MIDTERM DISCUSSIONS
Fundamental Concepts of Engineering Mechanics
MODULE I
Fundamental Concepts of
Engineering Mechanics
I
Objectives:
•
To develop a clear understanding of the fundamental principles and laws of mechanics, such
as Newton's laws of motion, free body diagram, force and components, and moment of force.
•
To apply the fundamental concepts of mechanics to solve real-world problems, including
analyzing the motion of objects, predicting forces and accelerations, and understanding the
behavior of systems under various conditions.
Outline:
•
Engineering Mechanics
•
Fundamental Concepts in Mechanics
a) Force
b) Particle
c) Rigid Body
d) Newton’s Three Laws of Motion
e) Mechanical Quantities
f) Scalar and Vector
•
Fundamental Principles
g) Principle of Transmissibility
h) Parallelogram Law
•
Free Body Diagram
•
Force and Components
•
Moment of Force
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MODULE I
Chapter Introduction
Engineering mechanics is a branch of applied mechanics that deals with the study of the behavior of
physical bodies and systems under the action of forces and moments. It provides the foundation for
analyzing and designing various structures and mechanisms. Engineering mechanics encompasses
several branches, including: Statics which focuses on the equilibrium of bodies at rest or under the
influence of balanced forces. It deals with the analysis of forces acting on stationary objects, determining
reactions, and solving problems related to structures and systems in a state of equilibrium and Dynamics
which deals with the study of bodies in motion or under the influence of unbalanced forces. It involves
analyzing the causes of motion, calculating accelerations, determining forces, and studying the behavior
of systems in motion. This branch includes kinematics (study of motion without considering forces) and
kinetics (study of motion considering forces). Under dynamics is kinematics is the branch of mechanics
that deals with the study of motion without considering the forces causing the motion. It focuses on
describing the position, velocity, and acceleration of objects as they move through space and time, and
kinetics, also known as dynamics, is the branch of mechanics that focuses on the study of motion and the
forces that cause or influence motion. It involves analyzing the causes of motion, calculating
accelerations, determining forces, and studying the behavior of systems in motion.
Fundamental Concepts in Mechanics
• Length: used to locate the position of a point in space and thereby describe the size of a physical
system.
• Time: conceived as a succession of events. Although the principles of statics are time
independent, this quantity plays an important role in the study of dynamics.
• Mass: This property manifests itself as a gravitational attraction between two bodies and provides
a measure of the resistance of matter to a change in velocity.
• Force: considered as a “push” or “pull” exerted by one body on another. It represents the action
of one body on another characterized by its magnitude, direction of its action, and its point of
application.
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Fundamental Concepts of Engineering Mechanics
MODULE I
Force
Force is a fundamental concept in mechanics that represents a push or pull exerted on an object.
It is a vector quantity, meaning it has both magnitude and direction. Force is responsible for causing
changes in the motion or shape of an object.
Forces can be categorized into several types, including contact forces and non-contact forces.
Contact forces are exerted through direct physical contact between objects, such as when you push a door
or lift a book. Examples of contact forces include normal force, frictional force, and tension force. Noncontact forces, also known as field forces, act over a distance without direct physical contact. The most
familiar non-contact force is gravity, which attracts objects towards each other. Other examples of noncontact forces include electromagnetic forces (such as electrical and magnetic forces) and the nuclear
forces that hold atomic particles together.
REMEMBER:
• Mass is a property of matter that does not change from one location to another.
• Weight refers to the gravitational attraction of the earth on a body or quantity of mass. Its
magnitude depends upon the elevation at which the mass is located. (W= mg)
• Weight of a body is the gravitational force acting on it.
Particle
In mechanics, a particle refers to an object or a body that is treated as a point mass, meaning its
size and internal structure are not considered relevant for the analysis. It is an idealization used to simplify
the study of motion and forces acting on an object.
A particle is characterized by its position in space and can be represented as a mathematical point
with mass. The motion of a particle is described in terms of its position, velocity (rate of change of
position), and acceleration (rate of change of velocity).
Rigid Body
A rigid body is an idealized concept in mechanics that refers to an object in which the distances between
its constituent particles or points remain constant, regardless of external forces or moments applied to it.
It is a simplification used to analyze and study the motion and forces acting on solid objects.
MIDTERM DISCUSSIONS
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MODULE I
The concept of a rigid body allows for the application of simplified mathematical models and
principles to analyze the behavior of objects without considering the complexities of internal
deformations. By assuming rigidity, the focus is primarily on the body's overall motion, forces, and
moments acting on it.
Rigid body mechanics involves studying the motion of rigid bodies, including their translation
(linear motion), rotation (angular motion), and combined translational and rotational motion. Principles
such as Newton's laws of motion, conservation of momentum, conservation of energy, and torque are
applied to analyze and predict the behavior of rigid bodies.
Newton’s Three Laws of Motion
Newton's laws of motion are three fundamental principles that describe the relationship between
the motion of an object and the forces acting upon it. These laws form the basis of classical mechanics
and are essential for understanding the behavior of objects in motion. Here are the three laws of Newton:
1. Newton's First Law (Law of Inertia): Newton's first law
states that an object at rest will remain at rest, and an object in
motion will continue moving with a constant velocity in a
straight line, unless acted upon by an external force. This is
often referred to as the law of inertia. Inertia is the property of
an object to resist changes in its state of motion. This law
highlights the concept that objects tend to maintain their current
state of motion (whether at rest or moving uniformly) unless a
force is applied to change that state.
2.
Newton's Second Law (Law of Acceleration):
Newton's second law states that the acceleration of an object is
directly proportional to the net force applied to it and inversely
proportional to its mass. Mathematically, this law can be stated
as F = ma, where F is the net force applied to an object, m is the
mass of the object, and a is the resulting acceleration. This law
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Fundamental Concepts of Engineering Mechanics
MODULE I
quantifies how the motion of an object changes when a force is applied to it. The larger the force
or the smaller the mass, the greater the resulting acceleration.
3. Newton's Third Law (Law of Action-Reaction):
Newton's third law states that for every action,
there is an equal and opposite reaction. This law
states that when one object exerts a force on a
second object, the second object exerts an equal
and opposite force on the first object. In other
words, forces always occur in pairs. These forces
are known as action and reaction forces, and they act on different objects. Despite being equal in
magnitude and opposite in direction, action-reaction forces do not cancel each other out because
they act on different objects.
Four Fundamental Quantities
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Fundamental Concepts of Engineering Mechanics
MODULE I
Unit Prefixes
Scalar Quantities
A scalar is a physical quantity that has magnitude but no specific direction. Scalars are
completely described by their numerical value and units. Examples of scalar quantities include mass,
temperature, time, energy, and speed. Scalars can be added, subtracted, multiplied, and divided using
ordinary arithmetic operations.
Vector Quantities
A vector is a physical quantity that has both magnitude and direction. Vectors are represented by
arrows, where the length of the arrow corresponds to the magnitude of the vector, and the direction of
the arrow represents its direction. Examples of vector quantities include displacement, velocity,
MIDTERM DISCUSSIONS
Fundamental Concepts of Engineering Mechanics
MODULE I
acceleration, force, and momentum. Vectors cannot be simply added or subtracted using ordinary
arithmetic; instead, vector operations involve considering both magnitude and direction.
A vector is shown graphically by an arrow. The length of the arrow represents the magnitude of
the vector, and the angle θ between the vector and a fixed axis defines the direction of its line of action.
The head or tip of the arrow indicates the sense of direction of the vector.
Free Vectors: In mathematics, a free vector refers to a vector that is not associated with a specific
point or origin. Free vectors are defined solely by their magnitude and direction and can be moved
parallelly in space without changing their essential properties. They are often represented as directed line
segments or arrows. Free vectors are used to represent physical quantities such as displacement, velocity,
acceleration, and force.
Sliding Vectors: Sliding vectors is not a standard term in vector terminology. It is possible that
it refers to vectors that are allowed to move or slide along a specific path or direction. This could imply
a constrained vector that is restricted to a particular motion or direction.
Fixed Vectors: Fixed vectors may refer to vectors that are stationary or do not change their
position or direction in a given context. They are not affected by translations or rotations and remain
fixed in space relative to a specific coordinate system. Fixed vectors are commonly used to represent
reference frames, coordinate axes, or fixed points of interest.
Principle of Transmissibility
The conditions of equilibrium or of motion of a rigid body remain unchanged if a force acting at
a given point of the rigid body is replaced by a force of the same magnitude and same direction, but
acting at a different point, provided that the two forces have the same line of action.
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Fundamental Concepts of Engineering Mechanics
MODULE I
Parallelogram Law
The Parallelogram Law is a method used to add or combine two vectors geometrically.
According to this law, the sum of two vectors can be obtained by constructing a parallelogram with the
vectors as adjacent sides. The diagonal of the parallelogram represents the resultant vector, which is the
vector sum of the two original vectors.
1. Draw two vectors to be added as adjacent sides of a parallelogram. The starting points of the
vectors should coincide.
2. Complete the parallelogram by drawing the remaining sides.
3. The diagonal of the parallelogram, originating from the common starting point of the original
vectors, represents the resultant vector.
4. The magnitude and direction of the resultant vector can be determined by measuring the length
and direction of the diagonal using appropriate units and reference angles.
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Fundamental Concepts of Engineering Mechanics
MODULE I
The Parallelogram Law is based on the principle of vector addition, which states that vectors can
be added by combining their magnitudes and directions. The resulting vector represents the net effect of
the combined influences of the original vectors.
Vector Addition
We can also add B to A using the triangle rule, which is a special case of the parallelogram law,
whereby vector B is added to vector A in a “head-to-tail” fashion, i.e., by connecting the head of A to
the tail of B. The resultant R extends from the tail of A to the head of B. In a similar manner, R can also
be obtained by adding A to B. By comparison, it is seen that vector addition is commutative; in other
words, the vectors can be added in either order, i.e., R = A + B = B + A.
Vector Addition of Forces
Finding the Components of a Force. Sometimes it is necessary to resolve a force into two
components in order to study its pulling or pushing effect in two specific directions. For example, F is to
be resolved into two components along the two members, defined by the u and v axes. In order to
determine the magnitude of each component, a parallelogram is constructed first, by drawing lines
starting from the tip of F, one line parallel to u, and the other line parallel to v. These lines then intersect
with the v and u axes, forming a parallelogram. The force components Fu and Fv are then established by
simply joining the tail of F to the intersection points on the u and v axes. This parallelogram can then be
reduced to a triangle, which represents the triangle rule. From this, the law of sines can then be applied
to determine the unknown magnitudes of the components.
