1
BECCBEC
Introduction
The learning module in BEC 122 STATICS OF RIGID BODIES was designed based
on the 3 unit course syllabus used by the College of Architecture and was prepared to
assist course facilitators and students in teaching and learning the subject. Objectives per
module are stated to enable students to know what to expect in each of the lessons. Key
information in each topic or lesson are provided and are supplemented with work
examples for instructors and students to analyze and understand better. Supplementary
exercises are also included for students to do in order to strengthen their skills in such
lessons. Quizzes will be given at the end of each module to be uploaded in Google
Classroom to determine students’ level of mastery of the concepts and theories they have
learned.
Some of the problems presented were adapted from the different textbooks and
the internet and were modified by the writer based on the needs of architecture students.
Problems are related to objects of everyday experiences, like forces on cables or cords,
axially loaded beams, Trusses, and so on. This module presents not only the computations
of forces but to give emphasis the importance of forces acting on a certain body and
further the effect of these forces which will be discussed in the next subject that requires
such knowledge.
Activities and Test Pages
At the end of each lesson, there are Problem Sets provided for the students to
work independently on given problems with the given time to accomplish such task.
Number of problems were minimized into four to five problems only in order to lessen
the burden of the students considering their situation in this time of pandemic, and
thinking that it is not the only subject that they are studying at home. This will be
uploaded in classwork in their google classroom in order to set date and time of
submission and to easily check who submitted on time or not. In this course most
problems have illustrations but they have to analyze the action of forces on each system
and can draw only the isolated view of forces acting on it. These activities may be done in
hand written or type written form, in word document or pdf and will be uploaded in
classwork for the course facilitator to check their work. Also, quizzes and major
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
2
BECCBEC
examination will be done in google form to measure the level of understanding of the
students regarding the lessons.
MODULE I
What is this module all about?
This module presents the fundamental Concepts and definitions of different terms
that will be encountered all throughout the study of this course. with the accompanying
illustrations and derivations of formulas to be used in the computations of the magnitude
of a force, its direction, and location, moments of a force, and resultant of concurrent and
parallel force systems. This includes work examples, exercises, and self-test for the
mastery of the lessons.
This module is divided into five subtopics under each lesson as follows:
Lesson 1:
Fundamental Concepts and Definitions
Lesson 2:
Rectangular Components of a Force
Lesson 3:
Resultant of Three or More Concurrent Forces
Lesson 4:
Moment of a Force; The Varignon’s Theorem
Lesson 5:
Resultant of Parallel Force System
At the end of this module, the learner is expected to:
Understand the difference of statics and strength of material and the fundamental
concepts and definitions used in the study of this course.
Learn that Forces can be resolved into its components, the X and Y components
using the knowledge in Trigonometry
Learned that the Resultant of Con-current Forces can be solved through the use
of its rectangular components
Learned that the moments of forces can be solved using its components and that
is the application of “Varignon’s Theorem”
Learn how the resultant of Parallel Forces can easily be solved by just summing
up all the forces given and using the principle of moments of a Force in order to
locate this resultant.
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
3
BECCBEC
Solve problems independently involving resultant of concurrent and parallel
force system using the principles of rectangular components and moments of a
Force.
MODULE I: RESULTANT OF COPLANAR FORCE SYSTEM
FUNDAMENTAL CONCEPTS AND DEFINITIONS
Statics is a branch of mechanics that studies the effects and distribution of forces of rigid
bodies which are and remain at rest. In this area of mechanics, the body in which forces
are acting is assumed to be rigid. The deformation of non-rigid bodies is treated
in Strength of Materials.
A RIGID BODY is defined as the definite amount of matter the parts of which are fixed in
position relative to each other. Actually, solid bodies are never rigid; they deform under
the action of applied forces. In many cases, this deformation is negligible compared to the
size of the body and the body may be assumed rigid. Bodies that are made of steel or cast
iron, for example, are of this type. The study of the strength of materials, however, is
based on the deformation (however small) of such bodies.
FORCE. Force may be defined as that which changes, or tends to change the state of
motion of a body. This definition applies to the external effect of a force. The internal
effect of a force is to produce stress and deformation in the body on which the force acts.
External effects of forces are considered in engineering mechanics (statics of rigid
bodies); internal effects in the strength of materials.
The characteristics of a force are (1) it’s the magnitude, (2) the position of its line of
action, and (3) the direction (or sense) in which the force acts along its line of action .
The principle of transmissibility of a force states that the external effect of a force on a
body is the same for all points of application along its line of action; i.e., it is independent
of the point of application. The internal effect of a force, however, is definitely dependent
on its point of application.
FORCE SYSTEMS
A FORCE SYSTEM is any arrangement where two or more forces act on a body or on a
group of related bodies. When the lines of action of all the forces in a force system lie in
one plane, they are referred to as being coplanar; otherwise they are non-coplanar. The
coplanar system is obviously simpler than a non-coplanar system since all the action lines
of the forces lie in the same plane. We shall consider first a discussion of coplanar
systems; it will then be a relatively simple step to the discussion od non-coplanar or space
systems of forces.
The force system are further classified according to their line of action. Forces whose lines
of action pass through a common point are called con-current; those in which the lines of
action are parallel are called parallel force systems; and those in which the lines of action
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
4
BECCBEC
neither are parallel nor intersect in a common point are known as non-concurrent force
systems.
AXIOMS OF MECHANICS
The principles of mechanics are postulated upon several more or less self-evident facts
which cannot be proven mathematically but can only be demonstrated to be true. We
shall call these facts the fundamental axioms of mechanics. The axioms are discussed at
length in subsequent topics as they used. At this time we shall merely collate them for
reference and state them in the following form:
1. The parallelogram law: The resultant of two forces is the diagonal of the
parallelogram formed on the vectors of these forces
2. Two forces are in equilibrium only when equal in magnitude, opposite in direction,
and collinear in action.
3. A set of forces in equilibrium maybe added to any system of forces without
changing the effect of the original system.
4. Action and reaction forces are equal but oppositely directed.
LESSON 1:
RECTANGULAR COMPONENTS OF A FORCE
Forces acting at some angle from the coordinate axes can be resolved into mutually
perpendicular forces called components. The component of a force parallel to the x-axis
is called the x-component, parallel to y-axis the y-component, and so on.
Components of a Force in XY Plane
Figure 1
In order to solve for the X and Y components or the so called horizontal and vertical
components we can use the right triangle formed from the figure above and by using the
trigonometric functions Sine, Cosine, and Tangent, we can easily solved for the value of
Fx, Fy, and the inclination of the force from the horizontal and vertical axes.
