PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
LEARNING
MODULE 02: –
Resultant of Force
System
AMT 3102ENGINEERING MECHANICS
Prepared by:
ENGR. CARMELITA C. ARBOZO
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
TABLE OF CONTENTS
Title
Introduction
Component of Forces
➢ Example Solved Problems
Resultant of Concurrent Force System
➢ Example Solved Problems
Moment of a Force
➢ Mathematical Definition of a Moment
➢ Direction of Moment
➢ Resultant Moment
➢ Principle of Moments
➢ Example Solved Problems
Resultant of Parallel Forces
➢ Resultant of Distributed Loads
➢ Example Solved Problems
Couples
➢ Example Solved Problems
➢ Plate No. 2 Resultant of Force System
Page
Allotted time
4, 5
6
7,8,9
10
11,12,13,14
15
15
16
16
17
18, 19
20
21
21,22,23,24
25
25,26,27,28
29, 30
5minutes
10minutes
60minutes
10 minutes
75 minutes
10 minutes
30 minutes
15 minutes
15 minutes
15 minutes
45 minutes
10 minutes
15minutes
75 minutes
10minutes
60 minutes
140 minutes
600 minutes
TABLES OF REFERENCES
References
No.
Materials (textbooks, references, journals, online)
Singer, Ferdinand L., Engineering Mechanics: Statics and Dynamics
Mcgill, David J., Engineering Mechanics: Statics and an Intro to Dynamics
Singer, F.L (1954). Engineering Mechanics (2nd Ed.) New York: Harpens
and Row, Publishers, Inc.
Hibbeler, R.C. (2010) Engineering Mechanics (12th Ed.) New Jersey:
Pearson Prentice Hall
Beer,F.P. & Johnston, Jr.E.R.(2016) Vector Mechanics for Engineers Static
(11th Ed.) New York: McGraw-Hill Education
1
2
3
4
5
6
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
This module discusses Force and Components, Resultant of Coplanar Concurrent
Force System, Moment of a Force, Resultant of Parallel Forces, Couples, Resultant
of Non-Concurrent Forces.
LEARNING OUTCOMES
Course Learning Outcomes [CLO]
Module Learning Outcomes [MLO]
CLO1. Apply the knowledge of
mathematics to solve complex MLO 01. Solve problems of three or more
Topic Learning Outcomes [TLO]
mechanics problems;
concurrent forces by using the rectangular
CLO2. Utilize scientific concepts, laws components of a vector to determine
and theories in solving problems related magnitude and direction.
TLO 04.Resolve forces into their
MLO 02. Fofrmulate the moment of a force rectangular components and represent
to their field of specialization;
about a secified axis through hands-on
the net effect of different force systems
CLO3. Convey a general understanding problem solving.
into a resultant through problems
of engineering mechanics as a way to
03. Determine resultants of parallel solving.
associate their knowledge in their MLO
forces by demosntarting the proper
chosen course;
TLO 05. Compute moment of force
procedures.
CLO 10. Demonstrate honesty in doing MLO 04. Computing resolution of force into about a specified axis by demonstrating
individual/group work on this course.
a force and a couple through sample the proper method.
problems
TLO 06. Solve resultant force of nonconcurrent forces through practice
problems.
HONESTY CLAUSE
As a state college, students are expected to uphold and integrity, principle and selfrespect, using their knowledge and skills for the enhancement of human welfare and
environment; being honest and fair in their class activity, requirements and other
projects will not engage in cheating or plagiarism. The institution undertakes as a
modest and nominal ideal of behaviors in academic matters that students be
straightforward and that they distribute for deposit solely the produce of their particular
efforts.
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
INTRODUCTION
The force is an important factor in the field of Mechanics, which maybe broadly
*defined as an agent which produces or tends to produce, destroys or tends to destroy
motion. e.g., a horse applies force to pull a cart and to set it in motion. Force is also
required to work on a bicycle pump. In this case, the force is supplied by the muscular
power of our arms and shoulders.
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.
•
Figure 2.0(©Ferdinand P. Beer & E.
Russell Johnston, Jr.)
Force is a vector quantity.
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:
✓ Fixed or bound vectors have well defined of
application that cannot be changed without affecting
an analysis.
✓ 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.
Figure 2.1(©Ferdinand P. Beer &
E. Russell Johnston, Jr.)
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Addition of Vector
•
Trapezoid rule for vector addition
•
Triangle rule for vector addition
•
Law of Cosines,
•
Law of sines
•
Vector addition is commutative,
•
Vector subtraction
•
Addition of three or more vectors
through repeated application of the
triangle rule.
•
The polygon rule for the addition of
three or more vectors
•
Vector addition is associative,
•
Multiplication of a vector by a scalar
Figure 2.2(©Ferdinand P. Beer &
E. Russell Johnston, Jr.)
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Component of Forces:
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
𝑥 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑥 = 𝐹𝑠𝑖𝑛𝜃𝑦
𝐹 = √𝐹𝑥 2 + 𝐹𝑦 2
𝑦 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
𝐹𝑦 = 𝐹𝑠𝑖𝑛𝜃𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑦
𝐹𝑦
𝑡𝑎𝑛𝜃 =
𝐹𝑥
Figure 2-3 a
Given the slope of the line of action of the force as
𝑟 = √ℎ2 + 𝑣 2
ℎ
𝐹𝑥 = 𝐹 ( )
𝑟
𝑣
ℎ
𝑣
𝐹𝑦 = 𝐹 ( )
𝑟
Figure 2.3 b
Components of a Force in 3D Space
𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑥
𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦
𝐹𝑧 = 𝐹𝑐𝑜𝑠𝜃𝑧
𝐹 = √𝐹𝑥 2 + 𝐹𝑦 2 + 𝐹𝑧 2
𝐹𝑥
𝐹
𝐹𝑦
𝑐𝑜𝑠𝜃𝑦 =
𝐹
𝐹𝑧
𝑐𝑜𝑠𝜃𝑧 =
𝐹
𝑐𝑜𝑠𝜃𝑥 =
Figure 2-3 c
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example 2.1
The screw eye in figure 2.4a is subjected to two forces 𝐹1 𝑎𝑛𝑑 𝐹2 . Determine the
magnitude and direction of the resultant force.