MIDTERM DISCUSSIONS
Fundamental Concepts of Engineering Mechanics
SAMPLE PROBLEM 1
The screw eye is subjected to two forces, F1 and F2.
Determine the magnitude and direction of the resultant force.
SOLUTION:
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MIDTERM DISCUSSIONS
MODULE I
Fundamental Concepts of Engineering Mechanics
𝐹𝑟 = √(100𝑁)2 + (150𝑁)2 − 2(100𝑁)(150𝑁) cos 1 15°
= √10 000 + 22 500 − 30 000(−0.4226)
𝑭𝒓 = 𝟐𝟏𝟐. 𝟔 𝑵
150𝑁𝑠𝑖𝑛𝜃 212.6𝑁
=
𝑠𝑖𝑛𝜃
sin 1 15
𝜃 = 39.8°
𝜃 = 39.8° + 15°
𝜽 = 𝟓𝟒. 𝟖°
SAMPLE PROBLEM 2
Determine the magnitude of the component force F and the
magnitude of the resultant force FR if FR is directed along the
positive y axis.
SOLUTION:
𝐹
sin 60°
200𝑁
= sin 45°
𝑭 = 𝟐𝟒𝟓 𝑵
𝐹𝑟
sin 75°
200𝑁
= sin 45°
𝑭𝒓 = 𝟐𝟕𝟑 𝑵
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Fundamental Concepts of Engineering Mechanics
MODULE I
Free Body Diagram
Free body diagram is necessary to investigate the condition of equilibrium of a body or system.
While drawing the free body diagram all the supports of the body are removed and replaced with the
reaction forces acting on it. A free body diagram is a visual representation that isolates an object or a
system of interest and shows all the external forces acting on it. It simplifies the analysis of forces by
focusing on the object itself, disregarding its internal structure or other objects it may interact with. In a
free body diagram, forces are represented as arrows that indicate the magnitude and direction of each
force. FBDs are commonly used in physics and engineering to analyze the forces acting on objects and
solve problems related to motion, equilibrium, and interactions between objects.
1. Center of Mass: The center of mass (also known as the center of gravity) is a point within an
object or system where the entire mass can be concentrated. It is a concept used to simplify the
analysis of the motion of an object or system by treating it as a point mass located at the center
of mass. For symmetric objects, the center of mass coincides with the geometric center. The
center of mass is an important concept in dynamics, as the motion of an object can be described
as if all the forces and torques acting on the object were concentrated at its center of mass.
2. Normal Force: The normal force is the force exerted by a surface to support the weight of an
object resting on it. It acts perpendicular to the surface and prevents objects from sinking into or
passing through the surface. For example, when a book rests on a table, the table exerts an upward
normal force equal in magnitude but opposite in direction to the weight of the book. The normal
force is a reaction force that arises due to the contact between objects and surfaces.
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Fundamental Concepts of Engineering Mechanics
MODULE I
3. Applied Force: An applied force is a force that is externally applied to an object or system. It is
a force that is not due to gravity or any other inherent property of the object. Applied forces can
be generated by physical interactions, such as pushing, pulling, or hitting an object. These forces
can cause changes in the motion or deformation of the object.
4. Friction Force: Friction force is the force that opposes the relative motion or tendency of motion
between two surfaces in contact. It acts parallel to the surface and can prevent or reduce sliding
or slipping motion. Friction arises due to the interactions between the microscopic roughness and
interlocking of surfaces. There are two main types of friction: static friction, which prevents
motion between stationary surfaces, and kinetic friction, which opposes the relative motion
between moving surfaces.
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
the y-axis the y-component, and so on.
• F1 and F2 are components of R.
MIDTERM DISCUSSIONS
Fundamental Concepts of Engineering Mechanics
• Fa and Fb are perpendicular projections on axes a and b respectively.
• R ≠ Fa + Fb unless a and b are perpendicular to each other
Components of XY plane
• Given the slope of the line of action of the force as v/h
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MIDTERM DISCUSSIONS
MODULE I
Fundamental Concepts of Engineering Mechanics
SAMPLE PROBLEM 3
Determine the magnitude and direction of the resultant
force.
SOLUTION:
𝐹𝑥1 = 58 cos 3 0° = 50.23𝑘𝑁
𝐹𝑥2 = −50 cos 4 5° = −35.36𝑘𝑁
𝐹𝑦1 = 58 sin 3 0° = 29𝑘𝑁
𝐹𝑦2 = 50 sin 4 5° = 35.36𝑘𝑁
5
𝐹𝑥3 = −45 (13) = −17.31𝑘𝑁
12
13
𝐹𝑦3 = −45 ( ) = −41.54𝑘𝑁
𝐹𝑥4 = 40𝑘𝑁
𝐹𝑦4 = 0
Σ𝐹𝑥 = 58 cos 3 0° − 50 cos 4 5° − 45 (
Σ𝐹𝑦 = 58 sin 3 0 + 50 sin 4 5° − 45 (
5
) + 40 = 37.5664𝑘𝑁
13
12
) + 0 = 22.8169𝑘𝑁
13
𝐹𝑟 = √(37.5664)2 + (22.8169)2
𝑭𝒓 = 𝟒𝟑. 𝟗𝟓𝟐𝟖 𝒌𝑵
MIDTERM DISCUSSIONS
MODULE I
Fundamental Concepts of Engineering Mechanics
𝜃 = tan−1 (
22.8169
)
37.5664
𝜽 = 𝟑𝟏. 𝟐𝟕𝟑𝟓°
SAMPLE PROBLEM 4
Determine the magnitude and direction of the resultant
force.
SOLUTION:
2
)
√13
= 400.49𝑙𝑏
𝑄𝑥 = −200 cos 6 0° = −100𝑙𝑏
3
)
√13
= 600.74𝑙𝑏
𝑄𝑦 = 200 sin 6 0° = 173.20𝑙𝑏
𝑃𝑥 = 722 (
𝑃𝑦 = 722 (
2
𝐹𝑥 = −448 ( 5) = −400.70𝑙𝑏
√
1
𝐹𝑦 = −448 ( 5) = −200.35𝑙𝑏
√
Σ𝐹𝑥 = 722 (
Σ𝐹𝑦 = 722 (
2
√13
3
√13
𝑇𝑥 = 400 sin 2 0° = 136.81𝑙𝑏
𝑇𝑦 = −400 cos 2 0° = −375.88𝑙𝑏
) − 200 cos 6 0° − 448 (
) + 200 sin 6 0° − 448 (
2
√5
1
√5
) + 400 sin 2 0° = 36.5982 𝑙𝑏
) − 400 cos 2 0° = 197.7167 𝑙𝑏
MIDTERM DISCUSSIONS
MODULE I
Fundamental Concepts of Engineering Mechanics
𝐹𝑟 = √(36.5982)2 + (197.7167)2
𝑭𝒓 = 𝟐𝟎𝟏. 𝟎𝟕𝟓𝟒 𝒍𝒃
𝜃 = tan−1 (
197.7167
)
36.5982
𝜽 = 𝟕𝟗. 𝟓𝟏𝟑𝟎°
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.
• 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.
MIDTERM DISCUSSIONS
Fundamental Concepts of Engineering Mechanics
MODULE I
SAMPLE PROBLEM 5
For each case illustrated in the
figure, determine the moment of the
force about point O.
SOLUTIONS:
(a)
𝑀𝑜 = (100𝑁)(2𝑚) = 𝟐𝟎𝟎 𝑵𝒎 ↻
(b) 𝑀𝑜 = (50𝑁)(0.75𝑚) = 𝟑𝟕. 𝟓 𝑵𝒎 ↻
(c) 𝑀𝑜 = (40𝑁)(4𝑓𝑡 + 2 cos 3 0°) = 𝟐𝟐𝟗 𝒍𝒃 𝒇𝒕 ↻
(d) 𝑀𝑜 = (600𝑁)(1 sin 4 5°) = 𝟒𝟐𝟒 𝑵𝒎 ↺
MIDTERM DISCUSSIONS
Fundamental Concepts of Engineering Mechanics
MODULE I
SAMPLE PROBLEM 6
Determine the resultant moment of the four
forces acting on the rod shown about point O.
SOLUTION:
+↺ (𝑀𝑅 )𝑜 = Σ𝐹𝑑
(𝑀𝑅 )𝑜 = −50(2𝑚) + 60𝑁(0) + 20𝑁(3 sin 3 0°𝑚) − 40𝑁(4𝑚 + 3 cos 3 0°𝑚)
(𝑀𝑅 )𝑜 = −334 𝑁𝑚 = 𝟑𝟑𝟒 𝑵𝒎 ↻
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MODULE I
ACTIVITY 1
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places.
1. Determine the magnitude of the resultant force and its direction measured counterclockwise from
the positive x axis. Use the parallelogram law to solve this problem.
2. Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent
force R. Determine the magnitude of the resultant force and its direction. Use the parallelogram
law to solve this problem.
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Fundamental Concepts of Engineering Mechanics
MODULE I
3. Suppose that F1 = 770 N and F2 = 725 N. Determine the magnitude and the direction of the
resultant force, measured counterclockwise from the positive x axis. (Hint: Add all the
component forces in x and y axis and express it as ∑ Fx and ∑ Fy to get the resultant force.)
4. Determine the resultant moment produced by the forces about point O.
MIDTERM DISCUSSIONS
Couple Moment and Resultant Force Systems
MODULE II
Couple Moment and
Resultant Force
Systems
II
Objectives:
•
To discuss the concept of the moment of a force and show how to calculate it in
two and three dimensions.
•
To provide a method for finding the moment of a force about a specified axis.
•
To define the moment of a couple
•
To show how to find the resultant effect of a nonconcurrent force system
•
To indicate how to reduce a simple distributed loading to a resultant force acting
at a specified location.
Outline:
•
Moment of Force
•
Force System
•
Couples
•
Equivalent Force System
•
Resultant of Force Systems
o Coplanar Concurrent Force System
o Parallel Force System
o Non-Concurrent Force System
MIDTERM DISCUSSIONS
Couple Moment and Resultant Force Systems
MODULE II
Chapter Introduction
This chapter focuses on the study of bodies at rest or in equilibrium under the action of
forces requires the concept of force, moment, and couple. Also resultant of force systems together
with its subcomponents namely Coplanar Concurrent Force System, Parallel Force System, and
Non-Concurrent Force System.
