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1. OVERVIEW OF THE MODULE
In this module, we will be discussing the concept of Equilibrium of Force Systems.
Equilibrium is present when the resultant of a system of forces is zero. A body is said to be in
equilibrium when the force system acting upon it has zero resultant.
1. 1 Purpose of the Module
The purpose or objectives of this module is for you, my dear students, to learn to do
the following:
To introduce the concept of the free-body diagram for a particle.
Know how to solve particle equilibrium problems using the equations of
equilibrium.
1 .2 Module Title and Description
MODULE NO. 2 – EQUILIBRIUM OF FORCE SYSTEMS
1 .3 Module Guide
This module deals with equilibrium of force system. It is understood that you should
have learned the basic concept of statics which is the resultant of force system discussed
in the previous module. In this module, you should take note of the following:
A review of your previous topic in the first module to help you understand the topic
at hand.
Read the learning outcomes of this module for you to determine the focus of our
topic.
The illustrative problems discuss the technique of solution as well as the
application of principles.
Solve the exercises that are provided for you after every subtopic.
Studying the example problems helps, but the most effective way of learning the
principles of the topic is to solve problems.
To be successful at this, it is important to always present the work in a logical and
orderly manner, as suggested by the following sequence of steps:
o Read the problem carefully and try to correlate the actual physical situation
with the theory studied.
o Tabulate the problem data and draw to a large scale any necessary
diagrams.
o Apply the relevant principles, generally in mathematical form. When writing
any equations, be sure they are dimensionally homogeneous.
o Solve the necessary equations, and report the answer with no more than
three significant figures.
o Study the answer with technical judgment and common sense to determine
whether or not it seems reasonable.
1 .4 Module Outcomes
After studying this module, you must be able to:
To introduce the concept of the free-body diagram for a particle.
Know how to solve particle equilibrium problems using the equations of
equilibrium.
1 .5 Module Requirements
Solve the exercises and write them in a short bond paper with 1” x 1” borderline
for uniformity. Have it stapled at the upper left hand corner of the bond paper.
Please do not use folder, however you can place it inside an envelope to avoid
from getting lost during its transportation. Erasures are not allowed.
You will not be given the next module if you do not finish the preceding module.
Activities in this module will only be accepted for checking if you have submitted
the activities in the previous module.
2. KEY TERMS
Before we proceed with our topic, let us first define some important terms so that you will
better understand the lesson well.
Equilibrium A particle is said to be in equilibrium if it remains at rest if originally at rest or has
a constant velocity if originally in motion_._
Free Body Diagram (FBD) is essentially a sketch, or part of it, showing the forces exerted by
other bodies on the one being considered.
3. LEARNING PLAN
3 .1 Activating Student’s Schemata
As discussed in the previous module, there are six general types of force systems.
These force systems may be written in vector or algebraic equations to express the condition of
the force system. We have also studied that a resultant may be obtained by taking the sum of
forces along specified moment center or axis.
3.2 Learning Objectives
3.2.1 Condition for the Equilibrium of a Particle
A particle is said to be in equilibrium if it remains at rest if originally at rest or has
a constant velocity, if originally in motion. Most often, however, the term “equilibrium” or,
more specifically, “static equilibrium” is used to describe an object at rest. To maintain
equilibrium, it is necessary to satisfy Newton’s first law of motion , which requires the
resultant force acting on a particle to be equal to zero. This condition may be stated
mathematically as
ΣF = 0 (3-1)
Where ΣF is the vector sum of all the forces acting on the particle.
Not only is Eq. 3 - 1 a necessary condition for equilibrium, it is also a sufficient
condition. This follows from Newton’s second law of motion which can be written as ΣF =
ma. Since the force system satisfies Eq. 3 - 1, then ma = 0 , and therefore the particle’s
acceleration a = 0. Consequently, the particle indeed moves with constant velocity or
remains at rest.
3.2.2 The Free-Body Diagram
To apply the equation of equilibrium, we must account for all the known and
unknown forces ( ΣF ) which act on a particle. The best way to do this is to think of the
particle as isolated and “free” from its surroundings. A drawing that shows the particle with
all the forces that act on it is called a free-body diagram (FBD).