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
5
BECCBEC
Note: Free vector is defined as one which does not
show the point of application of the vector, as
distinguished from a localized vector which does.
Figure 1a
Solving for Fx, use the function of Cosine;
Cos 𝜃𝑥 =
𝐹𝑥
𝐹
𝑭𝒙 = 𝑭𝑪𝒐𝒔𝜽𝒙
Solving for Fy, we can use the function of Sine;
𝑆𝑖𝑛 𝜃𝑥 =
𝐹𝑦
𝐹
𝑭𝒚 = 𝑭𝑺𝒊𝒏𝜽𝒙
Now that we have know Fx and Fy we can easily solve the force F by Pythagorean
Theorem using the same right triangle with F as the hypothenuse and Fx and Fy
as the two legs;
𝐹 2 = (𝐹𝑥)2 + (𝐹𝑦)2
𝑭 = √(𝑭𝒙)𝟐 + (𝑭𝒚)𝟐
Solving for the inclination of force F we can use the function of Tangent;
𝑻𝒂𝒏 𝜽𝒙 =
𝑭𝒚
𝑭𝒙
Given the slope of the line of action of the force as v/h
Figure 1b
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
6
BECCBEC
It is more easier if given the slope of the action line line of the force. Take a look at the
small triangle, instead of using the formula Fx=F Cos θx for the horizontal component,
𝒉
we can now replace 𝐂𝐨𝐬 𝜽𝒙 into 𝒓 , thus we can now use
𝑭𝒙 = (𝑭)
𝒉
𝒓
So with the vertical component or Y component;
𝑭𝒚 = (𝑭)
𝒗
𝒓
WORK PROBLEMS #1
Problem 1001: Horizontal and vertical components of planar forces
Determine the x and y components of the forces shown in Fig 1001.
𝑭𝒙 = 𝑭𝑪𝒐𝒔𝜽𝒙
𝑭𝒚 = 𝑭𝑺𝒊𝒏𝜽𝒙
𝒉
𝒓
𝒗
𝑭𝒚 = (𝑭)
𝒓
𝑭𝒙 = (𝑭)
Solution 1001
(Start from the 1st quadrant up to the 4th quadrant)
𝐹𝑥1 = 58 𝑘𝑁 Cos 30° = 𝟓𝟎. 𝟐𝟑 𝒌𝑵
𝐹𝑦1 = 58 𝑘𝑁 Sin 30° = 𝟐𝟗 𝒌𝑵
𝐹𝑥2 = −50 𝑘𝑁 Cos 45° = −𝟑𝟓. 𝟑𝟔 𝒌𝑵
𝐹𝑦2 = 50 𝑘𝑁 Sin 45° = 𝟑𝟓. 𝟑𝟔 𝒌𝑵
5
𝐹𝑥3 = −45 𝑘𝑁 ( ) = −𝟏𝟕. 𝟑𝟏 𝒌𝑵
13
12
𝐹𝑦3 = −45 𝑘𝑁 ( ) = −𝟒𝟏. 𝟓𝟒 𝒌𝑵
13
𝐹𝑥4 = 40 𝑘𝑁 Cos 0° = 𝟒𝟎 𝒌𝑵
𝐹𝑦4 = 40𝑘𝑁 Sin 0° = 𝟎
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
7
BECCBEC
PROBLEM 1002: Components of a force in rotated axes
Find the components in the x, y, u, and v directions of the force P = 10kN shown in
Fig.1002.
𝑭𝒙 = 𝑭 𝐜𝐨𝐬 𝜽𝒙
𝑭𝒚 = 𝑭 𝐬𝐢𝐧 𝜽𝒚
Figure 1002
Solving Px and Py from the X-axis
𝑃𝑥 = 10𝑘𝑁 Cos 60° = 5𝑘𝑁
𝑃𝑦 = 10𝑘𝑁 Sin 60° = 8.66𝑘𝑁
Using the rotated axis (u -v)
𝑃𝑢 = 10𝑘𝑁 Cos 40° = 7.66𝑘𝑁
𝑃𝑣 = 10𝑘𝑁 𝑆𝑖𝑛40° = 6.43𝑘𝑁
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
8
BECCBEC
PROBLEM 1003: Components of a force parallel and perpendicular to the
incline
A block is resting on an incline of slope 5:12 as shown in Fig. P-003. It is subjected to a
force F = 500 N on a slope of 3:4. Determine the components of F parallel and
perpendicular to the incline.
Slope ?
Trigonometric
Functions
Figure 003
Solution 1003
θ = α+β
to solve for β;
5
tan β = 12 = 22.62°
to solve for α;
3
tan α = 4 = 36.87°
𝜃 =𝛽+𝛼
𝜃 = 22.62° + 36.87°
𝜽 = 𝟓𝟗. 𝟒𝟗°
Hint: Take one axis parallel to the incline
𝑭𝒙 = 𝑭 𝐂𝐨𝐬 𝜽𝒙
𝑭𝒙 = 500 𝐶𝑜𝑠 59.49°
𝐹𝑥 = 253.85𝑘𝑁
answer
𝐹𝑦 = −500 𝑆𝑖𝑛 59.49°
𝐹𝑦 = −430.77𝑘𝑁
answer
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
9
BECCBEC
PROBLEM: 1004 Components of force normal and tangent to the
hypotenuse of a triangle
The triangular block shown in Fig. P-004 is subjected to loads P = 1600 lb and F = 600
lb. If AB = 8 in. and BC = 6 in., resolve each load into components normal and tangential
to AC.
Figure 004
(a)
Solution 1004 (Hint: apply the principle of transmissibility of a force)
The principle of transmissibility of a force states that the external
effect of a force on a body is the same for all points of application
along its line of action, i.e., it is independent of the pt. of application.
(b)
(c)
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
10
BECCBEC
Figure (a) shows the slope of the incline, you can use that slope in
order to solve for the components of force F and P. But if you are
not familiar with the slope, you can use the angle of inclination (see
Fig. b) and proceed with your solution using the free body diagram
(FBD) in Figure (c).