Equivalent Force System
It is sometimes advantageous to simplify a system of forces and couple moments acting on
a body by replacing it with an equivalent system consisting of a single resultant force acting at a
specific point and a resultant couple moment. A system is equivalent if the external effects it has
on a body are the same as those caused by the original force and couple moment system.
An equivalent force system refers to a different arrangement of forces that produces the
same overall effect on a body as the original force system. It is a concept used in structural analysis
and engineering to simplify the analysis of complex force systems by replacing them with simpler
and more manageable configurations while maintaining the same resultant effect on the body.
To establish an equivalent force system, it is necessary to consider the magnitude, direction,
and line of action of the original forces, as well as their points of application. By carefully selecting
the appropriate combination of forces, their magnitudes and directions can be adjusted to match
those of the original forces, resulting in an equivalent force system. The equivalence of force
systems is based on the principle of equilibrium, which states that for a body to be in equilibrium,
the sum of all forces and moments acting on it must be zero. Therefore, an equivalent force system
preserves equilibrium conditions and ensures that the body remains in a balanced state.
In the case of coplanar force systems (forces acting in the same plane), an equivalent force
system can be achieved by replacing a system of concurrent forces with a single resultant force
acting at a specific point. This resultant force has the same magnitude, direction, and line of action
as the original forces, producing an equivalent effect on the body. For non-concurrent force systems
(forces not meeting at a common point), the equivalent force system may involve the introduction
of a fictitious force, such as a couple or a force couple. These fictitious forces are introduced to
counterbalance the effects of the original forces and create an equilibrium condition. The
magnitudes, directions, and locations of these fictitious forces are carefully determined to match
the characteristics of the original force system.
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
Moment of a Force
The tendency of a force to produce rotation about an axis is called the moment of a force about an axis.
Consider a wrench that is applied to a nut on a bolt. To obtain the maximum rotation or turning effect on
the nut, we know from common experience that the force should be applied perpendicular to the handle
as far away from the axis through the center of the bolt as possible. The following definition is consistent
with common experience: The moment of a force about an axis is defined as
the product of the force and the perpendicular distance from the line of action
of the force to that axis. With forces in a plane, the moment of a force about a
point is understood to mean the moment about an axis perpendicular to the
plane at that point. A concept often used in mechanics is the principle of
moment. Because it was originally developed by French mathematician Pierre
Varignon (1654–1722), it is sometimes called Varignon's theorem. It states
that the moment of force about a point is equal to the sum of the moment of
the force component about the point. This theorem is easy to prove using the vector cross product since
the cross products follow the distributive law. For example, consider the moments of the force F and two
of its components about point O, Fig. 2-1. Since F = F1 + F2 we have
MO = r x F = r (F1 + F2) = r x F1 + r x F2
Fig. 2-1
For two-dimensional problems, Fig. 2-2, 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,
𝑀𝑂 = 𝐹𝑥 𝑦 + 𝐹𝑦 𝑥
Fig. 2-2
This method is generally easier than finding the same
moment using
MO = Fd
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
EXAMPLE
Calculate the magnitude of the moment about the base point O of the 600-N force.
SOLUTION:
Replace the force by its rectangular components at A,
𝐹1 = 600𝑐𝑜𝑠40 = 𝟒𝟔𝟎 𝑵
𝐹2 = 600𝑠𝑖𝑛40 = 𝟑𝟖𝟔 𝑵
By Varignon’s Theorem, the moment becomes
𝑀𝑂 = 460(4) + 386(2) = 𝟐𝟔𝟏𝟎 𝑵. 𝒎
PRACTICE PROBLEM
Force F acts at the end of the angle bracket. Determine the
moment of the force about point O.
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
SOLUTION:
↺ +𝑀𝑂 = 400𝑠𝑖𝑛30 − 400𝑐𝑜𝑠30(0.4𝑚)
= −98.6𝑁. 𝑚 = 𝟗𝟖. 𝟔𝑵. 𝒎 ↻
Force Systems
When two bodies interact, they exert different forces on each other. A "force system" or "system of force" is formed
when multiple forces act on a body at the same time. In layman's terms, a force system or system of force is "a
group of forces acting on a body or a number of connected bodies." The study of the system of forces is critical for
analyzing the effect of the system of forces on the object and calculating the system of forces results.
A member is subject to various systems of forces. Different force systems on the same body result in different behaviors.
That is why studying different types of force systems is essential. In general, there are two types of force systems.
1. Coplanar Force System
The term coplanar denotes that the quantities are in the same plane. A
coplanar force system is one in which all of the force systems acting on a
body have their lines of action in the same plane as the body.
The types of coplanar systems of forces are as follows:
•
•
•
•
Collinear coplanar force system
Concurrent coplanar force system
Parallel coplanar force system
Non-concurrent coplanar force system
2. Non coplanar Force System
The term coplanar refers to the quantities should not be in the same plane.
A non-coplanar force system is one in which all of the force systems acting
on a body do not have the same plane of action.
The following are examples of non-coplanar force systems:
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
•
Concurrent non-coplanar force system
•
Parallel non-coplanar force system
•
Non-concurrent non-coplanar force system
Couples
A couple is defined as two parallel forces of equal magnitude. However, they face opposite directions and are separated
by a perpendicular distance d, Fig. 2-3. Because the resultant force is zero, the only effect of a couple is to create a rotation,
or if no movement is possible, there is a rotational tendency in a specific direction.
Fig. 2-3
The moment produced by a couple is called a couple moment. We can determine its value by finding the sum of the
moments of both couple forces about any arbitrary point. The moment of a couple, M, is defined as having a magnitude
of
𝑀 = 𝐹𝑑
where F is the magnitude of one of the forces and d is the perpendicular distance or moment arm between the forces.
The direction and sense of the couple moment are determined by the right-hand rule, where the thumb indicates this
direction when the fingers are curled with the sense of rotation caused by the couple forces. If two couples produce a
moment with the same magnitude and direction, then these two couples are equivalent. Since couple moments are
vectors, their resultant can be determined by vector addition. If more than two couple moments act on the body, we
may generalize this concept and write the vector resultant as
𝑀𝑅 = ∑ 𝐹𝑑
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
EXAMPLE
Replace the force and couple system shown by an equivalent
resultant force and couple moment acting at point O.
SOLUTION:
(→+) (𝐹𝑔 )𝑥 = ∑ 𝐹𝑥 ;
3
5
(𝐹𝑔 )𝑥 = (3𝑘𝑁) cos 30° + ( ) (5𝑘𝑁) = 𝟓. 𝟓𝟗𝟖 𝒌𝑵 →
(↑+) (𝐹𝑔 )𝑦 = ∑ 𝐹𝑦 ;
4
(𝐹𝑔 )𝑦 = (3𝑘𝑁) sin 30° − ( ) (5𝑘𝑁) − 4𝑘𝑁
5
= −𝟔. 𝟓𝟎 𝒌𝑵 𝒐𝒓 𝟔. 𝟓𝟎 𝒌𝑵 ↓
𝐹𝑅 = √(𝐹𝑅 )2𝑥 + (𝐹𝑅 )2𝑦
𝐹𝑅 = √(5.598)2 + (−6.50)2
𝐹𝑅 = 𝟖. 𝟓𝟖 𝒌𝑵
(𝐹𝑅 )𝑦
∅ = tan−1 ((𝐹
𝑅 )𝑥
)
6.50
∅ = tan−1 (5.598)
∅ = 𝟒𝟗. 𝟑°
(𝑀𝑔 )𝑜 = ∑ 𝑀𝑜 ;
(𝑀𝑔 )𝑜 = (3𝑘𝑁) sin 30° (0.2 𝑚) −
3
(3𝑘𝑁) cos 30° (0.1 𝑚) + ( ) (5𝑘𝑁)(0.1 𝑚) −
5
4
(5) (5𝑘𝑁)(0.5 𝑚)
= −𝟐. 𝟒𝟔 𝒌𝑵(𝒎)
MIDTERM DISCUSSIONS
Couple Moment and Resultant Force Systems
MODULE II
Resultant of Force System
A force system's resultant is a force or a couple that has the same effect on the body in both
translation and rotation if all the forces are removed and replaced by the resultant. The resultant of
a force system is a single force or a couple that represents the combined effect of all the individual
forces acting on a body. It is a vector quantity that summarizes the net force and the net moment
(torque) produced by the forces in the system. The resultant force and resultant moment have the
same effect on the body as the original system of forces, both in terms of translational motion (linear
displacement) and rotational motion (angular displacement).
To determine the resultant of a force system, the individual forces are vectorially summed,
taking into account their magnitudes, directions, and points of application. The resultant force is
obtained by adding all the forces together, considering both their magnitude and direction. It
represents the net force acting on the body, which produces an equivalent linear motion.
In addition to the resultant force, a force system may also generate a resultant moment or
torque. The resultant moment is the net torque produced by the forces in the system, and it
represents the rotational effect or tendency of the forces. It is obtained by summing the individual
torques generated by each force, considering their magnitudes, directions, and points of application.
The resultant moment influences the rotational motion of the body and determines its orientation
and angular displacement.
The principle of replacing a force system with its resultant is based on the concept of
equivalence. By replacing all the individual forces with the resultant force and resultant moment,
the overall effect on the body remains the same. This simplifies the analysis and calculations, as
the complex system of forces is reduced to a single force or couple that captures the combined
effect.
The equation involving the resultant of force system are the following:
1. 𝑅𝑥 = ∑ 𝐹𝑥 = 𝐹𝑥1 + 𝐹𝑥2 + 𝐹𝑥3 +. ..
The x-component of the resultant is equal to the summation of forces in the x- direction.
2. 𝑅𝑦 = ∑ 𝐹𝑦 = 𝐹𝑦1 + 𝐹𝑦2 + 𝐹𝑦3 +. ..
The y-component of the resultant is equal to the summation of forces in the y- direction.
3. 𝑅𝑧 = ∑ 𝐹𝑧 = 𝐹𝑧1 + 𝐹𝑧2 + 𝐹𝑧3 +. ..
MIDTERM DISCUSSIONS
Couple Moment and Resultant Force Systems
MODULE II
The z-component of the resultant is equal to the summation of forces in the z- direction.
Note that according to the type of force system, one or two or three of the equations above will be
used in finding the resultant.
