Fundamentals of Statics Chapter 2: Resultant Forces

Chapter 2

Fundamental of Statics

2-1

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.

2-2

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
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.

2-3

Vectors
• Vector: parameters possessing magnitude and direction
which add according to the parallelogram law. Examples:
displacements, velocities, accelerations.
• Scalar: parameters possessing magnitude but not
direction. Examples: mass, volume, temperature
• Vector classifications:
- Free vectors may be freely moved in space without
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.

2-4

External and Internal Forces
• Forces acting on rigid bodies are
divided into two groups:
- External forces
- Internal forces

• External forces are shown in a
free-body diagram.

• If unopposed, each external force can impart a motion of
translation or rotation, or both.

2-5

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.

• 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.

2-6

Classification of A Force System

2-7

Classification of a Force System

2-8

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.

2-9

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.

2 - 10

Resultant of Coplanar Forces

• The resultant of coplanar forces may be determined by
following two method:
1. Analytical method
2. Graphical method

2 - 11

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
opposite direction, then resulting is
obtained by subtracting that forces.

R= F1 + F2 - F3

2 - 12

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.
F1 F2 F3
a b c d
R= F1 + F2 + F3

2 - 13

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

2 - 14


• 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.
• The magnitude of Resultant force R

• The direction of Resultant force R with the force P

 Q sin α 
θ = tan 
 P + Q cos α 
−1

 
2 - 15


• (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.
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.

2 - 16

• When more than two forces act at a point.
• Analytical Method

R = (∑ H ) 2 + (∑ V ) 2 tan θ =
∑V
∑H
2 - 17


• (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.”
F2 c
F1 F3
d F2

o F4 b
F1
F3 e
F4 R a
2 - 18


Vector Addition: The Order Does NOT Matter

2 - 19

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.

• 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.

2 - 20

Coplanar Parallel Forces

• Moment of Forces
• The tendency of a force to produce
rotation of body about some axis or point
is called the moment of a force.
• moment of a force about a point
• moment of a force about an axis
• moment due to a couple

The moment (m) of the force F about O is given by,
M=F x d

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

2 - 21

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 is
known as anti clock wise moment.
• Sign conv. : clockwise (-ve),
counterclockwise (+ve)

2 - 24

Resultant Moment of Forces

• The resultant moment of F1, F2, and F3 about O
= -F1 x r1 – F2 x r2 + F3 x r3

2 - 25

Resultant Moment of Forces

20N
30N D C
• For Example:
• Find the resultant moment about point A.
• Soln:
• Forces at point A and B passes through Point 2m
A.
• 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. A 2m B 10N
• Moment of force at C about point A:
20 x 2 =40N(CCW) 40N
• Moment of force at D about point A :
30 x 2= 60N(CCW)

So resultant moment at point A = 40 + 60 = 100N(CCW)
2 - 26

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
u uu uu
r r r uu
r
R = F1 + F2 + K + Fn
• Similar resultant comes from the addition of moments
uuu
r u r uu
r uu
r uu
r
M 0 = R d R = F1 d1 + F2 d 2 + K + Fn d n

2 - 27

Principle of Moments

2 - 28

Types of Parallel Forces

• Two important types of parallel forces
1. Like parallel forces
2. Unlike 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.

2 - 29

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.

2 - 30


• 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.

• We have to calculate the resultant force acting on the body and
its position.
• From condition of static equilibrium;
R = F1 + F2 ...(1)

2 - 31


• 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 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.
2 - 32


• 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
acting on the body and its position.
• From condition of static equilibrium,
R = F1 – F2 …(1)

2 - 33


• 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 are unlike but equal in magnitude

• Two parallel forces having different lines of action, equal in
magnitude, but opposite in sense constitute a couple.

• A couple causes rotation about an axis
perpendicular to its plane.