Before presenting a formal procedure as to how to draw a free-body diagram, we
will first consider two types of connections often encountered in particle equilibrium
problems.
Springs. If a linearly elastic spring (or cord) of undeformed length l
o
is used to support a
particle, the length of the spring will change in direct proportion to the force F acting on it
The forces that are known should be labeled with their proper magnitudes
and directions. Letters are used to represent the magnitudes and directions of
forces that are unknown.
The bucket is held in equilibrium by the cable, and instinctively we know
that the force in the cable must equal the weight of the bucket. By drawing a free-
body diagram of the bucket we can understand why this is so. This diagram shows
that there are only two forces acting on the bucket , namely, its weight W and the
force T of the cable. For equilibrium, the resultant of these forces must be equal to
zero, and so T = W.
FBD:
The sphere in Fig. 3-3a has a mass of 6 kg and is supported as shown.
Draw a free body diagram of the sphere, the cord CE , and the knot at C.
Fig. 3-3a
Solution:
Sphere. By inspection, there are only two forces acting on sphere namely,
its, weight, 6kg (9.81 m/s2) = 58.9 N, and the force of cord CE. The free-body
diagram is shown in Fig. 3-3b.
Fig. 3-3b
Cord CE. When cord CE isolated from its surroundings, its free-body diagram
shows only two forces acting on it, namely, the force of the sphere and the force
of the knot, Fig. 3-3c. Notice that F CE shown here is equal but opposite to that
shown in Fig. 3-3b, a consequence of Newton’s third law of action-reaction. Also,
F
CE
and F EC
pull on the cord and keep it in tension so that it doesn’t collapse. For
equilibrium, F CE
= F
EC
Fig. 3-3c
Knot. The knot at C is subjected to three forces, Fig. 3-3d. They are caused by
the cords CBA and CE and the spring CD. As required the free-body diagram
shows all these forces labeled with their magnitudes and directions. It is important
to recognize that the weight of the sphere does not directly act on the knot. Instead,
the cord CE subjects the knot to this force.
Fig. 3-3d
3.2.3 Coplanar Force Systems
If a particle is subjected to a system of coplanar forces that lie in the x-y plane as in Fig.
3 - 4, then each force can be resolved into its I and j components. For equilibrium, these
forces must sum to produce a zero force resultant, i.e.,
ΣF = 0
ΣF
X
i = 0 + ΣF Y
j = 0
For these vector equation to be satisfied, the force’s x and y components must both be
equal to zero. Hence
ΣFx= 0 (3-3)
ΣF
Y
A system of cords knotted
together at A and B support the
weights shown in Fig. 3- 4. Compute
the tensions P, Q, F and T acting in
the various cords.
We begin by drawing a FBD of
knots A and B. Of these two coplanar
force systems, we must first solve that
at A. The force system at B is
temporarily indeterminate because it
contains three unknown forces and
has available only two independent
equations of equilibrium. Its solution must be postponed until one of the unknown,
P in the instance, has been determined from the coplanar forces acting at A, where
P , exerting an equal and opposite effect to its action on B, is only one of the
unknowns.
Method I – Using Horizontal and Vertical Axes.
This is a routine method requiring no
imagination. Selecting reference axes that
are horizontal and vertical as shown in Fig. 3-
5 , we apply the conditions of equilibrium, Eq.
(3-1), to obtain
ΣFx = 0
P cos 15° - Q cos 30° = 0 (a)
ΣF
Y
P sin 15°+ Q sin 30° - 300 = 0 (b)
Solving Equations (a) and (b) simultaneously yields
P = 367 lbs
Q = 410 lbs
Method II – Using Rotated Axes.
The disadvantage of Method I is the necessity of solving simultaneous
equations. Since the reference axes are arbitrarily selected in the first place, a
better choice of the reference axes will eliminate simultaneous equations; this
simplifies the numerical work and reduces the chance for error. For example, let
the X axis be selected to pass through one of the unknowns, say Q. In this case Q
will have no Y component and will not appear in a Y summation.