Coplanar Concurrent Force System
In a coplanar concurrent force system, all the forces involved lie in the same plane, and their
lines of action intersect at a common point. This arrangement allows for the determination of the
resultant force through specific formulas within the x-y plane.
By applying the previous formulas, the resultant force in a coplanar concurrent force system
can be accurately determined, providing valuable information about the net effect of all the forces
acting on the body. This resultant force represents the combined impact of the individual forces and
can be used to analyze the overall motion, stability, and equilibrium of the system.
𝑅𝑥 = ∑ 𝐹𝑥
𝑅𝑦 = ∑ 𝐹𝑦
𝑅 = √𝑅𝑥2 + 𝑅𝑦2
tan ∅𝑥 =
𝑅𝑦
𝑅𝑥
EXAMPLE 1
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. Find the net force on the ring due to the three applied forces.
𝑅𝑥 = ∑ 𝐹𝑥
𝑅𝑥 = 30 cos 37° − 50 cos 45° − 80 cos 60°
𝑅𝑥 = −51.40 𝑙𝑏
𝑅𝑥 = 51.40 𝑙𝑏 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
𝑅𝑦 = ∑ 𝐹𝑦
𝑅𝑦 = 30 sin 37° + 50 sin 45° − 80 sin 60°
𝑅𝑦 = −15.87 𝑙𝑏
𝑅𝑦 = 15.87 𝑙𝑏 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑
𝑅 = √𝑅𝑥2 + 𝑅𝑦2
𝑡𝑎𝑛 ∅𝑥 =
𝑅𝑦
𝑅𝑥
=
15.87
51.40
𝑅 = √51.402 + 15.872
∅𝑥 = 𝟏𝟕. 𝟏𝟔°
𝑅 = 𝟓𝟑. 𝟕𝟗 𝒍𝒃
Coplanar Parallel Force System
Parallel forces can be in the same or opposite directions. The sign of the direction can be
chosen arbitrarily, which means that taking one direction as positive makes the opposite direction
negative. The resultant is fully defined by its magnitude, direction, and line of action.
A coplanar parallel force system refers to a configuration of forces that lie in the same plane
and have parallel lines of action. In such a system, the forces do not intersect but run parallel to
each other. This arrangement allows for the analysis and determination of the resultant force and
its characteristics.
When dealing with coplanar parallel force systems, it is essential to understand that the forces
exerted on the body have the same direction or are collinear. They can act either in the same
direction, resulting in an additive effect, or in opposite directions, leading to a subtractive effect.
The forces can be of equal or unequal magnitudes, contributing to the overall force distribution on
the body.
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
𝑅 = ∑ 𝐹 = 𝐹1 + 𝐹2 + 𝐹3 +..
𝑅𝑑 = ∑ 𝐹𝑥 = 𝐹1 𝑥1 + 𝐹2 𝑥2 + 𝐹3 𝑥3 +..
EXAMPLE 2
A parallel force system acts on the lever shown. Determine the magnitude and position of
the resultant.
𝑅 = ∑𝐹
𝑀𝑎 = ∑ 𝑥𝐹
𝑅 = 30 + 60 − 20 + 40
𝑀𝑎 = 3(30) + 5(60) − 7(20) + 11(40)
𝑅 = 110 𝑙𝑏 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑
𝑀𝑎 = 660 𝑓𝑡(𝑙𝑏) 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝑅𝑑 = 𝑀𝑎
110𝑑 = 660
𝑑 = 𝟔 𝒇𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕 𝒐𝒇 𝑨
Thus, 𝑅 = 110 𝑙𝑏 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑 at 6 ft to the right of A.
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
Non-Concurrent Force System
Non-concurrent forces are two or more forces whose
magnitudes are equal but act in opposite directions with a common line
of action. This basically means that when the forces acting on a FBD
do not intersect at a common point, the system of forces is said to be
non- concurrent. The resultant of non-concurrent force system is
defined according to magnitude, inclination, and position.
The resultant of non-concurrent force system is defined according to magnitude, inclination,
and position. The magnitude of the resultant can be found as follows:
𝑅𝑥 = Σ𝐹𝑥
𝑅𝑦 = Σ𝐹𝑦
𝑅 = √𝑅𝑥2 + 𝑅𝑦2
The inclination from the horizontal is defined by:
𝑡𝑎𝑛𝜃𝑥 =
𝑅𝑦
𝑅𝑥
The position of the resultant can be determined according to the principle of moments.
𝑀𝑅 = Σ𝑀𝑜
𝑅𝑦 𝑖𝑥 = 𝑀𝑅
𝑅𝑑 = 𝑀𝑅
𝑅𝑥 𝑖𝑦 = 𝑀𝑅
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
Where,
Fx = component of forces in the x-direction
Mo = moment of forces about any point O
Fy = component pf forces in the y-direction
d = moment arm
Rx = component of the resultant in x-direction
MR = moment at a point due to resultant forces
Ry = component of the resultant in y-direction
Ix = x-intercept of the resultant R
R = magnitude of the resultant
Iy = y-intercept of the resultant
Θx = angle made by a force from the x-axis
EXAMPLE 1
Determine the resultant of the three forces acting on the
dam and locate its intersection with the base AB.
SOLUTION:
ΣFx = R x
45 − 25𝑐𝑜𝑠30° = 23.349𝑘𝑁
ΣFy = R y
−100 − 25𝑠𝑖𝑛30° = −112.5
𝑅 = √(23.349)2 + (112.5)2
𝑹 = 𝟏𝟏𝟒. 𝟖𝟗𝟕 𝒌𝑵
ΣMB = −45(2) + 100(6 − 2.3) + 25(1.3)
𝚺𝐌𝐁 = 𝟑𝟏𝟐. 𝟓 𝒌𝑵 ⋅ 𝒎
ΣMB = 𝑅𝑦 (𝑥)
(112.5)(𝑥) = 312.5
𝒙 = 𝟐. 𝟕𝟖 𝒎
MIDTERM DISCUSSIONS
MODULE II
Couple Moment and Resultant Force Systems
EXAMPLE 2
Compute the resultant of the three forces.
Locate its intersection with X and Y axes.
SOLUTION:
𝑅𝑥 = Σ𝐹𝑥
𝑅𝑥 = 390 (
12
3
) + 722 (
) − 300𝑠𝑖𝑛30°
13
√13
𝑅𝑥 = 810.74 𝑙𝑏 𝑡𝑜 𝑡ℎ𝑒 𝒓𝒊𝒈𝒉𝒕
𝑅𝑦 = Σ𝐹𝑦
5
13
2
)−
√13
𝑅𝑦 = 390 ( ) − 722 (
300𝑐𝑜𝑠30°
𝑅𝑦 = 510.30 𝑙𝑏 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑
𝑅𝑦
𝑅 = √𝑅𝑥2 + 𝑅𝑦2
𝑡𝑎𝑛𝜃𝑥 = 𝑅
𝑅 = √(810.74)2 + (510.30)2
𝑡𝑎𝑛𝜃𝑥 =
𝑅 = 957.97 𝑙𝑏
𝜃𝑥 = 32.19°
𝑥
510.30
810.74
𝑀𝑜 = Σ𝐹𝑑
𝑀𝑜 = −390 (
12
5
2
) (3) − 390 ( ) (5) − 722 (
) (4) + (300𝑐𝑜𝑠30°)(3)
13
13
√13
𝑀𝑜 = −1481.97 𝑙𝑏 ⋅ 𝑓𝑡
𝑀𝑜 = 1481.97 𝑙𝑏 ⋅ 𝑓𝑡 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
Intersection with y axis
Intersection with x axis
𝑅𝑥 𝑏 = 𝑀𝑜
𝑅𝑦 𝑎 = 𝑀𝑜
810.74𝑏 = 1481.97
510.30𝑎 = 1481.97
𝑏 = 1.83 𝑓𝑡 𝑎𝑏𝑜𝑣𝑒 𝑝𝑜𝑖𝑛𝑡 𝑂
𝑎 = 2.90 𝑓𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑜𝑓 𝑝𝑜𝑖𝑛𝑡 𝑂
𝑻𝒉𝒖𝒔, 𝑹 = 𝟗𝟓𝟕. 𝟗𝟕 𝒍𝒃 𝒅𝒐𝒘𝒏𝒘𝒂𝒓𝒅 𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕 𝒂𝒕 𝜽𝒙 = 𝟑𝟐. 𝟏𝟗°. 𝑻𝒉𝒆 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 𝒊𝒔 𝒂𝒕
𝟐. 𝟗𝟎 𝒇𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕 𝒐𝒇 𝑶 𝒂𝒏𝒅 𝒕𝒉𝒆 𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 𝒊𝒔 𝟏. 𝟖𝟑 𝒇𝒕 𝒂𝒃𝒐𝒗𝒆 𝒑𝒐𝒊𝒏𝒕 𝑶.
MIDTERM DISCUSSIONS
Couple Moment and Resultant Force Systems
MODULE II
ACTIVITY 2
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places.
1. Determine the resultant moment produced by the forces about point O.
2. Determine the equivalent resultant force and couple moment acting at point O by replacing the
force and couple system acting on the member.
MIDTERM DISCUSSIONS
Couple Moment and Resultant Force Systems
MODULE II
3. Find the value of P and F so that the four forces shown produce an upward resultant of 300 lb
acting at 4 ft from the left end of the bar.
4. Replace the loading of non-concurrent force system by an equivalent resultant force and
specify where the resultant’s line of action intersects the member AB measured from A.
MIDTERM DISCUSSIONS
Equilibrium of Force Systems
III
MODULE III
Equilibrium of Force
Systems
Objectives:
•
To introduce the concept of the free-body diagram for a particle.
•
To show how to solve particle equilibrium problems using the equations of
equilibrium.
Outline:
•
Conditions for Equilibrium of a Particle
•
Free-Body Diagram
•
Conditions for rigid body equilibrium
•
Support Reactions
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
Chapter Introduction
This chapter focuses on the conditions for equilibrium of a particle and a rigid body, what freebody diagram is and how it is created, and the support reactions often encountered in particle equilibrium
problems.
Conditions for Equilibrium of a Particle
A particle is said to be in equilibrium if:
• It remains at rest if originally at rest
• Has a constant velocity if originally in motion
Most often, the term “equilibrium” or “static equilibrium”, specifically, 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 is stated by the equation of equilibrium.
∑F = 0
Where ∑F is the vector sum of all the forces acting on the particle.