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

M=F*a

Unit: Nm
2 - 35

Moment of a Couple

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

2 - 36

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
B A B A B A

= d F = MA

MA = d F

2 - 37

Replace a Force-Couple System with Just Forces

F F
A A
F2
d2
MA =
F2
C C

d2 F2 = MA

2 - 38

Reducing a System of Forces to a Resultant Force-
Couple System (at a Chosen Point)

F1
R

r1
r2 F2
A
r3 = MA
r r
F3 R = ∑F
r r r
MA = ∑ r ×F
( )
2 - 39

Reducing a System of Forces to a Resultant Force-
Couple System (at a Chosen Point)

2 - 40

Reduce a System of Forces to a Single Resultant Force
F1
R R R

r1
r2 F2 B
A
r3 = MA
= MA

F3 R R

Using method B
from prior slide =
2 - 41

Reduce a System of Forces to a Single Resultant Force

R
R
Ry
B B
A
A Rx Rx
MA Ry

R
R
dx

dx R = –MA
2 - 42

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
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.

2 - 43

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
point.
– If
∑ M = 0 then system is in equilibrium .

– If ∑ M ≠ 0 the system reduce to a resultant couple.
– And the calculated moment gives the moment of that couple.

2 - 44

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 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

2 - 45

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

2 - 46

Difference between moment and couple
Moment Couple
• Moment = force x perpendicular • Two equal and opposite forces whose lines of
distance M = Fd action are different from a couple

• It is produced by a single force not • It is produced by the two equal and opposite
passing through Centre of gravity of parallel, non collinear forces.
the body.

• The force move the body in the • Resultant force of couple is zero. Hence,
direction of force and rotate the body. body does not move, but rotate only.
It is the resultant force.

• To balance the force causing moment, • Couple cannot be balanced by a single force,
equal and opposite force is required. it can be balanced by a couple only.

• For example, • For example,
• To tight the nut by spanner • To rotate the key in lock
• To open or close the door • To open or close the wheel valve of water line
• To rotate the steering wheel of car.
2 - 47

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.

2 - 48


• Principle of Equilibrium:

•
∑F = 0 …………(1)

• ∑M = 0 ………....(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.

∑ Fx = 0 ∑ Fy = 0
2 - 49


• 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.

∑ Fx = 0 ∑ Fy = 0 ∑ M = 0
• 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.

∑ Fx = 0 ∑ Fy = 0
2 - 50


• Force Law of Equilibrium:
• There are three main force Law of Equilibrium:
• Two force system
• Three force system
• Four or more force system

2 - 51


(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.
• Two condition is satisfied:
• (1) ∑ Fx = 0 (2) ∑ Fy = 0 as F1 = F2
• Third condition is not satisfied:
• (3) ∑ M ≠0 MA = -F2 X AB

• 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 - 52


• (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
• 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.
2 - 53


• 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:
• (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 = 0

2 - 55

• 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.

2 - 56


• 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 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.

2 - 57


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

2 - 58


• 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.
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.

2 - 59


• 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.

Support Reaction(s) Description Number of
of reaction(s) unknowns

Tension of unknown one
magnitude T in the
direction of the
cable

2 - 60


• 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.

Force of unknown one
magnitude N
directed normal to
the surface

2 - 61


• 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.


Force of unknown one
magnitude N normal
to the surface
supporting the roller

2 - 62


• 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
force.
Force of unknown
magnitude N normal Two
to the surface and a
friction force of
unknown magnitude
F parallel to the
surface
2 - 63


• 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).

2 - 64


• 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
commonly chosen to be two components of R, such as Rx and Ry )
– The magnitude C of the reactive couple.

Unknown force R
and a couple of Three
unknown magnitude
C

2 - 65


• 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) 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


2 - 70


2 - 71


• 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
that appear on the free-body diagram.
• Step 3: Solve the equilibrium equations for the unknowns.

2 - 72


• 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.
• 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.

2 - 73


• Statically indeterminate and Improper Constraints

2 - 74


2 - 75


2 - 76


• Statistical determinate and proper Constraints

2 - 77

Any question?

2 - 78

Fundamentals of Statics Chapter 2: Resultant Forces

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