Given:
Required: 𝐹𝑅 =? & 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
Figure 2.4 (a) & (b)(©Russell C. Hibbler)
Solution:
Parallelogram Law: The parallelogram is formed by drawing a line from the head of
𝐹1 that is parallel to 𝐹2 and another line from Ihe head of 𝐹2 that is parallel to 𝐹1 .The
resultant force 𝐹𝐵 extends to where these lines intersect at point A. Fig. 2.4b. The two
unknowns are the magnitude of.𝐹𝐵 and the angle 𝜃(theta).
Trigonometry. From the parallelogram. the vector triangle is constructed. Fig. 2.4c.
Using the law of cosines.
𝐹𝑅 = √𝐹1 2 + 𝐹2 2 − 2𝐹1 𝐹2 𝑐𝑜𝑠𝛽
𝜷 = 𝟏𝟓° + 𝟗𝟎° + 𝟏𝟎° = 𝟏𝟏𝟓°
𝐹𝑅 = √(100𝑁)2 + (150𝑁)2 − 2(100𝑁)(150𝑁)𝐶𝑜𝑠115°
𝐹𝑅 = √10,000 + 22,500 − 30,000(−0.4226)
𝐹𝑅 = 212. 55𝑁
Applying the law of sines to determine 𝜃
𝑭𝟐
𝑭𝑹
=
𝒔𝒊𝒏𝜽 𝒔𝒊𝒏𝜷
150𝑁 212.55𝑁
=
𝑠𝑖𝑛𝜃
𝑠𝑖𝑛 115°
Figure 2.4 c (©Russell C. Hibbler)
150𝑁(𝑠𝑖𝑛 115°)
212.55𝑁
𝜃 = 𝑎𝑟𝑐𝑠𝑖𝑛 0.6396
𝜽 = 𝟑𝟗. 𝟖°
The direction ∅ (𝒑𝒉𝒊)𝒐𝒇 𝑭𝒈 , measured the horizontal is,
∅ = 𝟑𝟗. 𝟖° + 𝟏𝟓. 𝟎° = 𝟓𝟒. 𝟖°
𝒔𝑖𝑛𝜃 =
∅ = 54.8°
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Another solution using the component method.
𝒙 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕
𝒚 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕
𝐹1𝑥 = 100𝑁 𝑐𝑜𝑠15°
𝐹1𝑦 = 100𝑁 𝑠𝑖𝑛15°
= 96.59𝑁
= 25.88𝑁
𝐹2𝑥 = 150𝑁𝑠𝑖𝑛10°
𝐹2𝑦 = 150𝑁𝑐𝑜𝑠10°
= 26.05𝑁
= 147.72𝑁
∑ 𝐹𝑥 = 96.59𝑁 + 26.05𝑁 = 122.64𝑁
∑ 𝐹𝑦 = 25.88𝑁 + 147.72𝑁 = 173.6𝑁
𝐹𝑅 = √(∑ 𝐹𝑥 )2 + (∑ 𝐹𝑦 )
2
= √(122.64𝑁)2 + (173.6𝑁 )2
= √(15040.5696 𝑁 2 ) + (30136.96 𝑁 2 )
= √45177.5296 𝑁 2
𝐹𝑅 = 212.55𝑁 = 212.6 𝑁
∑ 𝐹𝑦
173.6𝑁
=
= 1.4155
∑ 𝐹𝑥 122.64𝑁
𝜃 = 𝑡𝑎𝑛1 1.4155
𝑡𝑎𝑛𝜃 =
𝜃 = 54.8°
Example 2.2
Resolve the horizontal 600-lb force in Fig. 2.5 into components acting along the
𝑢 𝑎𝑛𝑑 𝑣 axes and determine the magnitudes of these components.
Solution:
Figure 2.5 (a) & (b)(©Russell C. Hibbler)
Parallelogram Law. The parallelogram is constructed by extending a line from the
head of 600 lbs. force parallel to the 𝑣 axis until it intersects the 𝑢 axis at a point B,
Figure 5b. The arrow from A to B represents 𝐹𝑢 . Similarly, the line extended from the
head of the 600-lb force drawn parallel to the 𝑢 axis intersects 𝑣 axis at point C, which
gives 𝐹𝑣 .
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
The vector addition using the triangle rule is shown in Fig. 2.5c. The two unknowns
are the magnitude of 𝐹𝑢 and 𝐹𝑣 . Applying the law of sines.
𝐹𝑢
600𝑙𝑏
=
𝑠𝑖𝑛 120° 𝑠𝑖𝑛30°
𝐹𝑢 (𝑠𝑖𝑛30°) = (600𝑙𝑏)(𝑠𝑖𝑛120°)
600𝑙𝑏(𝑠𝑖𝑛120°)
𝐹𝑢 =
𝑠𝑖𝑛30°
𝐹𝑢 = 1,039 𝑙𝑏
𝐹𝑣
600𝑙𝑏
=
𝑠𝑖𝑛 30° 𝑠𝑖𝑛30°
𝐹𝑣 (𝑠𝑖𝑛30°) = (600𝑙𝑏)(𝑠𝑖𝑛30°)
(600𝑙𝑏)(𝑠𝑖𝑛30°)
𝐹𝑣 =
𝑠𝑖𝑛30°
𝐹𝑣 = 600 𝑙𝑏
Figure 2.5c
NOTE: The result for 𝐹𝑢 that sometimes a component can have a greater magnitude
than the resultant.