If a particle is subjected to a system of coplanar forces that lie in the x–y plane, then each force
can be resolved into its x and y components. For equilibrium, the resultant force’s x and y components
must both be equal to zero.
∑F = 0
∑Fx = 0
∑Fy = 0
• If a force has an unknown magnitude, the arrowhead sense of the
force on the free-body diagram can be assumed.
• If the solution yields a negative scalar, this indicates that the sense of the force is opposite to that
which was assumed.
Free-Body Diagram
To apply the equation of equilibrium, we must account for all the known and unknown forces
(F) which act on the particle. The best way to do this is to think of the particle as isolated and “free” from
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
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 three types of supports often encountered in particle equilibrium problems.
Three types of supports often encountered in particle equilibrium problems:
1. Springs.
Springs are an elastic machine component able to deflect under load in a
prescribed manner and to recover its initial shape when unloaded.
If a linearly elastic spring (or cord) of undeformed length l0 is used to
support a particle, the length of the spring will change in direct proportion to the
force F acting on it.
A characteristic that defines the “elasticity” of a spring is the spring
constant or stiffness k. The magnitude of force exerted on a linearly elastic spring
which has a stiffness k and is deformed (elongated or compressed) a distance s = l - l0, measured from
its unloaded position, is
F = ks
2.
Cables and Pulleys.
Cables are flexible structures that support the applied transverse loads by
the tensile resistance developed in its members. On the other hand, pulleys, are a
wheel that carries a flexible rope, cord, cable, chain, or belt on its rim.
Unless otherwise stated, all cables (or cords) will be assumed to have
negligible weight and they cannot stretch. Also, a cable can support only a tension
or “pulling” force, and this force always acts in the direction of the cable. Hence,
for any angle, the cable is subjected to a constant tension T throughout its length.
MIDTERM DISCUSSIONS
Equilibrium of Force Systems
MODULE III
3. Smooth Contact.
If an object rests on a smooth surface, then the surface will exert a force
on the object that is normal to the surface at the point of contact.
An example of this is shown in the figure. In addition to this normal force
N, the cylinder is also subjected to its weight W and the force T of the cord. Since
these three forces are concurrent at the center of the cylinder, we can apply the
equation of equilibrium to this “particle,” which is the same as applying it to the
cylinder.
Procedure for drawing a Free-Body Diagram
To construct a free-body diagram, the following steps are necessary.
1. Draw outlined shape. Imagine the particle to be isolated or cut “free” from its
surroundings. This requires removing all the supports and drawing the particle’s
outlined shape.
2. Show all forces. Indicate on this sketch all the forces that act on the particle. These
forces can be:
• Active forces, which tend to set the particle in motion
• Reactive forces, which are the result of the constraints or supports that
tend to prevent motion.
To account for all these forces, it may be helpful to trace around the particle’s
boundary, carefully noting each force acting on it.
3. Identify each force. 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.
NOTE:
The first step in solving any equilibrium problem is to draw the particle’s free-body diagram. This requires removing all the
supports and isolating or freeing the particle from its surroundings and then showing all the forces that act on it.
Equilibrium means the particle is at rest or moving at constant velocity. In two dimensions, the necessary and sufficient
conditions for equilibrium require Fx = 0 and Fy = 0.
MIDTERM DISCUSSIONS
Equilibrium of Force Systems
MODULE III
Procedure for Analysis
Equations of Equilibrium
• Apply the equations of equilibrium, Fx = 0 and Fy = 0. For convenience, arrows can be written
alongside each equation to define the positive directions.
• Components are positive if they are directed along a positive axis, and negative if they are
directed along a negative axis.
• If more than two unknowns exist and the problem involves a spring, apply F = ks to relate the
spring force to the deformation s of the spring.
• Since the magnitude of a force is always a positive quantity, then if the solution for a force yields
a negative result, this indicates that its sense is the reverse of that shown on the free-body
diagram.
EXAMPLE 1
Determine the tension in cables BA and BC necessary to support the 60-kg cylinder.
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
SOLUTION:
+ → ∑Fx = 0
;
4
𝑇𝐶 𝑐𝑜𝑠45° − (5) 𝑇𝐴 = 0 → 𝑇𝐴 =
𝑇𝐴 =
+ ↑ ∑Fy = 0
;
5√2
8
𝑇𝑐 𝑐𝑜𝑠45°
4
5
→ eq 1
𝑇𝐶
3
Tc sin45° + (5) TA − 588.6 = 0
To get TC, plug in equation 1 to 2:
3 5√2
𝑇𝑐 𝑠𝑖𝑛45° + ( ) (
𝑇 ) − 588.6 = 0
5
8 𝐶
Using shift solve:
𝑇𝑐 = 475.6606302 𝑁
𝑻𝒄 = 𝟒𝟕𝟓. 𝟔𝟔𝟎𝟔 𝑵
To get TA, use eq. 1:
TA =
TA =
5√2
T
8 C
5√2
(475.6606302 )
8
TA = 420.4285714 N
𝑻𝑨 = 𝟒𝟐𝟎. 𝟒𝟐𝟖𝟔 𝑵
→ eq 2
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
EXAMPLE 2
Determine the required length of cord AC so that the 8-kg lamp can be suspended in the position
shown. The undeformed length of spring AB is l’AB = 0.4 m, and the spring has a stiffness of kAB = 300
N/m.
SOLUTION:
+ → ∑Fx = 0
;
TAB − TAC cos30° = 0
;
𝑇𝐴𝐶 𝑠𝑖𝑛30° − 78.5 = 0
→ eq. 1
+ ↑ ∑Fy = 0
78.5
𝑇𝐴𝐶 = 𝑠𝑖𝑛30°
𝑻𝑨𝑪 = 𝟏𝟓𝟕 𝑵
To get TAB, use eq. 1:
TAB − TAC cos30° = 0
TAB = 157 cos30°
TAB = 135.9659884 N
𝑻𝑨𝑩 = 𝟏𝟑𝟓. 𝟗𝟔𝟔𝟎 𝑵
Therefore, the stretch of spring AB is:
𝑻𝑨𝑩 = 𝒌𝑨𝑩 𝒔𝑨𝑩
;
135.9659884 N = 300 N/m (sAB )
sAB = 0.4532199613 m
Stretched length:
lAB = l′AB + sAB
lAB = 0.4 m + 0.4532199613 m
𝒍𝑨𝑩 = 𝟎. 𝟖𝟓𝟑𝟐𝟏𝟗𝟗𝟔𝟏𝟑 𝒎
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
To get lC:
2 m = lAC cos30° + 0.8532199613
lAC = 1.324186528 m
𝒍𝑨𝑪 = 𝟏. 𝟑𝟐𝟒𝟐 𝒎
EXAMPLE 3
A 300-lb box is held at rest on a smooth plane by a force P inclined at an angle θ with the plane.
If θ = 45°, determine the value of P and the normal pressure N exerted by the plane.
SOLUTION:
To get P:
∑Fx = 0
;
Pcosq = Wsin30°
Pcos45° = 300sin30°
P = 212.1320344 lb
𝐏 = 𝟐𝟏𝟐. 𝟏𝟑𝟐𝟎 𝐥𝐛
To get N:
∑Fx = 0 ;
N = Psinq + Wcos30°
N = 212.1320344sin45° + 300cos30°
N = 409.8076211 lb
𝑵 = 𝟒𝟎𝟗. 𝟖𝟎𝟕𝟔 𝒍𝒃
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
Conditions for Rigid Body Equilibrium
When the body is subjected to a system of forces, which all lie in the x–y plane, then the forces
can be resolved into their x and y components. Consequently, the conditions for equilibrium in two
dimensions are:
∑Fx = 0
∑Fy = 0
∑MO = 0
Where, ∑Fx = 0 and ∑Fy = 0 represents the algebraic sums of the x and y components of all the
forces acting on the body, and ∑MO = 0 represents the algebraic sum of the couple moments and the
moments of all the force components about the z axis, which is perpendicular to the x–y plane and passes
through the arbitrary point O.
Procedure for analysis
To construct a free-body diagram for a rigid body or any group of bodies considered as a single
coplanar force equilibrium problem for a rigid body can be solved using the following procedure:
• Apply the moment equation of equilibrium, ∑MO = 0, about a point (O) that lies at the intersection
of the lines of action of two unknown forces. In this way, the moments of these unknowns are
zero about O, and a direct solution for the third unknown can be determined
• When applying the force equilibrium equations, 𝐹𝑥=0 and 𝐹𝑦=0 orient the x and y axes along
lines that will provide the simplest resolution of the forces into their x and y components.
• If the solution of the equilibrium equations yields a negative scalar for a force or couple moment
magnitude, this indicates that the sense is opposite to that which was assumed on the free-body
diagram.
Support Reactions
Before presenting a formal procedure as to how to draw a free-body diagram, it is necessary to
first consider the various types of reactions that occur at supports and points of contact between bodies
subjected to coplanar force systems.
As a general rule:
MIDTERM DISCUSSIONS
Equilibrium of Force Systems
MODULE III
• A support prevents the translation of a body in a given direction by exerting a force on the
body in the opposite direction.
• A support prevents the rotation of a body in a given direction by exerting a couple moment
on the body in the opposite direction.
Free-Body Diagram of Support Reactions
1
2
3
4
5
Supports for Rigid Bodies Subjected to Two-Dimensional Force Systems
Type of Connection
Reaction
Number of Unknowns
• One unknown.
The reaction is a tension
force which acts away
from the member in the
direction of the cable.
• One unknown.
The reaction is a force
which acts along the
axis of the link.
• One unknown.
The reaction is a force
which
acts
perpendicular to the
surface at the point of
contact.
• One unknown.
The reaction is a force
which
acts
perpendicular to the
surface at the point of
contact
• One unknown.
The reaction is a force
which
acts
perpendicular to the
surface at the point of
contact
MIDTERM DISCUSSIONS
Equilibrium of Force Systems
6
7
8
9
10
MODULE III
• One unknown.
The reaction is a force
which
acts
perpendicular to the
slot.
• One unknown.
The reaction is a force
which
acts
perpendicular to the rod.
• Two unknowns.
The reactions are two
components of force, or
the magnitude and
direction ∅ of the
resultant force. Note
that ∅ and θ are not
necessarily
equal
[usually not, unless the
rod shown is a link as in
(2)]
• Two unknowns.
The reactions are the
couple moment and the
force which acts
perpendicular to the rod.
• Three unknowns.
