction are known as like unparallel forces.

• The magnitude of a forces may be equal or unequal.

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Resultant of Two Parallel forces

• The resultant of following two parallel forces will be considered:

– Two parallel forces are like.

– Two parallel forces are unlike and are unequal in magnitude.

– Two parallel forces are unlike but equal in magnitude.

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• Two parallel forces are like

• Suppose that two like but unequal parallel

forces act on a body at position A and B as

shown in figure.

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• We have to calculate the resultant force acting on the body and

its position.

• From condition of static equilibrium;

R = F1 + F2 ...(1)


• The position of R can be obtained by using Varignon’s theorem. To use the

theorem consider a point O along the line AB, such that

• Algebraic sum of moments of F1 and F2 about O = Moment of resultant about O

• Now,

�Moment of F1 about O = F1 × AO (clockwise)

�Moment of F about O = F × BO (anti-clockwise)�Moment of F2 about O = F2 × BO (anti-clockwise)

�Moment of R about O = R × CO (anti-clockwise)

� – F1 × AO + F2 × BO = + R × CO …(2)

� – F1 × AO + F2 × BO = (F1 + F2) × CO

� F1 (AO + CO) = F2 (BO – CO)

� F1 × AC = F2 × BC

� F1 / F2 = BC / AC

• Therefore, it can be observed that R acts at a point C which divides the length

AB in the ratio inversely proportional to the magnitudes of F1 and F2.

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• Two parallel forces are unlike and are unequal in magnitude

• Suppose that two unlike and unequal

parallel forces act on a body at position A

and B as shown in figure.

• We have to calculate the resultant force

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• We have to calculate the resultant force

acting on the body and its position.

• From condition of static equilibrium,

R = F1 – F2 …(1)


• Once again, the position of R can be obtained by using Varigonon’s theorem.

Consider a point O along the line AB, such that

• Algebraic sum of moments of F1 and F2 about O = Moment of resultant about O

• Now, Moment of F1 about O = F1 × AO (clockwise)

• Moment F2 about O = F2 × BO (anti-clockwise)

• Moment of R about O = R × CO (anti-clockwise)

⇒

⇒

⇒ – F1 × AO – F2 × BO = – R × CO …(2)

⇒ F1 × AO + F2 × BO = R × CO

⇒ F1 × AO + F2 × BO = (F1 – F2) × CO

⇒ F2 (BO + CO) = F1 (CO – AO)

⇒ F1 / F2 = BC / AC …(3)

• Since F1 > F2, BC will be greater than AC. Hence point C will lie outside AB on

the same side of F1. Thus, it can be observed that R acts at C which externally

divide length AB in the ratio inversely proportional to the magnitude of F1 and

F2.2 - 34

Moment of a Couple

• Two parallel forces having different lines of action, equal in

magnitude, but opposite in sense constitute a couple.

• Two parallel forces are unlike but equal in magnitude

• A couple causes rotation about an axis

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• A couple causes rotation about an axisperpendicular to its plane.

• The perpendicular distance between theparallel forces is known as arm of the couple.

M = F * a

Unit: Nm

Moment of a Couple

• Two couples are equivalent if they cause the same moment:

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Resolution of a Force into a Force and a Couple

• A force, F, acting at point B can be replaced by the force, F,

and a moment, MA, acting at point A.

F F F F

ABAB AB

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FMA= =d

MA = d F

ABAB AB

Replace a Force-Couple System with Just Forces

F F

AA

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MA =

F2

d2

F2

d2 F2 = MA

A

C

A

C

Reducing a System of Forces to a Resultant Force-

Couple System (at a Chosen Point)

F1

Fr1 r2

R

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A F2

F3

r2

r3=

MA

( )A

R F

M r F

=

= ×

∑∑

r r

r rr

Reducing a System of Forces to a Resultant Force-

Couple System (at a Chosen Point)

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Reduce a System of Forces to a Single Resultant Force

F1

A F2

r1 r2

r3=

MA

R

=MA

R R

B

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F3

3 MA MA

R

=

R

BUsing method

from prior slide

Reduce a System of Forces to a Single Resultant Force

R

R

Rx Rx

Ry

B BA

A

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RR

Rx Rx

Ry

dx

MA

dx R = –MA

A

General Case of parallel forces in a plane

• R1= Resultant of (F1, F2, F4) and R2= Resultant of (F3, F5)

• The resultant R1 and R2 are acting in opposite direction and parallel to each • The resultant R1 and R2 are acting in opposite direction and parallel to each

other.