Example 2.3
Determine the magnitude of the component force F in Fig. 2.6a and the magnitude of
the resultant force 𝐹𝑅 and if 𝐹𝑅 is directed along the positive y axis.
Figure 2.6 (a, b, c) (©Russell C. Hibbler)
Solution:
The parallelogram law is shown in Fig. 2.6b, and the triangle rule is shown in Fig. 2.6c.
The magnitudes of 𝐹𝑅 𝑎𝑛𝑑 𝐹 are the two unknowns. The can be determined by
applying the law of sines.
𝐹
200𝑙𝑏
=
𝑠𝑖𝑛60° 𝑠𝑖𝑛45°
𝐹𝑅
200𝑙𝑏
=
𝑠𝑖𝑛75°
𝑠𝑖𝑛45°
𝐹(𝑠𝑖𝑛45°) = (200𝑙𝑏)(𝑠𝑖𝑛60°)
𝐹𝑅 (𝑠𝑖𝑛45°) = (200𝑙𝑏)(𝑠𝑖𝑛75°)
𝐹=
(200𝑙𝑏)(𝑠𝑖𝑛60°)
𝑠𝑖𝑛45°
𝐹 = 244.95𝑙𝑏
𝐹𝑅 =
(200𝑙𝑏)(𝑠𝑖𝑛75°)
𝑠𝑖𝑛45°
𝐹𝑅 = 273.21𝑙𝑏
Figure 2.6c (©Russell C.
Hibbler)
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Resultant of concurrent forces
Figure 2.7𝑴𝒂𝒕𝒉𝒂𝒍𝒊𝒏𝒐. 𝒄𝒐𝒎
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.
The following equation involving resultant of force system are the following.
1. Determine the x and y components of each forces
𝑥 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
𝐹1𝑥 = 𝐹1 𝑐𝑜𝑠𝜃1
𝐹2𝑥 = −𝐹2 𝑐𝑜𝑠𝜃2
𝐹3𝑥 = −𝐹3 𝑐𝑜𝑠𝜃3
𝐹4𝑥 = 𝐹4 𝑐𝑜𝑠𝜃4
𝑦 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
𝐹1𝑦 = 𝐹1 𝑠𝑖𝑛𝜃1
𝐹2𝑦 = 𝐹2 𝑠𝑖𝑛𝜃2
𝐹3𝑦 = −𝐹3 𝑠𝑖𝑛𝜃3
𝐹4𝑦 = −𝐹4 𝑠𝑖𝑛𝜃4
2. The x component of the resultant is equal to the summation of forces in
the x-direction.
∑ 𝐹𝑥 = 𝐹1𝑥 − 𝐹2𝑥 − 𝐹3𝑥 + 𝐹4𝑥 = 𝐹𝑅𝑥
3. The y-components of the resultant force is equal to the summation of the
forces in the y direction.
∑ 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 − 𝐹3𝑦 − 𝐹4𝑦 = 𝐹𝑅𝑦
4. The magnitude of the resultant force system is found by using the
Pythagorean Theorem
𝑭𝑹 = √𝑭𝟐𝑹𝒙 + 𝑭𝟐𝑹𝒚
And the direction of the resultant force is given by:
𝑭𝑹𝒚
𝒕𝒂𝒏𝜽 =
𝑭𝑹𝒙
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example 2- 4. Determine magnitude and direction of the concurrent forces of the
figure 2.8.
𝒙 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔
𝐹1𝑥 = 58𝑘𝑁𝑐𝑜𝑠30°
= 50.23𝑘𝑁
𝐹2𝑥 = −50𝑘𝑁𝑐𝑜𝑠45°
= −35.36𝑘𝑁
5
𝐹3𝑥 = −45𝑘𝑁 ( )
13
= −17.32𝑘𝑁
𝐹4𝑥 = 40𝑘𝑁
𝒚 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔
𝐹1𝑦 = 58𝑘𝑁𝑠𝑖𝑛30°
= 29𝑘𝑁
𝐹2𝑦 = 50𝑘𝑁𝑠𝑖𝑛45°
= 35.36𝑘𝑁
12
𝐹3𝑦 = −45𝑘𝑁 ( )
13
= −41.54𝑘𝑁
𝐹4𝑦 = 0
∑ 𝐹𝑥 = 50.23𝑘𝑁 − 35.36𝑘𝑁 − 17.32𝑘𝑁 + 40𝑘𝑁
∑ 𝐹𝑥 = 37.55𝑘𝑁
∑ 𝐹𝑦 = 29𝑘𝑁 + 35.36𝑘𝑁 − 41.54𝑘𝑁 + 0
∑ 𝐹𝑦 = 22.82𝑘𝑁
2
Figure 2.8 Mathalino.com
2
𝑅𝐹 = √∑ 𝐹𝑥 + ∑ 𝐹𝑦 = √(37.55𝑘𝑁)2 + (22.82𝑘𝑁)2
𝑅𝐹 = √1410.0025𝑘𝑁 2 + 520.7524𝑘𝑁 2
𝑅𝐹 = √1930.7549𝑘𝑁 2
𝑅𝐹 = 43.94𝑘𝑁
∑ 𝐹𝑦 22.82𝑘𝑁
𝑡𝑎𝑛𝜃 =
=
= 0.6077
∑ 𝐹𝑥 37.55𝑘𝑁
𝜃 = 𝑡𝑎𝑛−1 0.6077
∑ 𝐹𝑦 = 22.82𝑘𝑁
𝜃
∑ 𝐹𝑥 = 37.55𝑘𝑁
𝜃 = 31.29° 𝑁 𝑜𝑓 𝐸
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example 2-5
The link in Figure 2.7a is subjected to two
forces 𝐹1 𝑎𝑛𝑑 𝐹2 . Determine the magnitude
and direction of the resultant force.
Solution:
Scalar Notation: First resolve each force into
its x and y components, Fig. 2.7b, then sum
these components algebraically.