The reactions are the
couple moment and the
two force components,
or the couple moment
and the magnitude and
direction ∅ of the
resultant force.
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
EXAMPLE 1
Determine the horizontal and vertical components of reaction on the beam caused by the pin at
B and the rocker at A. Neglect the weight of the beam.
SOLUTION:
+ → ∑Fx = 0 ;
600cos45° − BX = 0
BX = 424. 2640687 N
𝑩𝑿 = 𝟒𝟐𝟒. 𝟐𝟔𝟒𝟏 𝑵
+ ↑ ∑Fy = 0 ;
319 − 600 𝑠𝑖𝑛45° − 100 − 200 + 𝐵𝑦 = 0
𝐵𝑦 = 405.2640687 𝑁
𝑩𝒚 = 𝟒𝟎𝟓. 𝟐𝟔𝟒𝟏 𝑵
+ ∑MB = 0
;
100 (2 m) + (600𝑠𝑖𝑛45°)(5) − (600𝑐𝑜𝑠45°)(0.2) − 𝐴𝑦 (7) = 0
𝐴𝑦 = 319.4953614 𝑁
𝑨𝒚 = 𝟑𝟏𝟗. 𝟒𝟗𝟓𝟒 𝑵
MIDTERM DISCUSSIONS
MODULE III
Equilibrium of Force Systems
PRACTICE PROBLEM 1
Determine the horizontal and vertical components of reaction on the member at the pin A, and
the normal reaction at the roller B.
SOLUTION:
+ ∑MB = 0:
(NB 𝑐𝑜𝑠30°)(6) − (𝑁𝐵 𝑠𝑖𝑛30°)(2) − 750(3) = 0
𝑁𝐵 = 536.2054981 𝑁
𝑵𝑩 = 𝟓𝟑𝟔. 𝟐𝟎𝟓𝟓 𝑵
+ → ∑Fx = 0 ;
AX − (536.2054981𝑠𝑖𝑛30°) = 0
AX = 268.1027491 N
𝑨𝑿 = 𝟐𝟔𝟖. 𝟏𝟎𝟐𝟕 𝑵
+ ↑ ∑Fy = 0 ;
𝐴𝑦 + (536.2054981𝑐𝑜𝑠30°) − 750 = 0
𝐴𝑦 = 285.632417 𝑁
𝑨𝒚 = 𝟐𝟖𝟓. 𝟔𝟑𝟐𝟒 𝑵
MIDTERM DISCUSSIONS
Equilibrium of Force Systems
MODULE III
ACTIVITY 3
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places.
1. Determine the tension in cables AB, BC, and CD, necessary to support the 10-kg and 15-kg
traffic lights at B and C, respectively. Also, find the angle.
2. The springs BA and BC each have a stiffness of 500 N/m and an unstretched length of 3 m.
Determine the horizontal force F applied to the cord which is attached to the small ring B so that
the displacement of the ring from the wall is d = 1.1 m.
3. The box wrench is used to tighten the bolt at A. If the wrench does not turn when the load is
applied to the handle, determine the torque or moment applied to the bolt and the force of the
wrench on the bolt. (Hint: The bolt act as a fixed support.)
MIDTERM DISCUSSIONS
Friction
IV
MODULE IV
Analysis of Simple
Trusses
Objectives:
•
To show how to determine the forces in the members of a truss using the method
of joints and the method of sections.
•
To analyze the forces acting on the members of frames and machines composed
of pin-connected members.
Outline:
•
Simple Trusses
•
The Method of Joints
•
Zero-Force Members
•
The Method of Sections
•
Frames and Machines
MIDTERM DISCUSSIONS
MODULE IV
Friction
Chapter Introduction
Chapter 4 focuses on analyzing the internal forces present in different types of structures, namely trusses,
frames, and machines. The analysis specifically considers statically determinate structures, which have
the minimum necessary support constraints to maintain equilibrium. As a result, the equilibrium
equations alone are sufficient to determine all unknown reactions.
The analysis of trusses, frames, machines, and beams under concentrated loads is a straightforward
application of the concepts covered in the previous two chapters. The fundamental procedure introduced
in Chapter 3, which involves isolating a body and constructing an accurate free-body diagram, is vital
for analyzing statically determinate structures. This approach forms the basis for understanding and
solving the internal force distribution in these structures.
Simple Trusses
A truss is a structure composed of slender members joined together at their end points. The members
commonly used in construction consist of wooden struts or metal bars. Planar trusses lie in a single plane
and are often used to support roofs and bridges. Shown below is an example of a typical roof-supporting
truss.
Assumption for Design. In order to design the members and connections of a truss, it is
essential to initially determine the forces generated in each member when the
truss is subjected to a specific load. To accomplish this, we will rely on two
significant assumptions:
•
All loadings are applied at the joints.
In most situations, such as for bridge and roof trusses, this assumption is true.
Frequently the weight of the members is neglected because the force
supported by each member is usually much larger than its weight.
•
The members are joined together by smooth pins.
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
The joint connections are usually formed by bolting or welding the ends of the members
to a common plate, called a gusset plane. We can assume these connections act as pins
provided the center lines of the joining members are concurrent.
Because of these two assumptions, each truss member will act as a two
force member, and therefore the force acting at each end of the member
will be directed along the axis of the member. If the force tends to
elongate the member, it is a tensile force (T); whereas if it tends to
shorten the member, it is a compressive force (C).
A simple truss is a basic structural framework composed of straight
members connected at joints. It is a form of a truss system that consists
of triangular elements, which are the most stable and efficient in terms
of load distribution. In a simple truss, the members are typically
subjected to axial forces only, and the joints are assumed to be perfectly pinned or hinged, allowing for
rotation but not translation.
If three members are pin connected at their ends, they form a triangular truss that will be rigid. Attaching
two more members and connecting these members to a new joint D forms a larger truss. If a truss can be
constructed by expanding the basic triangular truss in this way, it is called a simple truss. Figure shown
below shows the addition of two more members to form a larger truss.
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
Method of Joints
To analyze or design a truss, it is necessary to determine the forces present in its individual
members. One approach to achieve this is by utilizing the method
of joints. This method is based on the principle that if the entire truss
is in equilibrium, then each joint within the truss is also in
equilibrium. By constructing free-body diagrams for each joint and
applying force equilibrium equations, the member forces acting on
each joint can be determined.
Since the members of a planar truss are straight and only experience
two forces, they lie within a single plane. Consequently, each joint
in the truss is subject to a concurrent and coplanar force system. As a result, achieving equilibrium at
each joint only requires satisfying the equations ΣFx = 0 and ΣFy = 0.
In the figure shown, three forces are acting upon the pin. It is the 500-N force and the forces exerted by
members BA and BC.
The method of joints is a technique used in structural analysis to determine the internal forces in each
member of a truss. It involves analyzing the equilibrium of forces at each joint in the truss. Here's a stepby-step outline of the method of joints:
1. Start by examining the entire truss and identifying all the external loads acting on the truss
members, including applied loads and reactions at the supports.
2. Begin the analysis at a joint where only two unknown forces act. This is typically a joint where
one member is connected to the support or where two members intersect.
3. Isolate the joint you've chosen and draw a free-body diagram of the joint. Include all the forces
acting on the joint, including any known forces and any forces in the members meeting at that
joint.
4. Apply the equilibrium equations (sum of forces in the x-direction and sum of forces in the ydirection equal to zero) to the joint. This allows you to determine the unknown forces in the
members connected to the joint.
5. Move to the next joint in the truss where two unknown forces act. Repeat steps 3 and 4 to
determine the forces in the members connected to that joint.
6. Continue analyzing each joint with two unknown forces until you have determined the forces in
all the members of the truss.
7. Finally, check the calculated forces against the strength and load-bearing capacity of the truss
members to ensure they are within acceptable limits.
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
By systematically analyzing each joint in the truss using the method of joints, you can determine the
internal forces in all the members. This information is essential for assessing the structural integrity,
designing reinforcements, or evaluating the load-carrying capacity of the truss.
Things to remember when using the method of joints.
• Always start at a joint having at least one known force and at most two unknown
forces.
In this way, application of ΣFx = 0 and ΣFy = 0
yields two algebraic equations which can be solved
for the two unknowns.
•
Always assume the unknown member forces acting on the joint’s freebody diagram is in tension.
B
500 N
With this, the forces in the joints are pulling on the pin. If this is
done, then the numerical solution of the equilibrium equations will
yield positive scalars for members in tension and negative scalers
FBC (tension)
for members in compression. Once an unknown member force is
FBA (tension)
found, use its correct magnitude and sense (T or C) on subsequent
joint free-body diagrams.
SAMPLE PROBLEM
The structure shown is a truss which is
pinned to the floor at point A and
supported by roller at point D. Determine
the force to all members of the truss.
MIDTERM DISCUSSIONS
MODULE IV
Analysis of Simple Trusses
SOLUTION:
First, determine the support
reactions acting on the truss.
No joint can be analyzed until
the support reactions are
determined
Ax
Dy
Ay
Using the summation of the forces in the x-axis, the value of Ax can be determined.
∑ 𝐹𝑥 = 0
𝐴𝑥 + 80 = 0
𝐴𝑥 = −80 𝑘𝑁
𝑨𝒙 = 𝟖𝟎 𝒌𝑵 ⃪
Assuming that the moment is positive at clockwise,
+ ∑ 𝑀𝐴 = 0
(50 𝑘𝑁)(2 𝑚) + (80 𝑘𝑁)(0.75 𝑚) + (𝑫𝒚 )(3 𝑚) = 0
𝑫𝒚 = 𝟓𝟑. 𝟑𝟑𝟑𝟑 𝒌𝑵
Through the summation of forces at the y-axis, the value of Ay can be computed.
∑ 𝐹𝑦 = 0
𝐴𝑦 + 53.3333 𝑘𝑁 − 50 𝑘𝑁 = 0
𝑨𝒚 = −𝟑. 𝟑𝟑𝟑𝟑 𝒌𝑵
𝑨𝒚 = 𝟑. 𝟑𝟑𝟑𝟑 𝒌𝑵 ↓
MIDTERM DISCUSSIONS
MODULE IV
Friction
With all values of the forces at support reactions determined, we can now proceed to the
determination of forces at individual joints.