• Two important case are possible.

• 1. R1 may not be equal to R2.

– Then two unequal parallel forces acting in opposite direction.

– The resultant R= R1-R2

– The point of application easily found with the help of Varignon's

Theorem or moments of forces.

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General Case of parallel forces in a plane

• 2. R1 is equal to R2.

– Then two equal parallel forces acting in opposite direction.

– The resultant R= R1-R2=0

– Now the system may be reduce to a couple or a system is in equilibrium.

– The algebraic sum moment of all forces(F1, F2,…, F5) taken about any 1 2 5

point.

– If then system is in equilibrium .

– If the system reduce to a resultant couple.

– And the calculated moment gives the moment of that couple.

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∑ = 0M

∑ ≠ 0M

Equivalent System

• Two force systems that produce the same external effects on a rigid body are

said to be equivalent.

• An equivalent system for a given system of coplanar forces, is a

combination of a force passing through a given point and a moment about

that point.

• The force is the resultant of all forces acting on the body.• The force is the resultant of all forces acting on the body.

• The moment is the sum of all the moments about that point.

• Equivalent system consists of :

• (1) a single force R passing through the given point P

• (2) a single moment MR

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Equivalent System

• For Examples:

• Determined the equivalent system through

point O.

• These means find:

• (1) a single resultant force, R

• (2) a single moment through, O

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Difference between moment and couple

Moment Couple

• Moment = force x perpendicular

distance M = Fd

• It is produced by a single force not

passing through Centre of gravity of

the body.

• Two equal and opposite forces whose lines of

action are different from a couple

• It is produced by the two equal and opposite

parallel, non collinear forces.

• The force move the body in the

direction of force and rotate the body.

It is the resultant force.

• To balance the force causing moment,

equal and opposite force is required.

• For example,

• To tight the nut by spanner

• To open or close the door

• Resultant force of couple is zero. Hence,

body does not move, but rotate only.

• Couple cannot be balanced by a single force,

it can be balanced by a couple only.

• For example,

• To rotate the key in lock

• To open or close the wheel valve of water line

• To rotate the steering wheel of car.

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Equilibrium of Rigid Bodies

External forces �� Body start moving or rotating.

• If the body does not start moving and also does not start rotating about any

point, then body is said to be in equilibrium.

• For a rigid body in static equilibrium, the external forces and moments are

balanced and will impart no translational or rotational motion to the body.

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• Principle of Equilibrium:

• …………(1)

• ………....(2)

∑ = 0F

∑ = 0M• ………....(2)

• Eq. (1) is known as the force law of equilibrium and Eq. (2) is known as the

moment law of equilibrium.

• The forces are generally resolved into horizontal and vertical components.

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∑ = 0M

∑ = 0Fx ∑ = 0Fy


• Equilibrium of non-concurrent forces system:

• A non-concurrent forces system will be in equilibrium if the resultant of all

forces and moment is zero.

∑ = 0Fx ∑ = 0Fy ∑ = 0M

• Equilibrium of concurrent forces system:

• For the concurrent forces, the line of actions of all forces meet at a point, and

hence the moment of those forces about that point will be zero

automatically.

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∑ ∑ ∑

∑ = 0Fx ∑ = 0Fy


• Force Law of Equilibrium:

• There are three main force Law of Equilibrium:

• Two force system

• Three force system

• Four or more force system• Four or more force system

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(1) Two force system:

• According to this principle, if a body is in equilibrium under the action of

two forces, then they must be equal, opposite and collinear.

• If the two forces acting on a body are equal and

opposite but are parallel, as shown in fig., then the

body will not be in equilibrium.

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• Two condition is satisfied:

• (1) (2) as F1 = F2

• Third condition is not satisfied:

• (3) MA = -F2 X AB

∑ = 0Fx ∑ = 0Fy

∑ ≠ 0M

• A body will not be in equilibrium under the action of two equal and

opposite parallel forces.