𝑥 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
𝑦 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
𝐹1𝑥 = 600𝑁𝑐𝑜𝑠30°
𝐹1𝑦 = 600𝑁𝑠𝑖𝑛30°
𝐹1𝑥 = 519.6𝑁
𝐹1𝑦 = 300𝑁
𝐹2𝑥 = −400𝑁𝑐𝑜𝑠45° 𝐹2𝑦 = 400𝑁𝑠𝑖𝑛45°
𝐹2𝑥 = −282.8𝑁
𝐹2𝑦 = 282.8𝑁
∑ 𝐹𝑥 = 300𝑁 − 282.8𝑁 = 236.8𝑁
∑ 𝐹𝑦 = 300𝑁 + 282.8𝑁 = 582.8𝑁
The resultant force is shown in Fig. 2.7c,
has magnitude of
𝑅𝐹 = √(236.8𝑁)2 + (582.8𝑁)2
𝑅𝐹 = √56074.24𝑁 2 + 339655.84𝑁 2
𝑅𝐹 = √395730.08𝑁 2
𝑅𝐹 = 609.1𝑁
The direction of the resultant force
𝑡𝑎𝑛𝜃 =
∑ 𝐹𝑦 582.8𝑁
=
∑ 𝐹𝑥 236.8𝑁
𝑡𝑎𝑛𝜃 = 2.64
𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛2.64
Figure 2.7a,b &c(©Russell C. Hibbler
𝜃 = 67.9°
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example 2-6
The end of the boom O in Fig.2.8a is subjected to three
concurrent and coplanar forces. Determine the magnitude
and direction of the resultant force.
Figure 2.8a(©Russell C. Hibbler)
Solution:
Each force is resolved into its x and y
components,
Fig.
2.8b.
Summing
the
x
components and y components:
𝑥 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
𝐹1𝑥 = −400𝑁
𝐹2𝑥 = 250𝑁𝑠𝑖𝑛45°
𝐹2𝑥 = 176.8𝑁
4
𝐹3𝑥 = −200𝑁 ( )
5
𝐹3𝑥 = −160𝑁
𝑦 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠
𝐹1𝑦 = 0
𝐹2𝑦 = 250𝑁𝑐𝑜𝑠45°
𝐹2𝑦 = 176.8𝑁
3
𝐹3𝑦 = 200𝑁 ( )
5
𝐹3𝑦 = 120𝑁
∑ 𝐹𝑥 = −400𝑁 + 176.8𝑁 − 160𝑁 = −𝟑𝟖𝟑. 𝟐𝑵
∑ 𝐹𝑦 = 176.8𝑁 + 120𝑁 = 𝟐𝟗𝟔. 𝟖𝑵
The resultant force, shown in Fig. 2.8c, has
magnitude of
𝑅𝐹 = √(−383.2𝑁)2 + (296.8𝑁)2
𝑅𝐹 = 485𝑁
The direction angle 𝜃 is
296.8
tan 𝜃 = (
)
383.2
𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛 0.7745
Figure 2.8b,2.8c (©Russell C. Hibbler)
𝜃 = 37.8°
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example 2-7 Three ropes are tied to a small metal ring. At the end of each rope three
students are pulling, each trying to move the ring in their direction. If we look down
from above, the forces and directions they are applying are shown in Fig. shown below.
Find the net force on the ring due to the three applied forces.
Required: R and 𝜽
Given:
Fig. 2-7a Mathalino.com
Solution:
𝒙 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔
𝒚 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕𝒔
𝐹1𝑥 = 30𝑙𝑏𝑐𝑜𝑠37°
𝐹1𝑦 = 30𝑙𝑏𝑠𝑖𝑛37°
𝐹1𝑥 = 𝟐𝟑. 𝟗𝟔𝒍𝒃
𝐹1𝑦 = 18.05𝑙𝑏
𝐹2𝑥 = −50𝑙𝑏𝑐𝑜𝑠45°
𝐹2𝑦 = 50𝑙𝑏𝑠𝑖𝑛45°
𝐹2𝑥 = −35.36𝑙𝑏
𝐹2𝑦 = 35.36𝑙𝑏
𝐹3𝑥 = −80𝑙𝑏𝑐𝑜𝑠60°
𝐹3𝑦 = −80𝑙𝑏𝑠𝑖𝑛60°
𝐹3𝑥 = −40𝑙𝑏
𝐹3𝑦 = −69.28𝑙𝑏
Fig. 2-7b@ Mathalino.com
∑ 𝐹𝑥 = 23.96𝑙𝑏 − 35.36𝑙𝑏 − 40𝑙𝑏 = −51.40𝑙𝑏
∑ 𝐹𝑦 = 18.05𝑙𝑏 + 35.36𝑙𝑏 − 69.28𝑙𝑏 = −15.87𝑙𝑏
𝑅𝐹 = √(−51.40)2 + (−15.87)2 = 53.79𝑙𝑏
𝑅𝐹 = 53.79𝑙𝑏
𝑡𝑎𝑛𝜃 =
15.87
= 0.3088
51.40
𝜃 = 𝑎𝑟𝑐𝑡𝑎𝑛0.3088 = 17.16° 𝑆 𝑜𝑓 𝑊
Fig. 2-7c @Mathalino.com
𝜃 = 17.16° 𝑆 𝑜𝑓 𝑊
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Moment of a Force
•
•
What is the Moment of Force?
A force can have two effects on a rigid body:
➢ Translation - tends to move it linearly along its line of action.
➢ Rotation - tends to rotate it about an axis.
Moment of a Force: the measure of a force’s
tendency to rotate the body about an axis.
The moment of a force (moment) is also
called torque (Ex: torque wrench).