The method of determining the angles between the forces and the rotated
reference axes is shown in Fig. 3-6a; the final values of the angles are shown in
Fig. 3-6b. When actually solving the problems, only the X axis need be drawn, as
in Fig. 3-6a. The Y axis can be omitted; it is understood to be perpendicular to the
X axis.
Since the X axis was chosen to coincide with Q , it is evident that Q has no
Y component. Hence by applying the condition of equilibrium, ΣFY = 0 , we
automatically eliminate Q from the equation. Thus we have
ΣF
Y
P sin 45° - 300 sin 60° = 0
P = 367 lb
Having determined P, we readily find the second unknown Q by applying
the second equation of equilibrium:
ΣF
X
367 cos 45° + 300 cos30° - Q = 0
P = 410 lb
Note carefully the technique used. When the X axis is chosen so that it coincides
with one of the unknowns, the Y summation determines the other unknown. Then the
X summation determines the remaining unknown.
Method III – Using Force Triangle
When three forces are in equilibrium, the easiest solution is generally
obtained by applying the sine law to the triangle representing the polygon of forces.
Since forces in equilibrium have a zero resultant, the tip of the last vector must
touch the tail of the first vector. This tip to tail addition gives the closed polygon of
forces shown in Fig. 3 - 7. Applying the law of sines to this triangle, we obtain
sin 45°
sin 60°
sin 75°
Whence as before
P = 367 lb and Q = 410 lb
If this solution is compared with ΣF Y
= 0 in Method II, it will be seen to give
the same equation. Indeed, the solution from a force triangle is equivalent to using
two sets of rotated axes, one set drawn so as to pass through one unknown force,
whereas the other set coincide with the other unknown force.
We are now ready to determine the forces F and T holding the coplanar
system of forces at B in equilibrium. A closed polygon of forces for this system
forms a quadrilateral so that the sine law cannot be applied. Although a diagonal
of this quadrilateral can be drawn that will subdivide it into two triangles to which
the sine law can be applied, this procedure is more cumbersome than the method
of using rotated axes described above in Method II.
Applying the method of rotated axes to
the FBD of B , we draw the X axis to coincide
with T as in Fig. 3 - 8, thereby eliminating T from
a Y summation. Hence we obtain F from
ΣF
Y
F sin 45° - 367 sin 45° - 200 sin 60° = 0
F = 612 lb answer
The remaining unknown T is now determined
from
ΣF
X
T + 200 cos 60° - 367 cos 45° - 612 cos 45° = 0
T = 593 lb answer
- The system of knotted cords shown
support the indicated weights. Compute
the tensile force in each cord.
From the knot where 400 lb load is hanging
ΣFH=
Dsin75°=Csin30°
D=0.5176C
ΣFV=
Dcos75°+Ccos30°=
(0.5176C) cos75°+Ccos30°=
C=400 lb answer
D=0.5176(400)
D=207.06 lb answer
From the knot where 300-lb load is hanging
ΣFV=
Bsin45°=300+Ccos30°
Bsin45°=300+400cos30°
B=914.16 lb answer
ΣFH=
A=Bcos45°+Csin30°
A=914.16cos45°+400sin30°
A=846.41 lb answer
FBD of knot w/ 400 lb load:
FBD of knot w/ 300 lb load:
3.2.1.1b Exercises (Activity 3)
- & 2. A 300-lb box is held at rest on a smooth plane by a force P inclined at an
angle θ with the plane as shown in the Figure below. If θ = 45°, determine the value
of P and the normal pressure N exerted by the plane.
- Determine the magnitude of P and F necessary to keep the concurrent force system
in Figure shown in equilibrium.
- Figure below represents the concurrent force system acting at a joint of a bridge truss.
Determine the value of P and E to maintain equilibrium of the forces.
- The 300-lb force and the 400-lb force shown below are to be held in equilibrium by a
third force F acting at an unknown angle θ with the horizontal. Determine the values
of F and θ.