𝑭𝒕 = 𝑭 𝐂𝐨𝐬 𝜽𝒙
𝑭𝒏 = −𝑭 𝑺𝒊𝒏 𝜽𝒙
𝐹𝑡 = 600𝑙𝑏 Cos 36.87 ᵒ = 480lb
𝐹𝑛 = −600lb Sin 36.87ᵒ = -360lb
𝑃𝑡 = −1600𝑙𝑏 Cos 53.13 ᵒ = -960lb
𝑃𝑛 = −1600𝑙𝑏 Sin 53.13 ᵒ = -1280lb
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
11
BECCBEC
PROBLEM SET #1:
Instruction: Solve the following problems completely using the theories and principles discussed
in our previous topics. Completely means, the problem, illustrations, and solutions to the
problems are included in your output. Submission is at 8:00 o’clock on the following day the task
was given.
PS1001 (Start your solutions from the 1st quadrant up to the 4th quadrant)
Compute the x and y components of each of the four forces shown in Fig. below.
PS 1002
A force P = 800 N is shown in Figure.
a. Find the y-component of P with respect to x
and y axis.
b. Find the y'-component of P with respect to x'
and y' axis
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
12
BECCBEC
PS 1003
The components of forces are defined by Fx = 300lb and Fy = 200lb, Px = -100lb and Py
= 300lb, Tx = -250lb and Ty = -145lb, and Qx = 150lb and Qy = -350lb. Determine the
magnitude, inclination with the X-axis, and pointing of each forces.
PS 1004
Rework Problem 004 in WORK PROBLEBS #1 if θ = 60ᵒ
PS 1005
The body on the 30° incline in Fig. P-004 is acted upon by a force P inclined at 20° with
the horizontal. If P is resolved into components parallel and perpendicular to incline and
the value of the parallel component is 1800 N, compute the value of the perpendicular
component and that of P. (Hint: Consider X-axis parallel to the incline and the one normal
or perpendicular to the incline is the Y-axis)
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
13
BECCBEC
LESSON 2: RESULTANT OF CONCURRENT FORCE SYSTEM
https://www.youtube.com/watch?v=ColDxGlux3Q
https://www.youtube.com/watch?v=g2XZM2LhfBY
The determination of the resultant of three or more concurrent forces that are not
collinear requires determining the sum of three or more vectors. There are two ways of
accomplishing the addition of three or more vectors; graphically and analytically.
Graphically. Two vectors can be added to give a resultant; this resultant in turn
can be added to a third vector, etc., until all the vectors have been added together to give
an overall resultant. These vectors can be added in any order.
Consider the system of three concurrent forces shown in figure (a). If the
parallelogram method of vector addition is used, forces F and P may be combined to give
a resultant R1 as shown in figure (b) since R1 is equivalent to and replaces F and P, the
original system of three forces now consists of only two: R1 and Q. These may also be
combined by the parallelogram method to give the final resultant R. if the original system
consists of more than three forces, this same technique can be extended to include the
additional forces.
Figure 2001
The same resultant be more readily obtained by the use of free vectors and the
application of the triangle law. Thus in figure 401, by using the free vector P, the resultant
of F and P (i. e., R1) is easily obtained. To this resultant and the free vector Q is added to
give the final resultant R. Observe that R1 need not to be drawn at all, the total resultant
of the system being obtained by joining the tail of the first vector (F) with the tip of the
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
14
BECCBEC
last vector (Q). the same result would be obtained if the order of addition had been P, F,
and Q. In fact, any convenient order of tip-to-tail vector addition may be used.
Figure 2002
Analytically. The vectors can be resolved into components that coincide with
arbitrary chosen axes. The components of each vector with respect to these axes can be
added algebraically, and the resulting additions will be the components of the overall
resultant vector.
Figure 2003
The resultant of a force system is a force or a couple that will have the same effect
to the body, both in translation and rotation, if all the forces are removed and replaced by
the resultant.
Figure 201 can be redrawn as in Figure 202 to show the X and Y components of each force
by projection upon the reference axes. It is apparent that Rx, the X component of R, is
equivalent to the algebraic sum of the X components of F, P, and Q; also that Ry is
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
15
BECCBEC
equivalent to the algebraic sum of the Y components of F, P, anf Q. Denoting such algebraic
summations of the components of the forces by ƩX and ƩY respectively, we have
Rx = ƩX
Ry = ƩY
Having thereby computed the components of the resultant R, we can now determine its
magnitude and inclination.
𝟐
𝑹 = √(∑ 𝑿) + (∑ 𝒀)
𝐭𝐚𝐧 𝜽 =
𝟐
∑𝒀
∑𝑿
The pointing of R is determined by the signs of its rectangular components ∑ 𝑿 and
∑ 𝒀.
The equation involving the resultant of force system are the following
1. Rx = ΣX = Fx + Px + Qx
The x-component of the resultant is equal to the summation of forces in the xdirection.
2. Ry = ΣY = Fy + Py + Qy
The y-component of the resultant is equal to the summation of forces in the ydirection.
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
16
BECCBEC
WORK PROBLEMS #2
Problem 2001: Resultant of Concurrent Force System
Determine completely the resultant of concurrent force system shown
Fig. 2001
Solution 2001 (Start the solutions from the 1st quadrant up to the 4th quadrant)
X – COMPONENTS
𝐹𝑥1 = 58 𝑘𝑁 Cos 30° = 𝟓𝟎. 𝟐𝟑 𝒌𝑵
𝐹𝑥2 = −50 𝑘𝑁 Cos 45° = −𝟑𝟓. 𝟑𝟔 𝒌𝑵
5
𝐹𝑥3 = −45 𝑘𝑁 (13) = −𝟏𝟕. 𝟑𝟏 𝒌𝑵
𝐹𝑥4 = 40 𝑘𝑁 Cos 0° = 𝟒𝟎 𝒌𝑵
∑ 𝑿 = 𝟑𝟕. 𝟓𝟔 𝒌𝑵
Y-COMPONENTS
𝐹𝑦1 = 58 𝑘𝑁 Sin 30° = 𝟐𝟗 𝒌𝑵
𝐹𝑦2 = 50 𝑘𝑁 Sin 45° = 𝟑𝟓. 𝟑𝟔 𝒌𝑵
12
𝐹𝑦3 = −45 𝑘𝑁 (13) = −𝟒𝟏. 𝟓𝟒 𝒌𝑵
𝐹𝑦4 = 40𝑘𝑁 Sin 0° = 𝟎
∑ 𝒀 = 𝟐𝟐. 𝟖𝟐
𝑅 = √(𝟑𝟕. 𝟓𝟔2 + (𝟐𝟐. 𝟖𝟐2
𝑹 = 𝟒𝟑. 𝟗𝟓 𝒌𝑵
∑𝑌
22.82
tan 𝜃 = ∑ 𝑋 ; tan 𝜃 = 37.56 ; 𝜽 = 𝟑𝟏. 𝟐𝟖°
Therefore the magnitude of the resultant is 43.95kN located on the 1st quadrant with an
angle of inclination from the horizontal of 31.28Type equation here.