Figure 2-7a (©Russell C. Hibbler)
Figure 2-7b (©Russell C. Hibbler)
Figure 2-7(©Russell C. Hibbler)
Mathematical Definition of a Moment: The moment Mo of a force F about a point
O is equal to the magnitude of the force multiplied by the perpendicular distance d
from O to the line of action of the force.
Magnitude of 𝑀𝑜
𝑀𝑜 = 𝑑𝐹
•
•
•
Where:
d = is the moment arm or the perpendicular distance
from the axis at point o to the line of action of the force.
Point O is referred to as the Moment Center
Units- in terms of force & distance.
➢ SI Units: N·m or kN·m
➢ English Units: ft·lb or in·lb
Figure 2-8a &b (©Russell C. Hibbler)
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Direction of Moment
• A moment is a vector quantity just like a force vector. The direction of a
moment vector is determined by the right-hand rule.
• Positive Moment – the force tends to cause a counter clockwise rotation about
the moment center.
• Negative Moment – the force tends to cause a clockwise rotation about the
moment center.
• Summation of Moments: In 2D case (planar forces), moments can be added
algebraically just like forces with the same line of action. (𝑅𝑥 = 𝐹1𝑥 + 𝐹2𝑥 +…)
Two ways to calculate the moment of a force about a point:
1. Transmissibility Method:
• Extend the line of action to find the moment arm length
d (perpendicular distance from center to line of action)
• Calculate the moment using; of action).
𝑀𝑜 = 𝐹 ∙ 𝑑
where d is the moment arm ≠ L
2. Rectangular Component Method
• Find the component of the force find the moment arm
length d perpendicular to rod L
• Calculate the moment using;
𝑴𝒐 = 𝑭𝒙 ∙ d
is the normal component of F
d is the moment arm = L
Resultant Moment.
For two dimensional problems where all the forces lie within the x-y plane, Figure
2.9, the resultant moment (𝑀𝑅 )𝑜 about point O (the z axis) can be determined by
finding the algebraic sum of the moments caused by all the forces in the system.
Figure 2-9(©Russell C. Hibbler)
(𝑀𝑅 )𝑜 = ∑ 𝐹𝑑 = 𝐹1 𝑑1 − 𝐹2 𝑑2 + 𝐹3 𝑑3
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Principle of Moments
A concept frequently used in mechanics is the principle of moments, which is
occasionally referred to as Varignon’s theorem since it was formerly developed by the
French mathematician Varignon’s (1654-1722).
Varignon’s Theorem: The moment of a force about any point is equal to the
sum of moments produced by the components of the force about the same point.
In Figure 2-12, consider the
moments of force F and two of its
components about point O.
Since 𝐹 = 𝐹1 + 𝐹2
𝑴𝒐 = 𝒓 𝒙 𝑭 = 𝒓(𝑭𝟏 + 𝑭𝟐 )
𝑴𝒐 = 𝒓𝑭𝟏 + 𝒓𝑭𝟐
Figure 2-12(©Russell C. Hibbler)
For two-dimensional problems,
Figure 2-12, we can use the
principle of moments by resolving
the force into its rectangular
components and then determine
the moment using a scalar
analysis. Thus,
Thus
𝑴𝒐 = 𝑭𝒙 𝒚 − 𝑭𝒚 𝒙
This method is generally easier
than finding the same moment
using 𝑀𝑜 = 𝐹𝑑.
Figure 2-13(©Russell C. Hibbler)
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example 2-8
For each case illustrated in Figure 2-10, determine the moment of the force about
point O.
Solution:
The line of action of each force is extended as a dashed line in order to establish the moment
arm d. Also illustrated is the tendency of rotation of the member as a caused by the force.
Furthermore, the orbit of the force about O is shown as a colored curl. Thus,
Figure 2-10a (©Russell C. Hibbler)
𝑎) 𝑀0 = (100𝑁)(2𝑚)
𝑀𝑜 = 200𝑚 ∙ 𝑁, 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
Figure 2-10b (©Russell C. Hibbler)
𝑏) 𝑀0 = (100𝑁)(2𝑚) = 200𝑁 ∙ 𝑚
𝑀𝑜 = 200𝑚 ∙ 𝑁, 𝑐𝑙𝑜𝑐𝑤𝑖𝑠𝑒
Figure 2-10c (©Russell C. Hibbler)
Figure 2-10d(©Russell C. Hibbler)
𝑐) 𝑀𝑜 = (40𝑙𝑏)(4𝑓𝑡 + 2𝑓𝑡𝑐𝑜𝑠30°)
𝑀𝑜 = 229𝑙𝑏 ∙ 𝑓𝑡 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑑) 𝑀0 = (60𝑙𝑏)(1𝑓𝑡𝑠𝑖𝑛45°)
𝑀𝑜 = 42.4𝑙𝑏 ∙ 𝑓𝑡 𝑐𝑜𝑢𝑛𝑡𝑒𝑟 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑀𝑜 = (7𝑘𝑁)(4𝑚 − 1𝑚) = 21.0𝑚 ∙ 𝑘𝑁, 𝑐𝑜𝑢𝑛𝑡𝑒𝑟 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
Figure 2-10e (©Russell C. Hibbler
18 | P a g e
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example Problem 2-9
Determine the moment of the force at point O of the figure shown below.
Given:
Required: Moment at point O
Figure 2-14a(©Russell C. Hibbler)
Solution 1
The moment arm d in figure shown above can be found from trigonometry.