• Two equal and opposite parallel forces produce a couple.


• (2) Three force system:


three forces then the resultant of any two forces must be equal, opposite and

collinear with the third force.

• Three forces acting on a body either concurrent or parallel

• Case (a) When three forces are concurrent

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• Case (a) When three forces are concurrent

• The resultant of F1 and F2 is given by R.

• If the force F3 is collinear equal, opposite to the

resultant R, then the body will be in equilibrium.

• The force F3 which is equal and opposite to resultant R is known as

equilibrant.

• Hence for three concurrent forces acting on a body when the body is in

equilibrium, the resultant of the two forces should be equal and opposite to

the third force.


• Case (2): When three forces are parallel

• If the three parallel forces F1, F2, and F3 are acting in the same direction,

then there will be a resultant R= F1 + F2 + F3 and body will not be in

equilibrium.

• If the three forces are acting in opposite direction and their magnitude is so

adjusted that there is no resultant forces and body is in equilibrium.

• Apply the three condition of equilibrium:• Apply the three condition of equilibrium:

• (1) Σ Fx = 0,(No horizontal forces) (2) Σ Fy = 0, (F1+ F3=F2)

• (3) Σ M = 0 about any point.

Σ MA= -F2 X AB + F3 X AC

• For equilibrium Σ MA should be zero.

-F2 X AB + F3 X AC= 0

• If the distance AB and AC are such that the above equation

is satisfied, then the body will be in equilibrium under the action of three

parallel forces.2 - 54


• (3) Four or more force system:


four forces then the resultant of any two forces must be equal, opposite and

collinear with the resultant of the other two forces.

Σ Fx = 0, Σ Fy = 0, Σ M = 0x y

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• Two moment equations.

• Σ Fy = 0

• Σ MA = 0,

• Σ MB = 0

• where A and B are any two points in the xy-plane, provided that the line AB

is not parallel to the y-axis.

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• Free Body Diagram of a Body:

• The first step in equilibrium analysis is to identify all the forces that act on

the body. This is accomplished by means of a free-body diagram.

• The free-body diagram (FBD) of a body is a sketch of the body showing all

forces that act on it. The term free implies that all supports have beenforces that act on it. The term free implies that all supports have been

removed and replaced by the forces (reactions) that they exert on the body.

• Free-body diagrams are fundamental to all engineering disciplines that are

concerned with the effects that forces have on bodies.

• The construction of an FBD is the key step that translates a physical problem

into a form that can be analyzed mathematically.

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• Forces that act on a body can be divided into two general categories—

� Reactive forces (or, simply, reactions) and

� Applied forces (action)

• Reactions are those forces that are exerted on a body by the supports to

which it is attached.which it is attached.

• Forces acting on a body that are not provided by the supports are called

applied forces.

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• The following is the general procedure for constructing a free-body

diagram.

1. A sketch of the body is drawn assuming that all supports (surfaces of

contact, supporting cables, etc.) have been removed.

2. All applied forces are drawn and labeled on the sketch. The weight of the

body is considered to be an applied force acting at the center of gravity.body is considered to be an applied force acting at the center of gravity.

3. The support reactions are drawn and labeled on the sketch. If the sense of a

reaction is unknown, it should be assumed. The solution will determine the

correct sense: A positive result indicates that the assumed sense is correct,

whereas a negative result means that the correct sense is opposite to the

assumed sense.

4. All relevant angles and dimensions are shown on the sketch.

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• The most difficult step to master in the construction of FBDs is the

determination of the support reactions.

• Flexible Cable (Negligible Weight).

• A flexible cable exerts a pull, or tensile force, in the direction of the cable.

With the weight of the cable neglected, the cable forms a straight line. If its

direction is known, removal of the cable introduces one unknown in a free-

body diagram—the magnitude of the force exerted by the cable.body diagram—the magnitude of the force exerted by the cable.

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Support Reaction(s) Description

of reaction(s)

Number of

unknowns

Tension of unknown

magnitude T in the

direction of the

cable

one


• Frictionless Surface: Single Point of Contact.