@ Joint A
3
∑Fy = 0 ; −3.3333 + 𝐹𝐴𝐵 (5) = 0
𝑭𝑨𝑩 = 𝟓. 𝟓𝟓𝟓𝟔 𝒌𝑵, 𝑻 (𝒕𝒆𝒏𝒔𝒊𝒐𝒏)
4
∑Fx = 0 ; −80 + 5.5556 (5) + 𝐹𝐴𝐸 = 0
𝑭𝑨𝑬 = 𝟕𝟓. 𝟓𝟓𝟓𝟔 𝒌𝑵, 𝑻 (𝒕𝒆𝒏𝒔𝒊𝒐𝒏)
@ Joint B
4
∑Fx = 0 ; 𝐹𝐵𝐶 − 5.5556 (5) = 0
𝑭𝑩𝑪 = 𝟒. 𝟒𝟒𝟒𝟒 𝒌𝑵, 𝑻 (𝒕𝒆𝒏𝒔𝒊𝒐𝒏)
3
∑Fy = 0 ; −𝐹𝐵𝐸 − 5.5556 (5) = 0
𝑭𝑩𝑬 = 𝟑. 𝟑𝟑𝟑𝟒 𝒌𝑵, 𝑪 (𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏)
@ Joint E
∑Fx = 0 ;
3
−3.3334 + 𝐹𝐶𝐸 (5) = 0
𝑭𝑪𝑬 = 𝟓. 𝟓𝟓𝟓𝟔 𝒌𝑵, 𝑻 (𝒕𝒆𝒏𝒔𝒊𝒐𝒏)
∑Fy = 0 ;
4
−75.5556 + 5.5556 (5) − 𝐹𝐸𝐹 = 0
𝑭𝑬𝑭 = 𝟕𝟏. 𝟏𝟏𝟏𝟏 𝒌𝑵, 𝑻 (𝒕𝒆𝒏𝒔𝒊𝒐𝒏)
MIDTERM DISCUSSIONS
MODULE IV
Analysis of Simple Trusses
@ Joint F
∑Fx = 0 ;
−71.1111 + 𝐹𝐷𝐹 = 0
𝑭𝑫𝑭 = 𝟕𝟏. 𝟏𝟏𝟏𝟏 𝒌𝑵, 𝑻 (𝒕𝒆𝒏𝒔𝒊𝒐𝒏)
∑Fy = 0 ;
𝐹𝐶𝐹 − 50 = 0
𝑭𝑪𝑭 = 𝟓𝟎 𝒌𝑵, 𝑻 (𝒕𝒆𝒏𝒔𝒊𝒐𝒏)
@ Joint C
4
4
∑Fx = 0 ; −4.4444 − 5.5556 (5) + 𝐹𝐶𝐷 (5) + 80 = 0
𝐹𝐶𝐷 = −88.8889 𝑘𝑁
𝑭𝑪𝑫 = 𝟖𝟖. 𝟖𝟖𝟖𝟗 𝒌𝑵, 𝑪 (𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏)
Summary:
𝐹𝐴𝐵 = 5.5556 𝑘𝑁, 𝑇
𝐹𝐴𝐸 = 75.5556 𝑘𝑁, 𝑇
𝐹𝐵𝐶 = 4.4444 𝑘𝑁, 𝑇
𝐹𝐵𝐸 = 3.3334 𝑘𝑁, 𝐶
𝐹𝐶𝐷 = 88.8889 𝑘𝑁, 𝐶
𝐹𝐶𝐸 = 5.5556 𝑘𝑁, 𝑇
𝐹𝐶𝐹 = 50 𝑘𝑁, 𝑇
𝐹𝐸𝐹 = 71.1111 𝑘𝑁, 𝑇
𝐹𝐷𝐹 = 71.1111 𝑘𝑁, 𝑇
Zero Force Members
Zero force members are structural members within a truss or framework that carry no load or
force. These members exist due to the configuration of the structure and the forces applied to it. They
are typically found in trusses where the external loads and support conditions create a balanced condition
that results in certain members having zero forces.
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
Truss analysis using the method of joints is greatly simplified if we can first identify those
members which support no loading. These zero-force members are used to increase the stability of the
truss during construction and to provide added support if the loading is changed.
If a free-body diagram of the pin at joint A is drawn, Fig. 6–11b, it is seen that members AB and
AF are zero-force members. Here again it is seen that DC and AF are zero-force members. From these
observations, we can conclude that if only two non-collinear members form a truss joint and no
external load or support reaction is applied to the joint, the two members must be zero-force
members. Shown below are illustrations of zero-force members.
The free-body diagram of the pin at joint D is shown in the figure below. By orienting the y axis
along members DC and DE and the x axis along member DA, it is seen that DA is a zero-force member.
Note that this is also the case for member CA. In general then, if three members form a truss joint for
which two of the members are collinear, the third member is a zero-force member provided no
external force or support reaction has a component that acts along this member.
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
SAMPLE PROBLEM
Determine all the zero-force members.
Assume all joints are pin connected.
ANSWER:
Therefore, FBG, FGC, FCF, and FFD are the zero
force members.
Method of Sections
When we need to find force in only a few members of a truss, we can analyze the truss using the method
of sections. It is based on the principle that if the truss is in equilibrium, then any segment of the truss is
also in equilibrium. The method of sections can also be used to “cut” or section the members of an entire
truss. If the section passes through the truss and the free-body diagram of either of its two parts is drawn,
we can apply the equations of equilibrium of that part to determine the member forces at the “cut section.”
Since only three independent equilibrium equations (ΣFx = 0, ΣFy = 0, and ΣMO= 0) can be applied to
the free-body diagram of any segment, then we should try to select a section that, in general, passes
through not more than three members in which the forces are unknown.
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
The method of sections is a technique used in structural analysis to determine the internal forces (such
as axial forces or bending moments) within a specific section of a truss or framework. This method
involves cutting through the structure along a desired section and analyzing the equilibrium of forces on
that section.
Here's a step-by-step outline of the method of sections:
1. Identify the section of the truss or framework for which you want to determine the internal forces.
This section should typically include a maximum of three unknown forces to simplify the
analysis.
2. Cut the structure along the selected section, separating it from the rest of the structure.
3. Isolate the portion of the structure that has been cut. Treat this isolated section as a free-body
diagram.
4. Identify all the forces acting on the isolated section, including the externally applied loads and
any reactions at the supports.
5. Apply the equilibrium equations (sum of forces and moments equal to zero) to analyze the forces
acting on the isolated section.
6. Solve the equilibrium equations to determine the unknown forces within the section. This may
involve solving a system of equations.
By applying the method of sections to multiple sections of a truss or framework, you can determine the
internal forces in various parts of the structure. This information is crucial for evaluating the structural
integrity, designing reinforcements, or assessing the load-carrying capacity of the structure.
Things to remember when using the method of sections.
• Always assume that the unknown member forces at the cut section are tensile
forces.
• Before isolating the appropriate section, it may first be necessary to determine the
truss’s support reactions. If this is done, then the three equilibrium equations will
be available to solve for member forces at the section.
MIDTERM DISCUSSIONS
MODULE IV
Analysis of Simple Trusses
• Draw the free-body diagram of that segment of the sectioned truss which has the
least number of forces acting on it.
SAMPLE PROBLEM
Determine the force in members GE,
GC, and BC of the truss shown in the
figure. Indicate whether the members
are in tension or compression.
SOLUTION:
Like the method of joints, the determination of the forces in the support reactions comes first.
Support forces:
∑𝐹𝑥 = 0
𝐴𝑥 + 400 = 0
𝐴𝑥 = −400𝑁
𝑨𝒙 = 𝟒𝟎𝟎 𝑵
Q + ∑MA = 0
−1200(8) − 400(3) + Dy (12) = 0
𝑫𝒚 = 𝟗𝟎𝟎 𝑵
MIDTERM DISCUSSIONS
MODULE IV
Analysis of Simple Trusses
∑ 𝐹𝑦 = 0
𝐴𝑦 + 𝐷𝑦 − 1200 = 0
𝐴𝑦 + 900 − 1200 = 0
𝑨𝒚 = 𝟑𝟎𝟎 𝑵
G
GE
Force in members:
5
3
4
∑ 𝐹𝑦 = 0 ;
GC
3
𝐴𝑦 − 𝐹𝐺𝐶 (5) = 0
3
300 = 𝐹𝐺𝐶 (5)
A
BC
𝑭𝑮𝑪 = 𝟓𝟎𝟎 𝑵, 𝑻
B
Ax = 400 N
𝑄 + ∑𝑀𝐺 = 0;
Ay = 300 N
−400(3) − 300(4) + 𝐹𝐵𝐶 (3) = 0
𝑭𝑩𝑪 = 𝟖𝟎𝟎𝑵, 𝑻
∑ 𝐹𝑋 = 0
4
−𝐴𝑋 + 𝐹𝐵𝐶 + 𝐹𝐺𝐸 + 𝐹𝐺𝐶 ( ) = 0
5
4
−400 + 800 + 𝐹𝐺𝐸 + 500( ) = 0
5
𝑭𝑮𝑬 = 𝟖𝟎𝟎 𝑵, 𝑪
SUMMARY:
𝐹𝐺𝐶 = 500 𝑁, 𝑇
𝐹𝐵𝐶 = 800 𝑁, 𝑇
𝐹𝐺𝐸 = 800 𝑁, 𝐶
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
Frames and Machines
Frames and Machines are two types of structures which are often composed of pin-connected multi-force
members. Frames are used to support loads, whereas machines contain moving parts and are designed to
transmit and alter the effect of forces. Provided a frame or machine contains no more supports or
members than are necessary to prevent its collapse, the forces acting at the joints and supports can be
determined by applying the equations of equilibrium to each of its members. Once these forces are
obtained, it is then possible to design the size of the members, connections, and supports using the theory
of mechanics of materials and an appropriate engineering design code.
Free-Body Diagrams. In order to determine the forces acting at the joints and supports of a frame or
machine, the structure must be disassembled, and the free-body diagrams of its parts must be drawn. The
following important points must be observed:
• Isolate each part by drawing its outlined shape. Then show all the forces and/or couple moments
that act on the part.
• Identify all the two-force members in the structure and represent their free-body diagrams as
having two equal but opposite collinear forces acting at their points of application.
• Forces common to any two contacting members act with equal magnitudes but opposite sense on
the respective members. If the two members are treated as a “system” of connected members,
then these forces are “internal” and are not shown on the free-body diagram of the system;
however, if the free-body diagram of each member is drawn, the forces are “external” and must
be shown as equal in magnitude and opposite in direction on each of the two free-body diagrams.