𝑑 = (3𝑚)𝑠𝑖𝑛75° = 2.898 𝑚
𝑀𝑜 =𝐹𝑑 = (5𝑘𝑁)(2.898𝑚)
𝑀𝑜 = 14.5𝑘𝑁 ∙ 𝑚 , 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
Figure 2-14b (©Russell C. Hibbler)
Solution 2
The x and y components of the force are indicated in Fig.2-14b
Considering counterclockwise moments as positive, and applying the principle of
moments, we have
𝑀𝑜 = −𝐹𝑥 𝑑𝑦 − 𝐹𝑦 𝑑𝑥
𝑀𝑜 = −(5𝑘𝑁𝑐𝑜𝑠45°)(3𝑚𝑠𝑖𝑛30°) − (5𝑠𝑘𝑁𝑖𝑛45°)(3𝑀𝑐𝑜𝑠30°𝑚)
𝑀𝑜 = −14.5𝑘𝑁 ∙ 𝑚
𝑀𝑜 = 14.5𝑘𝑁 ∙ 𝑚, 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
19 | P a g e
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Solution 3
The x and y axes can be set parallel and
perpendicular to the rod’s axis
𝑀𝑜 = −𝐹𝑦 𝑑𝑥
= −(5𝑠𝑖𝑛75°𝑘𝑁)(3𝑚)
𝑀𝑜 = −14.5𝑘𝑁 ∙ 𝑚
𝑀𝑜 = 14.5𝑘𝑁 ∙ 𝑚, clockwise
Figure 2-14c(©Russell C. Hibbler)
Resultant of Parallel Forces
•
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.
Figure 2-15(©Mathalino.com)
𝑅 = ∑ 𝐹 = 𝐹1 + 𝐹2 + 𝐹3 + ⋯
𝑅𝑑 = ∑ 𝐹𝑥 =𝐹1 𝑥1 + 𝐹2 𝑥2 + 𝐹3 𝑥3 + ⋯
20 | P a g e
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
•
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.
Rectangular Load
𝑅 = 𝑤𝑜 𝐿
Triangular Load
𝑅 = 1/2𝑤𝑜 𝐿
Trapezoidal Load
1
𝑅 = 𝑤𝑜1 𝐿 + (𝑤𝑜2 − 𝑤𝑜1 )𝐿
2
Figure 2-16(©Mathalino.com)
Example Problem 2-10
A parallel force system acts on the lever
shown in Figure 2-17a Determine the
magnitude and position of the resultant.
Figure 2-17a(©Mathalino.com)
𝑅 = ∑ 𝐹𝑣
𝑅 = −30𝑙𝑏 − 60𝑙𝑏 + 20𝑙𝑏 − 40𝑙𝑏
𝑅 = −110 𝑙𝑏
𝑹 = 𝟏𝟏𝟎𝒍𝒃 (𝒅𝒐𝒘𝒏𝒘𝒂𝒓𝒅)
𝑀𝐴 = ∑ 𝑥 ∙ 𝐹
Figure 2-17b(©Mathalino.com)
𝑀𝐴 = 2𝑓𝑡(30𝑙𝑏) + 5𝑓𝑡(60𝑙𝑏) − 7𝑓𝑡(20𝑙𝑏) + 11𝑓𝑡(40𝑙𝑏)
𝑀𝐴 = 60𝑙𝑏 ∙ 𝑓𝑡 + 300𝑙𝑏 ∙ 𝑓𝑡 − 140𝑙𝑏 ∙ 𝑓𝑡 + 440𝑙𝑏 ∙ 𝑓𝑡
𝑴𝑨 = 660 𝒍𝒃 ∙ 𝒇𝒕 (clockwise)
𝑅𝑑 = 𝑀𝐴
110𝑑 = 660𝑓𝑡 ∙ 𝑙𝑏
𝑑=
660𝑓𝑡 ∙ 𝑙𝑏
= 𝟔. 𝟔𝒇𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕 𝒐𝒇 𝑨.
110𝑙𝑏
Thus, R=110lb downward a.6.6ft to the right of A.
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example Problem 2-11
Determine the resultant of the four parallel forces acting on the rocker arm of Figure
shown below.
Given:
Req’d: R=?
Figure 2-18a(©Mathalino.com)
Solution:
𝑅 = ∑ 𝐹𝑣
𝑅 = −50𝑙𝑏 + 40𝑙𝑏 + 20𝑙𝑏 − 60𝑙𝑏
𝑅 = −50 𝑙𝑏 (𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑)
𝑅 = 50 𝑙𝑏 (𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑)
Figure 2-18b(©Mathalino.com)
𝑀𝑜 = 𝐹𝑑
𝑀𝑜 = −50𝑙𝑏(6𝑓𝑡) + 40𝑙𝑏(2𝑓𝑡) − 20𝑙𝑏(3𝑓𝑡) + 60𝑙𝑏(8𝑓𝑡)
𝑀𝑜 = (−300 + 80 − 60 + 480)𝑙𝑏 ∙ 𝑓𝑡
𝑀𝑜 = 200 𝑙𝑏 ∙ 𝑓𝑡 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑅𝑑 = 𝑀𝑜
50𝑙𝑏(𝑑) = 200𝑙𝑏 ∙ 𝑓𝑡
𝑑=
200𝑙𝑏 ∙ 𝑓𝑡
50𝑙𝑏
𝑑 = 4𝑓𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝑂
22 | P a g e
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example problem 2-12
The weights of the various components of the truck are shown. Replace this system
of forces by an equivalent resultant force and specify its location measured from a) B
and b) A.
Given:
Required:
𝑎) 𝐹𝑅 =? 𝑑 =? 𝑓𝑟𝑜𝑚 𝐵
𝑏) 𝐹𝑅 =? ; 𝑑 =? 𝑓𝑟𝑜𝑚 𝐴
Figure 2-19(©Russell C. Hibbler)
Solution:
a. 𝐹𝑅 = ∑ 𝐹𝑦
𝐹𝑅 = −1,750 − 5,500 − 3,500
𝐹𝑅 = −10,750 𝑙𝑏
𝐹𝑅 = 10,750𝑙𝑏(𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑)
Location of resultant Force from point B
+ 𝑴𝑹𝑩 = ∑ 𝑴𝑩
−(10,750𝑑) = −3,500(3) − 5,500(17) − 1,750(25)
−(10,750𝑙𝑏)(𝑑) = (−10,500 − 93,500 − 43,750)𝑙𝑏 ∙ 𝑓𝑡
−(10,750𝑙𝑏)(𝑑) = −147,750𝑙𝑏 ∙ 𝑓𝑡
−147750𝑙𝑏 ∙ 𝑓𝑡
𝑑=
−10,750 𝑙𝑏
𝑑 = 13.74 𝑓𝑡 𝑓𝑟𝑜𝑚 𝐵
b.