• When a body is in contact with a frictionless surface at only one point, the

reaction is a force that is perpendicular to the surface, acting at the point of

contact.

• This reaction is often referred to simply as the normal force.

• Therefore, removing such a surface introduces one unknown in a free-body

diagram—the magnitude of the normal force.diagram—the magnitude of the normal force.

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of reaction(s)

Number of

unknowns

Force of unknown

magnitude N

directed normal to

the surface

one


• Roller Support.

• A roller support is equivalent to a frictionless surface: It can only exert a

force that is perpendicular to the supporting surface.

• The magnitude of the force is thus the only unknown introduced in a free-

body diagram when the support is removed.

2 - 62


of reaction(s)

Number of

unknowns

Force of unknown

magnitude N normal

to the surface

supporting the roller

one


• Surface with Friction: Single Point of Contact.

• A friction surface can exert a force that acts at an angle to the surface.

• The unknowns may be taken to be the magnitude and direction of the force.

• However, it is usually advantageous to represent the unknowns as N and F,

the components that are perpendicular and parallel to the surface,

respectively.

• The component N is called the normal force, and F is known as the friction• The component N is called the normal force, and F is known as the friction

force.

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of reaction(s)

Number of

unknowns

Force of unknown

magnitude N normal

to the surface and a

friction force of

unknown magnitude

F parallel to the

surface

Two


• Pin Support.

• Neglecting friction, the pin can only exert a force that is normal to the

contact surface, shown as R in Fig.(b).

• A pin support thus introduces two unknowns: the magnitude of R and the

angle α that specifies the direction of R (α is unknown because the point

where the pin contacts the surface of the hole is not known).

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• Built-in (Cantilever) Support.

• A built-in support, also known as a cantilever support, prevents all motion

of the body at the support. Translation (horizontal or vertical movement) is

prevented by a force, and a couple prohibits rotation.

• Therefore, a built-in support introduces three unknowns in a free-body

diagram:

– The magnitude and direction of the reactive force R (these unknowns are– The magnitude and direction of the reactive force R (these unknowns are

commonly chosen to be two components of R, such as Rx and Ry )

– The magnitude C of the reactive couple.

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of reaction(s)

Number of

unknowns

Unknown force R

and a couple of

unknown magnitude

C

Three


• You should keep the following points in mind when you are drawing

free-body diagrams.

1. Be neat. Because the equilibrium equations will be derived directly from the

free-body diagram, it is essential that the diagram be readable.

2. Clearly label all forces, angles, and distances with values (if known) or2. Clearly label all forces, angles, and distances with values (if known) or

symbols (if the values are not known).

3. Show only forces that are external to the body (this includes support

reactions and the weight). Internal forces occur in equal and opposite pairs

and thus will not appear on free-body diagrams.

2 - 66


• Sample Problem:

The mass of the bar is 50 kg. Take g = 9.81 m/s2

2 - 67


• Sample Problem:

• Neglecting the weights of the members.

2 - 68


2 - 69


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• Equilibrium analysis of a body

• The three steps in the equilibrium analysis of a body are:

• Step 1: Draw a free-body diagram (FBD) of the body that shows all of the

forces and couples that act on the body.

• Step 2: Write the equilibrium equations in terms of the forces and couples• Step 2: Write the equilibrium equations in terms of the forces and couples

that appear on the free-body diagram.

• Step 3: Solve the equilibrium equations for the unknowns.

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• Statically determinate and Statically indeterminate

• The force system that holds a body in equilibrium is said to be statically

determinate if the number of independent equilibrium equations equals the

number of unknowns that appear on its free-body diagram

• If the number of unknowns exceeds the number of independent equilibrium

equations, the problem is called statically indeterminate.equations, the problem is called statically indeterminate.

• The solution of statically indeterminate problems requires the use of

additional principles.

• When the support forces are sufficient to resist translation in both the x and

y directions as well as rotational tendencies about any point, the rigid body

is said to be completely constrained, otherwise the rigid body is unstable or

partially constrained.

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• Statically indeterminate and Improper Constraints

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• Statistical determinate and proper Constraints

2 - 77

Any question?

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Thank youThank you

2 - 79

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