BEC 122-STATICS OF RIGID BODIES
Engr. Maila R. Pedrigal-Ragasa
17
BECCBEC
Problem 2002: Resultant of two velocity vectors
Find the resultant vector of vectors A and B shown in Fig. 2002.
2
Fig. 2002
Solution 2002: Component Method
Rx = ΣX
Rx = (44 m/sec) Cos 50ᵒ − (17 m/sec) Cos 30ᵒ
Rx = 13.56 m/sec
to the right
Ry = ΣY
Ry = (−44 m/sec) Sin 50ᵒ − (17 m/sec) sin 30ᵒ
Ry = −42.21 m/sec downward
𝟐
𝟐
𝑹 = √(∑ 𝑿) + (∑ 𝒀)
𝑅 = √(13.56)2 + (42.21)2
𝑅 = 44.33 𝑚/𝑠𝑒𝑐
∑𝒀
𝐭𝐚𝐧 𝜽 = ∑ 𝑿
𝒕𝒂𝒏 𝜽 =
42.21
13.56
𝜽 = 𝟕𝟐. 𝟏𝟗ᵒ
The resultant vector R = 44.33 m/sec downward to the right at θx = 72.19°
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
17
18
BECCBEC
Problem 2003: Solving for a force and its angle and angle of two forces with
given resultant
Forces F, P, and T are concurrent and acting in the direction as shown in Fig. 2003.
Fig. 2003
a. Find the value of F and α if 𝑇 = 450𝑁, 𝑃 = 250𝑁, 𝛽 = 30° and the resultant is
300 N acting up along the y-axis.
b. Find the value of F and α if 𝑇 = 450𝑁, 𝑃 = 250𝑁, 𝛽 = 30° and the resultant is
zero.
c. Find the value of α and β if 𝑇 = 450𝑁, 𝑃 = 250𝑁, 𝐹 = 350𝑁, and the resultant is
zero.
Solution 2003
Part a: Unknown force and direction with non-zero resultant
Rx = 0 and Ry = 300 N
Rx = ƩX
Ry = ƩY
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
18
19
BECCBEC
Rx = 0 and Ry = 300 N
Rx = ΣX 𝜶𝐴 = 𝜋𝑟 2
Since Rx = 0, we have;
0 = 𝐹 cos 𝛼 + 250 cos 30° − 450
𝐹 cos 𝛼 = 233.49
𝑭=
𝟐𝟑𝟑.𝟒𝟗
𝐜𝐨𝐬 𝜶
equation 1
Ry = ΣY
300 = 𝐹 sin 𝛼 − 250 sin 30°
𝐹 sin 𝛼 = 425 equation 2
Substitute the value of F in equation 1 to equation 2 in order to solve for α
𝐹 sin 𝛼 = 425
233.49
sin 𝛼 = 425
cos 𝛼
233.49 tan 𝛼 = 425
425
tan 𝛼 = 233.49
𝛼 = 𝑇𝑎𝑛−1 1.8202
𝜶 = 𝟔𝟏. 𝟐𝟐° answer
Go back to equation 1 in order to solve for F
𝟐𝟑𝟑.𝟒𝟗
𝑭 = 𝑪𝒐𝒔 𝟔𝟏.𝟐𝟐ᵒ
equation 1
𝑭 = 𝟒𝟖𝟒. 𝟗𝟐𝑵answer
Part b: Unknown force and direction with zero resultant
Rx = 0 and Ry = 0
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
19
20
BECCBEC
0 = 𝐹 cos 𝛼 + 250 cos 30° − 450
𝐹 cos 𝛼 = 233.49
𝑭=
𝟐𝟑𝟑.𝟒𝟗
equation 1
𝑪𝒐𝒔𝜶
Ry = ΣY
0 = 𝐹 sin 𝛼 − 250 sin 30°
𝑭 𝐬𝐢𝐧 𝜶 = 𝟏𝟐𝟓
equation 2
Substitute the value of F in equation 1 to equation 2 in order to solve for 𝜶
𝑭 𝐬𝐢𝐧 𝜶 = 𝟏𝟐𝟓
233.49
sin 𝛼 = 125
cos 𝛼
233.49 tan 𝛼 = 125
𝛼 = 𝑇𝑎𝑛−1 0.5354
answer
Go back to equation 1 in order to solve for F
𝜶 = 𝟐𝟖. 𝟏𝟔°
𝑭=
𝑭=
𝟐𝟑𝟑.𝟒𝟗
𝑪𝒐𝒔𝜶
equation 1
𝟐𝟑𝟑. 𝟒𝟗
𝑪𝒐𝒔 𝟐𝟖. 𝟏𝟔ᵒ
F = 264.85 N
answer
Part c: Unknown direction of two forces with zero resultant
Rx = 0 and Ry = 0
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
20
21
BECCBEC
Ry = ΣY
0 = 350 sin 𝛼 − 250 sin 𝛽
Since 350 and 250 is divisible by 50
Then,
7 sin 𝛼 − 5 sin 𝛽 = 0
7 sin 𝛼 = 5 sin 𝛽
49𝑆𝑖𝑛2 𝛼 = 5𝑆𝑖𝑛2 𝛽
equation 1
Rx = ΣX
0 = 350 cos 𝛼 + 250 cos 𝛽 − 450
7 cos 𝛼 + 5 cos 𝛽 − 9 = 0
7 cos 𝛼 = 9 − 5 cos 𝛼
(7 cos 𝛼 = 9 − 5 cos 𝛼)²
49𝐶𝑜𝑠 2 𝛼 = (9 − 5 cos 𝛼)²
49 Cos²α = 81 – 90 Cosβ + 25 Cos²β
→ Equation (2)
Remember some Trigonometric
Identities
Sin²α + Cos²α = 1
Equation (1) + Equation (2)
49 sin² α=25 sin² β
49 Cos²α = 81 – 90 Cosβ + 25 Cos²β
→ Equation (1)
→ Equation (2)
49 Sin² α + 49 Cos² α = 25 Sin² β + (81 – 90 Cosβ + 25 Cos² β)
49(Sin² α + Cos² α) = 25(Sin² β + Cos² β) + 81 − 90 cosβ
49(1)=25(1)+81−90 Cosβ
90 Cosβ = 25+81−49
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
21
22
BECCBEC
57
90
−1
𝛽 = 𝐶𝑜𝑠 (0.633333)
𝜷 = 𝟓𝟎. 𝟕𝟎ᵒ
answer
𝐶𝑜𝑠 𝛽 =
From Equation (1)
49 Sin 2α =25 Sin 250.70°
7 Sinα = 5 sin 50.70°
5𝑆𝑖𝑛 50.70°
sin 𝛼 =
7
−1
𝛼 = 𝑆𝑖𝑛 (0.552743)
α = 33.56°
answer
Another Solution for Part c
By Cosine Law
(250)² = (350)² + (450)² −
2(350)(450)cosα
(350)2 + (450)2 − (250)²
𝐶𝑜𝑠𝛼 =
2(350)(450)
α = 33.557ᵒ answer
(350)² = (250)² + (450)² − 2(250)(450)cosβ
𝐶𝑜𝑠𝛽 =
(250)2 + (450)2 − (350)²
2(250)(450)
Β = 50.704°
answer
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
22
23
BECCBEC
LESSON 3: MOMENT OF A FORCE
Moment is the measure of the capacity or ability of the force to produce twisting or
turning effect about an axis. This axis is perpendicular to the plane containing the line of
action of the force. The magnitude of moment is equal to the product of the force and the
perpendicular distance from the axis to the line of action of the force. The intersection of
the plane and the axis is commonly called the moment center, and the perpendicular
distance from the moment center to the line of action of the force is called moment arm.