𝐹𝑅 = ∑ 𝐹𝑦
𝐹𝑅 = −1,750 − 5,500 − 3,500
𝐹𝑅 = −10,750 𝑙𝑏
𝐹𝑅 = 10,750𝑙𝑏(𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑)
Location of Resultant Force from Point A:
+ 𝑴𝑹𝑨 = ∑ 𝑴𝑨
{10,750(𝑑)} = 3,500(20) + 5,500(6) − 1,750(2)
10,750𝑙𝑏(𝑑) = (70,000 + 33,000 − 3500)𝑙𝑏 ∙ 𝑓𝑡
10,750𝑙𝑏(𝑑) = 99,500𝑙𝑏 ∙ 𝑓𝑡
99,500𝑙𝑏 ∙ 𝑓𝑡
𝑑=
10,750𝑙𝑏
𝑑 = 9.26 𝑓𝑡 𝑓𝑟𝑜𝑚 𝐴
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example problem 2-13
The beam AB in Fig. 2-20a supports a load which varies an intensity of 220 N/m to
890 N/m. Calculate the magnitude and position of the resultant load.
Figure 2-20a(©Mathalino.com)
Solution:
Figure 2-20b(©Mathalino.com)
220𝑁
𝐹1 = 6𝑚 (
)
𝑚
𝐹1 = 1,320𝑁
1
𝑁
𝐹2 = (6𝑚) (670 )
2
𝑚
𝑅𝑑 = ∑ 𝑀𝐴
𝑅𝑑 = 3𝐹1 + 4𝐹2
3330N(d) = 3(1320) + 4(2010)
3330N(d) = 3960N·m + 8040N·m
𝐹2 = 2,010𝑁
3,330𝑁(𝑑) = 12000𝑁 ∙ 𝑚
𝑅 = ∑ 𝐹𝑣
𝑑=
𝑅 = −𝐹1 − 𝐹2
𝑅 = −1,320𝑁 − 2010𝑁
𝑅 = −3,330𝑁
12000𝑁 ∙ 𝑚
3,330𝑁
𝑑 = 3.6𝑚
Thus, R = 3,330 N downward at 3.6
m to the left of A.
𝑅 = 3,330𝑁 (downdward)
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PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Couples
• Couple is a system of forces whose magnitude of the resultant is zero and yet
has a moment sum. Geometrically, couple is composed of two equal forces
that are parallel to each other and acting in opposite direction. The magnitude
of the couple is given by
Where F are the two forces and d is the moment arm, or the perpendicular
distance between the forces.
Couple is independent of the moment center; thus, the effect is unchanged in the
following conditions.
• The couple is rotated through any angle in its plane.
• The couple is shifted to any other position in its plane.
• The couple is shifted to a parallel plane.
In a case where a system is composed entirely of couples in the same plane or
parallel planes, the resultant is a couple whose magnitude is the algebraic sum of
the original couples.
Example Problem 2-14
Refer to Figure shown below. A couple consists of two vertical forces of 60 lb each.
One force acts up through A and the other acts down through D. Transform the
couple into an equivalent couple having horizontal forces acting through E and F.
Figure 2-21a, b &c(©Mathalino.com)
𝑪 = 𝟐(𝟔𝟎) = 𝟏𝟐𝟎𝒍𝒃 ∙ 𝒊𝒏
3𝑃 = 𝐶
3𝑃 = 120
𝑃 = 40𝑙𝑏
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Example 2-15
Two couples act on the beam as shown. Determine the magnitude of F so that
resultant couple moment is 300lb.ft counterclockwise. Where on the beam does the
resultant couple act?
Figure 2-22(©Russell C. Hibbler)
Solution:
3
4
𝐹(4𝑓𝑡) + 𝐹(1.5𝑓𝑡) − 200𝑙𝑏(1.5𝑓𝑡)
5
5
300𝑙𝑏 ∙ 𝑓𝑡 = 2.4𝑓𝑡(𝐹) + 1.2𝑓𝑡(𝐹) − 300𝑙𝑏 ∙ 𝑓𝑡
300𝑙𝑏 ∙ 𝑓𝑡 = 3.6𝑓𝑡(𝐹) − 300𝑙𝑏 ∙ 𝑓𝑡
3.6𝑓𝑡(𝐹) = 300𝑙𝑏 ∙ 𝑓𝑡 + 300𝑙𝑏 ∙ 𝑓𝑡
3.6𝑓𝑡(𝐹) = 600𝑙𝑏 ∙ 𝑓𝑡
600𝑙𝑏 ∙ 𝑓𝑡
𝐹=
3.6𝑓𝑡
𝐹 = 166.67𝑙𝑏
+(𝑀𝑐 )𝑅 =
Example Problem 2-16
Replace the force system acting on the truss by a resultant force and couple moment
at point C.