- Determine the values of α and θ so that the forces shown in the Figure will be in
equilibrium.
3.2.1.2 Conditions for Equilibrium of Parallel Forces
The sum of all the forces is zero.
ΣF=
The sum of moment at any point O is zero.
ΣM
O
3.2. 1 .2a Sample Problem:
- Determine the reactions for the beam shown.
Solution:
𝐑
𝟐
10R
1
+4(400) =16(300) +9[14(100)]
R
1
=1580 lb answer
𝐑 𝟏
10R2+6(300)=14(400)+1[14(100)]
R2=520 lb answer
- The roof truss shown is supported by a roller at A and a hinge at B. Find the values of the
reactions.
Replace the 3-20 kN and 2-10 kN forces by a single 80 kN force
ΣM
B
15R
A
R
A
=96.67 kN answer
100 lb/ft
ΣM
A
15R
B
R
B
=93.33 kN answer
3.2. 1. 2 b Exercises (Activity 4):
- Determine the reactions for the beam shown.
- Determine the reactions R 1
and R 2
of the beam in Figure as shown loaded with a
concentrated load of 1600 lb and a load varying from zero to an intensity of 400 lb per
ft.
- Determine the reactions for the beam loaded as shown.
- The wheel loads on a jeep are given in the Figure. Determine the distance x so that
the reaction of the beam at A is twice as great as the reaction at B.
- For the system of pulleys shown below, determine the ratio of W to P to maintain
equilibrium. Neglect axle friction and the weights of the pulleys.
- Using the same figure, if each pulley weighs 36 kg and W = 720 kg, find P to maintain
equilibrium.
3.2.1 Equilibrium of Non-Concurrent Force System
(Coplanar Force Systems)
There are three equilibrium conditions that can be used for non-concurrent,
non-parallel force system.
The sum of all forces in the x-direction or horizontal is zero.
ΣF
x
=0 or ΣF H
The sum of all forces in the y-direction or vertical is zero.
ΣF
y
=0 or ΣF V
The sum of moment at any point O is zero.
ΣM
O
The three equilibrium conditions can solved up to three unknowns in the system.
3.2.3.1 Sample Problem:
- The frame shown below is supported in pivots at A and B. Each member weighs 5 kN/m.
Compute the horizontal reaction at A and the horizontal and vertical components of the
reaction at B.
Solution:
Length of DF
L
DF
2
2
2
L
DF
2
L
DF
=5 m
Weights of members
W
AB
=6(5)=30 kN
W
CE
=6(5)=30 kN
W
DF
=5(5)=25 kN
ΣM
B
6 A
H
= 3 W
CE
+ 2 W
DF
6 A
H
A
H
=223.33 kN answer
ΣF
H
B
H
=A
H
BH=223.33 kN answer
ΣF
V
B
V
=W
AB
+W
CE
+W
DF
B
V
B
V
= 285 kN answer
- Compute the total reactions at A and B on the truss shown.
- Compute the total reactions at A and B for the truss shown.
- The beam shown is supported by a hinge at A and a roller on a 1 to 2 slope at B. Determine
the resultant reactions at A and B.
- A pulley 4 ft in diameter and supporting a load 200 lb is mounted at B on a horizontal beam
as shown in the figure. The beam is supported by a hinge at A and rollers at C. Neglecting
the weight of the beam, determine the reactions at A and C.
- The forces acting on a 1-m length of a dam are shown in the figure below. The upward
ground reaction varies uniformly from an intensity of p 1
kN/m to p 2
kN/m at B. Determine
p1 and p2 and also the horizontal resistance to sliding.
- The uniform rod in Fig. P-357 weighs 420 lb and has its center of gravity at G. Determine
the tension in the cable and the reactions at the smooth surfaces at A and B.
- A bar AE is in equilibrium under the action of the five forces shown in Fig. P-358.
Determine P, R, and T.
- A 4-m bar of negligible weight rests in a horizontal position on the smooth planes shown.
Compute the distance x at which load T = 10 kN should be placed from point B to keep
the bar horizontal.