Figure 3001
From the figure above, O is the moment center and d is the moment arm. The moment M
of force F about point O is equal to the product of F and d.
M=Fd
The principle of Moments. Varignon’s Theorem
Varignon’s Theorem states that, “the moment of a force is equivalent to the sum of the
moment of its components.”
Applications. In some cases it is more convenient tom determine the moment of a force
from the sum of the moments of its components rather than from the force itself. For
example, in figure 2, suppose a force F making an angle θ with the X axis, passes through
a point A having the coordinates (x, y). In this case it is inconvenient to calculate the
moment arm d. By resolving the force into its components Fx and Fy at A, the moment
arm of Fx about O is the coordinate distance y, and the moment arm of Fy about O is the
coordinate distance x. Then the moment of F is expressed by
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
23
24
BECCBEC
Figure 3002
+ 𝑴𝒐 = 𝑭 ∗ 𝒅 = 𝑭𝒙 ∗ 𝒚 − 𝑭𝒚 ∗ 𝒙
= 𝑭𝒙 ∗ 𝒚 − 𝑭 𝒚 ∗ 𝒙
from which the value of the moment arm d may be computed if desired
The intercepts of the line of action of F with the X and Y axes may also be computed
from the principle of moments. Replacing F by its components at B and at C in Fig. 2, we
have
+ 𝑀𝑜 = 𝐹𝑥 ∗ 𝑖𝑦
and
+ 𝑀𝑜 = 𝐹𝑦 ∗ 𝑖𝑥
Note that Fy at B and Fx at C both have zero moment about O since they both pass through
O and therefore have zero moment arms. Having already determined the moment of F by
means of Eq. (a), the intercepts iy and ix are now readily computed from Eq. (b)
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
24
25
BECCBEC
WORK PROBLEMS #3
Problem 3001: Moment of force about different points
In Figure 3001 assuming clockwise moments as positive, compute the moment of force
F = 450kN and force P = 300 kN about points A, B, C, and D.
Figure 3001
Solution 3001
Moment of force F about points A, B, C, and D:
4
3
Fh = 450kN ⌊5⌋
𝐹𝑉 = 450𝑘𝑁 [5]
𝑭𝒉 = 𝟑𝟔𝟎 𝒌𝑵
𝑭𝑽 = 𝟐𝟕𝟎𝒌𝑵
+ 𝐌𝐀 = −Fx ∗ y − Fy ∗ x
𝐌𝐀 = −360kN (0.90m) −
270kN (0.30m)
𝑀𝐴 = −405 𝑘𝑁𝑚
+ 𝐌𝐁 = Fx ∗ y + Fy ∗ x
𝐌𝐁 = 360 kN (0.90m) +
270 kN (1.20m)
𝑴𝑩 = 𝟔𝟒𝟖 𝒌𝑵 − 𝒎
+ 𝑀𝐶 = 𝐹𝑦 ∗ 𝑥
MC = 270kN (1.50m)
𝑴𝑪 = 𝟒𝟎𝟓𝒌𝑵 − 𝒎
Moment @ C is due to Fy only because Fx intersect @ the moment
center
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
25
26
BECCBEC
+ MD = Fx ∗ y − Fy ∗ x
MD = 360kN(0.90m) − 270kN (0.30m)
𝑴𝑫 = 𝟑𝟐𝟑. 𝟏𝟗 𝒌𝑵 − 𝒎
Note: Force P will be part of your Problem Set #2
Problem 3002: Moment of resultant force about a point
Two forces P and Q pass through a point A which is 4 m to the right of and 3 m above a moment
center O. Force P is 890 N directed up to the right at 30° with the horizontal and force Q is 445 N
directed up to the left at 60° with the horizontal. Determine the moment of the resultant of these
two forces with respect to O.
Solution 3002
𝑹𝒙 = 𝑷𝒙 − 𝑸𝒙
𝑅𝑥 = 890 cos 30° − 445 cos 60°
Rx = 548.26 N (to the right)
𝑹𝒚 = 𝑷𝒚 + 𝑸𝒚
𝑅𝑦 = 890 sin 30° + 445 sin 60°
Ry = 830.38 N (upward)
𝑴𝑶 = 𝟒𝑹𝒚 − 𝟑𝑹𝒙
𝑀𝑂 = 4(830.38) − 3(548.26)
Mo = 1676.74 N⋅m (counterclockwise)
answer
The moment of resultant about O can be solved actually without the use of Rx and
Ry. The moment effect of the components of R is the same as the combined moment effect
of the components P and Q. Thus, MO = 4Py + 4Qy + 3Qx − 3Px. Try it.
You can also find Mo by finding the magnitude of R and its moment arm about point
O. Moment arm is the perpendicular distance between the line of action of R and point O.
Problem 3003: Intercepts of the resultant force
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
26
27
BECCBEC
Without computing the magnitude of the resultant, compute where the resultant of the
forces shown in Fig. 3003- intersects the x and y axes.