Figure 2-23(©Russell C. Hibbler)
Solution:
∑(𝐹𝑅 )𝑥 = ∑ 𝐹𝑥
(𝐹𝑅 )𝑦 = ∑ 𝐹𝑦
4
(𝐹𝑅 )𝑥 = 500𝑙𝑏 ( )
5
(𝐹𝑅 )𝑥 = 400 𝑙𝑏
3
(𝐹𝑅 )𝑦 = −200𝑙𝑏 − 150𝑙𝑏 − 100𝑙𝑏 − 500𝑙𝑏 ( )
5
(𝐹𝑅 )𝑦 = −750𝑙𝑏
(𝐹𝑅 )𝑦 = 750𝑙𝑏 , downward
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
The magnitude of the resultant force
𝐹𝑅 is given by:
2
𝐹𝑅 = √(𝐹𝑅 )𝑥 + (𝐹𝑅 )𝑦
2
𝐹𝑅 = √4002 + 7502
𝐹𝑅 = 850𝑙𝑏
The direction of resultant force 𝐹𝑅
(𝐹𝑅 )𝑦 750𝑙𝑏
=
(𝐹𝑅 )𝑥 400 𝑙𝑏
𝑡𝑎𝑛𝜃 = 1.875
𝜃 = arctan 1.875
𝑡𝑎𝑛𝜃 =
𝜃 = 61.93°
Equivalent Couple Moment:
+ (𝑀𝑅 )𝐶 = ∑ 𝑀𝐶
3
4
(𝑀𝑅 )𝑐 = −200𝑙𝑏(2𝑓𝑡) − 150𝑙𝑏(4𝑓𝑡) − 100𝑙𝑏(6𝑓𝑡) − 500𝑙𝑏 ( ) (8𝑓𝑡) − 500𝑙𝑏 ( ) (6𝑓𝑡)
5
5
(𝑀𝑅 )𝑐 = −400𝑙𝑏 ∙ 𝑓𝑡 − 600𝑙𝑏 ∙ 𝑓𝑡 − 600𝑙𝑏 ∙ 𝑓𝑡 − 2,400𝑙𝑏 ∙ 𝑓𝑡 − 2,400𝑙𝑏 ∙ 𝑓𝑡
(𝑀𝑅 )𝐶 = −6,400𝑙𝑏 ∙ 𝑓𝑡
(𝑀𝑅 )𝐶 = 6,400𝑙𝑏 ∙ 𝑓𝑡 (𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒)
Example 2-17
Replace the force and couple system acting on the member in Fig. 2-27a by an
equivalent resultant force and couple moment acting at point O.
Figure 2-24a & b (©Russell C. Hibbler)
Solution:
Force Summation: Since the couple forces of 200N are equal but opposite, they
produce a zero-resultant force, and so it is not necessary to consider them in the
force summation. The 500N force is resolved into its x and y components, thus,
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
(𝐹𝑅 )𝑥 = ∑ 𝐹𝑥
3
(𝐹𝑅 )𝑥 = ( ) (500𝑁)
(𝐹𝑅 )𝑦 = ∑ 𝐹𝑥
4
(𝐹𝑅 )𝑦 = (500𝑁) ( ) − 750 = −350𝑁
5
5
(𝐹𝑅 )𝑥 = 300𝑁 →
(𝐹𝑅 )𝑦 = 350𝑁 ↓
The magnitude of the resultant force 𝐹𝑅 is
2
𝐹𝑅 = √(𝐹𝑅 )𝑥 + (𝐹𝑅 )𝑦
2
𝐹𝑅 = √(300𝑁)2 + (350𝑁)2
𝐹𝑅 = 46𝑁
And the angle 𝜃 𝑖𝑠
𝜃 = 𝑡𝑎𝑛−1 [
𝜃 = 49.4°
350𝑁
] = 1.167
300𝑁
Figure 2-24a (©Russell C. Hibbler)
4
3
+ (𝑀𝑅 )𝑜 = (500𝑁) (5) (2.5𝑚) − (500𝑁) (5) (1𝑚) − (750𝑁)(1.25𝑚) + 200𝑁 ∙ 𝑚
(𝑀𝑅 )𝑜 = −37.5𝑁 ∙ 𝑚
(𝑀𝑅 )𝑜 = 37.5𝑁 ∙ 𝑚𝑚 (𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒)
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
Name:
Date:
Rating: _________
Course, Year & Section: __________________ Schedule: ____________
Plate No. 2
Resultant of Force System
Instruction:
➢ Copy the problem and draw the diagram on a clean sheet of short bond
papers. One problem per page.
➢ Write legibly and avoid erasures.
➢ Box your final answer.
➢ If your done with your problem-solving activity take pictures and submit it in
the Google Classroom assigned to your section.
2.1 Replace the force system acting on the beam by an equivalent force and
couple moment at point A.
Figure 2-25(©Russell C. Hibbler)
2.2 The Howe roof truss shown in Figure shown below carries the given loads. The
wind loads are perpendicular to the inclined members. Determine the
magnitude of the resultant, its inclination with the horizontal, and where it
intersects AB.
Figure 2-26(©Mathalino.com)
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Version 2, Revision 2, August 2022
PHILIPPINE STATE COLLEGE OF AERONAUTICS
INSTITUTE OF ENGINEERING AND TECHNOLOGY
AIRCRAFT MAINTANCE AND TECHNOLOGY DEPARTMENT
Learning Module 02: Resultant of Force System
2.3 Determine the resultant of the coplanar concurrent force system shown.
Compute the magnitude and direction with the x-axis. Use the component
method.
𝐹2 = 300𝑙𝑏
𝐹1 = 200𝑙𝑏
𝐹3 = 400𝑙𝑏
𝐹4 = 50𝑙𝑏
𝐹5 = 100𝑙𝑏
Figure 2-25(©Russell C. Hibbler)
2.4 Replace the loading on the frame by a single resultant force. Specify where
the force acts, measured from end A. Given the values of the following:
𝐹1 = 450𝑁
𝜃 = 60°
𝑎 = 2𝑚
𝐹2 = 300𝑁
∅ = 30°
𝑏 = 4𝑚
𝐹3 = 700𝑁
𝑀 = 1500𝑁 ∙ 𝑚
𝑐 = 3𝑚
Figure 2-26(©Russell C. Hibbler)
2.5 Replace the two forces by an equivalent resultant force and couple moment at
point O. Set F = 20 lb.
Figure 2-27(©Russell C. Hibbler)
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