Figure 3003
Solution 3003
1
𝑃𝑥 = 500 ( ) = 353.55 lb
to the right
𝑃𝑦 = 500 ( ) = 353.55 lb
upward
√2
1
𝑄𝑥 =361(
𝑄𝑦 =361(
√2
3
) = 300.37 lb
to the right
) = 200.25 lb
downward
√13
12
√13
𝑀𝑂 = 5𝑄𝑦 = (200.25𝑙𝑏)5𝑖𝑛
𝑀𝑂 = 1001.25 𝑙𝑏. 𝑖𝑛
clockwise
𝑅𝑥 = 𝑃𝑥 + 𝑄𝑥
𝑅𝑥 = 353.55 + 300.37
𝑅𝑥 = 653.92 𝑙𝑏
to the right
𝑅𝑦 = 𝑃𝑦 − 𝑄𝑦
𝑅𝑦 = 353.55 − 200.25
𝑅𝑦 = 153.3 𝑙𝑏
𝑢𝑝𝑤𝑎𝑟𝑑
x-intercept of the resultant
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
27
28
BECCBEC
𝒂𝑹𝒚 = 𝑴𝑶
𝑎(153.3) = 1001.25
a = 6.53 in to the left of point O
answer
y- intercept of the resultant
𝒃𝑹𝑿 = 𝑴𝑶
𝒃(𝟔𝟓𝟑. 𝟗𝟐) = 𝟏𝟎𝟎𝟏. 𝟐𝟓
b =1.53 in
above point O
answer
Problem 3004: Y-coordinate of the point of application of the force
In Fig.3004, find the y-coordinate of point A so that the 361-lb force will have a clockwise
moment of 400 ft-lb about O. Also determine the X and Y intercepts of the line of action
of the force.
Figure 3004
Solution 3004
𝑀𝑜 = 𝑦𝐴 𝐹𝑥 − 𝑥𝐴 𝐹𝑦
3
2
) − 2(361) (
)
400 = 𝑦𝐴 (361) (
√13
√13
𝒚𝑨 = 𝟐. 𝟔𝟔𝟓 𝐟𝐭
answer
Y - intercept of the line of action of force F
MO = Fxb
400 = 361
3
√13
𝑏
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
28
29
BECCBEC
𝑏=
400√13
(361)3
b =1.332 ft above point O
answer
X-intercept of the line of action of force F
𝑀𝑜 = 𝐹𝑦 . 𝑎
400 = (361
2
)a
√13
𝑎 = 1.998𝑓𝑡
to the left of point O
answer
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
29
30
BECCBEC
LESSON 4: RESULTANT OF PARALLEL FORCES
Resultant of Parallel Force System
Coplanar Parallel Force System
Parallel forces can be in the same or in opposite directions. The sign of the direction can
be chosen arbitrarily, meaning, taking one direction as positive makes the opposite
direction negative. The complete definition of the resultant is according to its magnitude,
direction, and line of action.
𝑅 = ∑ 𝐹 = 𝐹1 + 𝐹2 + 𝐹3 + ⋯
𝑅. 𝑑 = ∑ 𝑀𝐹
𝑅 . 𝑑 = ∑ 𝐹 𝑋 = 𝐹1 𝑋1 + 𝐹2 𝑋2 + 𝐹3 𝑋3 + ⋯
RESULTANT OF DISTRIBUTED LOADS
The resultant of a distributed load is equal to the area of the load diagram. It is acting at
the centroid of that area as indicated. The figure below shows the three common
distributed loads namely; rectangular load, triangular load, and trapezoidal load.
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
30
31
BECCBEC
Rectangular Load
𝑅 = 𝑤𝑜 𝐿
Triangular Load
1
𝑅 = 𝑤𝑜 𝐿
2
Trapezoidal Load
1
𝑅 = 𝑤𝑜 𝐿 + 2 𝑤𝑜 𝐿
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
31
32
BECCBEC
WORK PROBLEMS #4
Problem 4001: Computation of the resultant of parallel forces acting on the
lever
A parallel force system acts on the lever shown in Fig. P-3001. Determine the magnitude
and position of the resultant.
Figure 301
Solution 3001
R = ΣF
R= (- 30 - 60 + 20 – 40) lb
R= -110 lb (downward)
+ MA=ΣMA
R. d = 2(30) + 5(60) − 7(20) + 11(40)
(110)d = 660 ft⋅lb
d=6 ft
to the right of A
Thus, R = 110 lb downward at 6 ft to the right of A.
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
answer
32
33
BECCBEC
Problem 4002: Finding the resultant of parallel forces acting on both sides
of the rocker arm
Determine the resultant of the four parallel forces acting on the rocker arm of Fig. P3002
Figure 3002
Solution 3002
R = ΣF ( + , -)
R = (-50 + 40 + 20 - 60) lb
R = -50 lb
downward
R.d = Ʃ Mo
( 50lb) d = −50lb(6ft) + 40lb(2ft) – 20lb(3ft) + 60lb(8ft)
50d = 200 lb⋅ft
d=4 ft
to the right of O
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
33
34
BECCBEC
Problem 4003: Finding the resultant of vertical forces acting on the Fink
truss
Locate the amount and position of the resultant of the loads acting on the Fink truss in
Fig. 3003.
Figure 303
Solution 4003
Since the truss is symmetrical and the load applied at the top chord is also symmetrical,
you can replace those loads/forces by an equivalent single force F that act at the center
of the truss which is 4.50m at both supports.
Magnitude of resultant
R =ΣFv
R= (-6230 - 8900) N
R=-15130 N downward
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
34
35
BECCBEC
Location of resultant
𝑹 ∗ 𝒅 = Ʃ𝑴𝑨
(15130)d = 6230N(4.5m) + 8900N(3m)
d=3.62 md=3.62 m to the right of A
Thus, R = 15 130 N downward at 3.62 m to the right of left support.
answer
PROBLEM SET #4
PS 4001: Y-coordinate of the point of application of the force
In Fig. shown below, find the y-coordinate of point A so that the 361-lb force will have a
clockwise moment of 400 ft-lb about O. Also determine the X and Y intercepts of the line
of action of the force.
PS 4002: Finding the unknown two forces with given resultant
Find the value of P and F so that the four forces shown in Fig. below produce an upward
resultant of 400 lb acting at 4 ft from the left end of the bar.
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
35
36
BECCBEC
PS 4003: Finding the magnitude and position of the missing force
The resultant of three parallel loads (one is missing in Fig below) is 80kN acting up at 3.5
m to the right of A. Compute the magnitude and position of the missing load.
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
36
37
BECCBEC
MODULE 2: Equilibrium of Force System
Definition and Meaning of Equilibrium
The subject matter of statics, as its name implies, deals essentially with the action
of forces on bodies, which are at rest. Such bodies are said to be in equilibrium.
Specifically, equilibrium is the term used to designate the condition where 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 a zero resultant. The physical meaning of equilibrium, as applied to a
body, is that the body either is at rest or is moving in a straight line with constant velocity.
In this module, the students shall determine and apply the conditions necessary
to produce equilibrium for coplanar force systems. The principles and techniques
developed in this module are the basic fundamentals of static; the student is urged to
master them. Actually what is studied is a method of reasoning – of learning how to apply
the basic concepts of the components of a force and its moment effect in the most efficient
manner.
Free-Body Diagram
An isolated view of a body which shows only the external forces exerted on the body is
called a free-body diagram (frequently abbreviated as FBD).
LESSON 1: EQUILIBRIUM OF CONCURRENT FORCE SYSTEM
Conditions of Static Equilibrium of Concurrent Forces
The sum of all forces in the x-direction or horizontal is zero.
ΣX = 0 or ΣFx=0
The sum of all forces in the y-direction or vertical is zero.
ΣY = 0 or ΣFV=0
Important Points for Equilibrium Forces
•
Two forces are in equilibrium if they are equal and oppositely directed.
•
Three coplanar forces in equilibrium are concurrent.
•
Three or more concurrent forces in equilibrium form a close polygon when
connected in head-to-tail manner.
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
37
38
BECCBEC
WORK PROBLEMS #5
Problem 5001 | Equilibrium of Concurrent Force System
The cable and boom shown in Fig. P-5001 support a load of 600 lb. Determine the
tensile force T in the cable and the compressive for C in the boom.
Fig. P-5001
Solution 5001
ΣX = 0
Ccos45ᵒ - Tcos30ᵒ = 0
Ccos45ᵒ = Tcos30ᵒ
𝐶=
𝑇 cos 30°
cos 45°
eq. 1
ΣY = 0
Tsin30ᵒ + Csin45ᵒ - 600 = 0
equ. 2
Substituting the value of equ. 1 to equ. 2
Tsin30ᵒ +
𝑇 cos 30°
cos 45°
sin45ᵒ = 600
Tsin30ᵒ + T Cos 30ᵒTan 45ᵒ = 600
1.366T = 600
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
38
39
BECCBEC
T=439.24 lb
answer ; back to equation 1 then substitute the value
of T
𝐶=
𝑇 cos 30°
cos 45°
𝟒𝟑𝟗. 𝟐𝟒 cos 30°
𝐶=
cos 45°
C = 537.94 lb answer
Problem 5002-5003 | Equilibrium of Concurrent Force System
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 Figure P-502, determine the value of P and the normal
pressure N exerted by the plane.
Figure P-5002
Solution 5002
ΣX = 0
Pcosθ – WCos 60ᵒ = 0
P Cos 45ᵒ = 300Cos 60ᵒ
300Cos 60ᵒ
𝑃=
Cos 45ᵒ
P=212.13 lb answer
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
39
40
BECCBEC
To solve for the normal pressure N
ΣY = 0
N – P Sinθ – W Sin 60ᵒ = 0
N = 212.13 Sin45ᵒ + 300 Cos60ᵒ
N=409.81 lb
answer
Problem 5003
If the value of P in Fig. 5002 is 175 lb, determine the angle θ at which it must be inclined
with the smooth plane to hold 300-lb box in equilibrium.
Problem 5004 | Equilibrium of Concurrent Force System
The 300-lb force and the 400-lb force shown in Fig. P-5004 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.
Solution 5004
ƩX = 0
400𝑙𝑏 𝐶𝑜𝑠 30° − 300𝑙𝑏 + 𝐹𝐶𝑜𝑠 𝜃 = 0
𝐹𝐶𝑜𝑠 𝜃 = −346.41 + 300𝑙𝑏
𝐹𝐶𝑜𝑠 𝜃 = −46.41
−46.41
𝐹 = 𝐶𝑜𝑠 𝜃
equ. 1
Figure 504
ƩY = 0
400𝑙𝑏 𝑆𝑖𝑛 30° − 𝐹𝑆𝑖𝑛 𝜃 = 0
200𝑙𝑏 = 𝐹𝑆𝑖𝑛 𝜃 (substitute equation 1 in equ. 2)
−46.41
200𝑙𝑏 =
𝑆𝑖𝑛 𝜃
𝐶𝑜𝑠 𝜃
tan 𝜃 =
200𝑙𝑏
−46.41
= 4.31
𝛉 = 𝐭𝐚𝐧−𝟏 𝟒. 𝟑𝟏 = −𝟕𝟔. 𝟗𝟒°
(go back to equ. 1 and substitute the value of θ in order to solve for the magnitude of
force F)
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
40
41
BECCBEC
−46.41
𝐹 = −𝐶𝑜𝑠 76.94°
𝑭 = 𝟐𝟎𝟓. 𝟑𝟖𝒍𝒃
The correct position of F should be on the
third quadrant.
to further check the solution, the resultant
must be equal to zero
ƩX = 0
ƩX = 400 Cos30ᵒ - 300 -205.38 Cos 76.94ᵒ
= 346.41lb -300lb -46.41lb
ƩY = 0
ƩY = 400 Sin 30ᵒ - 205.38 Sin 76.94ᵒƩY = 0
Since ƩX and ƩY are equal to zero, therefore the resultant R is also equal to zero.
Problem 5005 | Equilibrium of Concurrent Force System
The system of knotted cords shown in Fig. P-5005 support the indicated weights.
Compute the tensile force in each cord.
Fig. P-5005
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
41
42
BECCBEC
Solution 5005
Since we have only two equations that satisfies the equilibrium of Concurrent Force
System solve first the forces where 400-lb load is hanging.
ΣFH = 0
𝐷𝐶𝑜𝑠15° − 𝐶𝐶𝑜𝑠60° = 0
𝐂 𝐜𝐨𝐬 𝟔𝟎°
𝑫 = 𝐜𝐨𝐬 𝟏𝟓° equ. 1
ΣFV = 0
𝐷𝑠𝑖𝑛15° + 𝐶𝑆𝑖𝑛60° − 400𝑙𝑏 = 0
equ.2
Subst. the value of D fr. equ. 1 to equ. 2 to
solve for C
𝐶𝐶𝑜𝑠60°
𝑆𝑖𝑛15° + 𝐶𝑆𝑖𝑛60° = 400
𝐶𝑜𝑠15°
𝐶𝑜𝑠60°𝑇𝑎𝑛15° + 𝐶𝑆𝑖𝑛60° = 400
𝑪 = 𝟒𝟎𝟎𝒍𝒃
𝒂𝒏𝒔𝒘𝒆𝒓
go back to equ. 1 to solve for D
𝐷=
400lb cos 60°
cos 15°
𝑫 = 𝟐𝟎𝟕. 𝟎𝟔𝒍𝒃
𝒂𝒏𝒔𝒘𝒆𝒓
From the knot where 300-lb load is hanging
ΣFV = 0
𝐵𝑆𝑖𝑛45° − 400𝑙𝑏𝑆𝑖𝑛60° = 300𝑙𝑏
𝐵𝑆𝑖𝑛45° = 346.41𝑙𝑏 + 300𝑙𝑏
𝑩 = 𝟗𝟏𝟒. 𝟏𝟔𝒍𝒃 𝒂𝒏𝒔𝒘𝒆𝒓
ΣFH = 0
−𝐴 + 𝐵𝐶𝑜𝑠45° + 𝐶𝐶𝑜𝑠60° = 0 equ.2
−𝐴 + 914.16𝑙𝑏𝐶𝑜𝑠45° + 400𝑙𝑏𝐶𝑜𝑠60° = 0
𝑨 = 𝟖𝟒𝟔. 𝟒𝟏𝒍𝒃 𝒂𝒏𝒔𝒘𝒆𝒓
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
42
43
BECCBEC
Problem Set #5 | Equilibrium of Concurrent Force System
PS 5001: From Problem 5002 in the examples; if the value of P is 260 lb, determine the
angle θ at which it must be inclined with the smooth plane to hold 450-lb box in
equilibrium.
PS 5002: In the figure below, determine the magnitude of P and F necessary to keep the
concurrent force system in equilibrium.
PS 5003
Determine the values of α and θ so that the forces shown in Fig. below will be in
equilibrium.
PS 5004: A system of cords knotted together at A and B support the weights shown in
Fig. Compute the tensions P, Q, F, and T acting in the various cords. Determine the values
of α and θ so that the forces shown in Fig. P-316 will be in equilibrium
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
43
44
BECCBEC
LESSON 2: EQUILIBRIUM OF PARALLEL FORCE SYSTEM
Conditions for Equilibrium of Parallel Forces
The sum of all the forces is zero.
ΣF = 0
The sum of moment at any point O is zero.
ΣMo = 0
Problem 6001 | Equilibrium of Parallel Force System
Determine the reactions for the beam shown in Fig. P-601
Fig. P-6001
Solution 6001
ΣMR2=0
+
10R1 + 4(400) - 16(300) – 9(1400) = 0
10R1 = - 4(400) + 16(300) + 9(1400)
R1=1580 lb
answer
Since R1 has already solved, it is more
easier to solve R2 by ƩFv = 0. Assuming all upward forces are positive and downward
forces are negative
ƩFv = 0
R2 +1580 lb – 400lb - 1400lb - 300lb = 0
R2 = -1580 lb + 2100lb
R2 = 520 lb
answer
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
44
45
BECCBEC
Problem 6002 | Equilibrium of Parallel Force System
Determine the reactions R1 and R2 of the beam in Fig. P-507 loaded with a concentrated
load of 1600 lb and a load varying from zero to an intensity of 400 lb per ft.
Fig. P-6002
Solution 6002
Solve first for the reactions of the upper beam
carrying a uniformly varying load (triangular
load) with an intensity of loading from zero to
400lb/ft.
∑ 𝑀𝑅4 = 0
12𝑅3 − 4(2400) = 0
12𝑅3 = 4(2400)
𝑹𝟑 = 800 lb
Therefore, by summing up forces vertical;
𝑹𝟒 = 1600 lb
Solving for the lower supports, we have;
∑ 𝑀𝑅2 = 0
16𝑅1 − 12(800) − 13(1600) = 0
16𝑅1 = 12(800) + 13(1600)
𝑹𝟏 =𝟏𝟗𝟎𝟎 𝒍𝒃
answer
Therefore, by summing up forces vertical;
𝑹𝟐 = 𝟐𝟏𝟎𝟎 𝒍𝒃
answer
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
45
46
BECCBEC
Problem 6003 | Equilibrium of Parallel Force System
The roof truss in Fig. 6003 is supported by a roller at A and a hinge at B. Find the values
of the reactions.
Fig. P-6003
Solution 6003
Note: Take the advantage of the symmetry of the truss and the loadings on the rafters
(top chord), the only unsymmetrical is the load on the bottom chord. Now, since the top
chord is symmetrically loaded, you can replace the 3-20 kN forces and 2-10 kN forces by
a single 80 kN force acting at the center of the truss.
ΣMB = 0
+
15𝑅𝐴 - 10(60) - 7.5(80) - 5(50) = 0
15𝑅𝐴 = 1450 lb
𝑹𝑨 = 𝟗𝟔. 𝟔𝟕𝒌𝑵
answer
Since 𝑹𝑨 has been solve and only 𝑹𝑩 is unknown you can use
ƩFv=0 (assuming upward forces positive and downward forces negative)
𝑅𝐵 + 96.67 − 80 − 60 − 50 = 0
𝑹𝑩 = −𝟗𝟑. 𝟑𝟑𝒌𝑵
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
46
47
BECCBEC
Problem 6004 | Equilibrium of Parallel Force System
etermine the reactions for the beam loaded as shown in Fig. 6004.
Fig. 6004
Solution 6004
ΣMR2 = 0
+
7.5R1 - 6(12) + 4.5(18) + 1(22.5) = 0
R1 = 23.4 kN
answer
∑ 𝐹𝑣 = 0 to solve for R2
𝑅1 + 𝑅2 − 12 − 18 − 22.5 = 0
23.4 + 𝑅2 − 12 − 18 − 22.5 = 0
𝑅2 = 12 + 18 + 22.5 − 23.4
R2 = 29.1 kN
answer
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
47
48
BECCBEC
Problem Set #6 | Equilibrium of Parallel Force System
Find the reactions of the following beams loaded as shown.
PS 6001
PS 6002
PS 6003
PS 6004
BEC 122 Statics of Rigid Bodies
Engr.Maila R. Pedrigal-Ragasa, MAME
48