SAMPLE PROBLEM
Determine the horizontal and vertical
components of force which the pin at C
exerts on member BC of the frame.
MIDTERM DISCUSSIONS
MODULE IV
Analysis of Simple Trusses
SOLUTION:
FAB
2000 N
Cx
FAB
FAB
Cy
𝑄 + 𝑀𝐶 = 0
2000(2) = 𝐹𝐴𝐵 sin 60(4) = 0
𝑭𝑨𝑩 = 𝟏𝟏𝟓𝟒. 𝟕𝟎𝟎𝟓 𝑵
∑ 𝐹𝑌 = 0
−2000 + 1154.7005 sin 60 + 𝐶𝑌 = 0
𝑪𝒀 = 𝟏𝟎𝟎𝟎 𝑵 ↑
∑ 𝐹𝑋 = 0
1154.7005 cos 60˚ + 𝐶𝑋 = 0
𝐶𝑋 = −577.3503𝑁
𝑪𝑿 = 𝟓𝟕𝟕. 𝟑𝟓𝟎𝟑 𝑵 ←
MIDTERM DISCUSSIONS
MODULE IV
Analysis of Simple Trusses
PRACTICE PROBLEM
The frame supports the 50-kg cylinder. Determine the
horizontal and vertical components of reaction at A and the
force at C.
SOLUTION:
∑ 𝐹𝑋 = 0 ;
𝐷𝑋 − 50(9.81) = 0
∑ 𝐹𝑋 = 0
𝐷𝑋 = 490.5 𝑁
𝐴𝑋 − 245.25𝑁 − 490.5𝑁 = 0
𝑨𝑿 = 𝟕𝟑𝟓. 𝟕𝟓 𝑵
∑ 𝐹𝑌 = 0 ;
𝐷𝑌 − 50(9.81) = 0
𝐷𝑌 = 490.5 𝑁
∑ 𝐹𝑌 = 0
𝐴𝑌 − 490.5 = 0
+ 𝑀𝐴 = 0;
𝐹𝐵𝐶 (0.6) − 𝐷𝑌 (1.2) + 𝐷𝑋 (0.9) = 0
𝐹𝐵𝐶 (0.6) − 490.5(1.2) + 490.5(0.9) = 0
𝑭𝑩𝑪 = 𝟐𝟒𝟓. 𝟐𝟓 𝑵
𝑨𝒀 = 𝟒𝟗𝟎. 𝟓 𝑵
MIDTERM DISCUSSIONS
Analysis of Simple Trusses
MODULE IV
ACTIVITY 4
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places.
1. Determine the force in each member of the truss using method of joints, and state if the members
are in tension or compression. At the end of your solutions, make a summary of all your answers.
2. Using method of sections, determine the force in members CD, CJ and KJ of the truss which serves
to support the deck of a bridge. State if these members are in tension or compression.
3.
Determine the horizontal and vertical components of force at pins B and C. The suspended cylinder
has a mass of 75 kg. (20 points)
MIDTERM DISCUSSIONS
MODULE IV
Friction
Friction
V
Objectives:
•
•
•
•
•
•
Understand the concepts of Friction
Solve sample problems regarding slipping and tipping.
Understand the concept of Wedges
Solve sample problems regarding wedges.
Understand the concept of Friction Force on Flat Belts
Solve Sample Problems Regarding Friction Force on Flat Belts
Outline:
•
•
•
•
Chapter Introduction
Friction
o Introduction to Friction
o Characteristics of Dry Friction
▪ Impending Motion
▪ Motion
▪ Coefficient of Friction
o Slipping and Tipping
o Real-life Applications of Friction
o Sample and Practice Problems on Tipping and Slipping
Wedges
o Introduction to Wedges
o Real-life Applications of Wedges
o Sample and Practice problems on Wedges
Friction Force on Flat Belts
o Introduction to Friction Force and Flat Belts
o Real-life Applications of Friction Force on Flat Belts
o Sample and Practice Problems on Friction Force and Flat Belts
MIDTERM DISCUSSIONS
Friction
MODULE V
Chapter Introduction
This chapter delves into the fundamental concept of friction and its wide-ranging applications in
the field of mechanics. We explore the characteristics of dry friction, analyzing its influence on motion,
stability, and control in mechanical systems. Additionally, we investigate the phenomena of slipping and
tipping, uncovering the risks associated with unstable equilibrium and the loss of traction. Moreover, we
delve into the mechanics of wedges, showcasing their mechanical leverage and practical applications.
Lastly, we examine the friction forces exerted on flat belts, highlighting their role in power transmission
and efficient operation. Through this exploration, we gain a comprehensive understanding of the
significance of friction in enhancing safety, stability, and performance across various mechanical
domains.
Friction
In the realm of mechanics, friction emerges as a
force that arises when two surfaces come into contact and
either slide or attempt to slide against each other. For
instance, when we endeavor to push a toy car along the
floor, friction comes into play. It is important to note that
friction always acts in the direction opposite to the
motion of the object, consistently parallel to the plane of
contact. The magnitude of friction is contingent upon the
materials comprising the surfaces in question.
Intriguingly, the roughness of a surface directly influences the amount of friction it produces. Thus, a
rougher surface generates greater friction. Throughout this chapter, we will delve into the intricacies of
friction, examining its characteristics and exploring its significance in various mechanical scenarios.
Characteristics of Dry Friction
Dry friction stands as a significant force between solid surfaces in direct contact, without any
lubrication or fluid. This type of friction manifests when two surfaces slide or attempt to slide against
each other, influencing motion, stability, and control in mechanical systems. Understanding the
characteristics and implications of dry friction is crucial in designing and optimizing mechanisms,
ensuring safe and efficient operation.
MIDTERM DISCUSSIONS
MODULE V
Friction
Impending Motion
In cases where the surfaces of contact are rather
“slippery,” the frictional force F may not be great enough to
balance P, and consequently the block will tend to slip. P is
slowly increased, F correspondingly increases until it attains
a certain maximum value Fs, called the limiting static
frictional force.
In this case, the static fictional force has reached its
upper limit. The point of impending motion is the point before an object. It can also be thought of as a
maximum static friction force before slipping.
When this value is reached, the block is in unstable equilibrium since any further increase in P
will cause the block to move. It has been determined that this limiting static frictional force Fs is directly
proportional to the resultant normal force N.
Fmax = μsN
This equation applies only to cases where motion is impending with the friction force at its peak value
or when it is on the verge of slipping
Fmax < μsN
This equation applies for conditions of equilibrium when motion is not impending or the body remains
at rest.
𝐹𝑘
𝜇𝑠 𝑁
𝜙𝑠 = tan−1 ( ) = tan−1 (
) = tan−1
𝑁
𝑁
This equation is used to determine the angle.
𝜇𝑠
Motion
Kinetic Frictional Force is explained through the diagram. If the magnitude
of P acting on the block is increased so that it becomes slightly greater than Fs, the
frictional force at the contacting surface will drop to a smaller value Fk. The block
will also begin to slide with increasing speed.
𝐹𝑘
𝜇𝑘
𝜙𝑘 = tan−1 ( ) = tan−1 ( ) = tan−1 𝜇𝑘
𝑁
𝑁
MIDTERM DISCUSSIONS
Friction
MODULE V
This equation is used to solve for the missing angle.
Coefficient of Friction
The Coefficient of Friction relates the magnitude of the frictional force to the magnitude of the
normal force and is given by the equation: 𝐹𝑓 = 𝜇𝑁. We use the coefficient of static friction 𝜇𝑠 as the
maximum possible coefficient of friction for bodies at rest. We use the coefficient of kinetic friction 𝜇𝑘
for bodies in motion.
Frictional force is categorized in three ways.
1. F is a static frictional force if equilibrium is maintained.
2. F is a limiting static frictional force 𝐹𝑠 when it reaches a maximum value needed to
maintain equilibrium.
3. F is a kinetic frictional force 𝐹𝑘 when sliding occurs at the contacting surface.
The maximum static frictional force is generally greater than the kinetic frictional force for any
two surfaces of contact. However, if one of the bodies is moving with a very low velocity over the surface
of another, 𝐹𝑘 becomes approximately equal to 𝐹𝑠 . When slipping at the surface of contact is about to
occur, the maximum static frictional force is proportional to the normal force, and is given by: 𝐹𝑠 =
𝜇𝑠 𝑁 When slipping at the surface of contact is occurring, the kinetic frictional force is proportional to
the normal force, such that 𝐹𝑘 = 𝜇𝑘 𝑁.
Slipping and Tipping
Slipping is a critical phenomenon in engineering
mechanics, characterized by the loss of traction between
surfaces. It poses significant challenges to the stability
and performance of mechanical systems. In this chapter,
we explore the causes and implications of slipping,
focusing on strategies to mitigate its effects. By
understanding and addressing slipping in engineering
applications, we can enhance safety and optimize performance in various mechanical systems.
The pushing force will exceed the maximum static friction force and the box will begin to slide
across the surface. The normal force must be located at a certain distance from the center of the body.
MIDTERM DISCUSSIONS
Friction
MODULE V
Tipping, a phenomenon of immense importance in engineering
mechanics, occurs when an object loses its balance and rotates around
a pivot point. It poses significant challenges to the stability and safety
of mechanical systems. In this chapter, we delve into the causes and
consequences of tipping, examining its effects on various engineering
applications. By exploring the principles and strategies to prevent
tipping, we can optimize the design and operation of mechanical
systems, ensuring stability, reliability, and mitigating potential risks.
When an object starts to tip, it starts to pivot around a point. As a result, the contact forces
(frictional and normal) must be applied at the pivot point or edge of the body.
Real-life application of Friction
Friction plays a crucial role in various real-life applications in mechanics. Here are a few
examples:
1. Transportation: Friction is essential for the movement of vehicles on roads. The friction between
the tires and the road surface provides the necessary grip and traction, allowing vehicles to
accelerate, decelerate, and navigate turns safely.
2. Braking systems: Friction is employed in braking systems to slow down or stop moving objects.
For instance, the friction between brake pads and rotors in a car's disc brakes converts kinetic
energy into heat, resulting in the vehicle's deceleration or complete halt.
3. Fasteners and joints: Friction is employed in mechanical fasteners such as screws, bolts, and nails
to hold objects together. The frictional force generated between the mating surfaces ensures that
the fasteners remain secure and prevent unintentional loosening or separation.
Here are some real-life applications of dry friction:
1. Automobile tires: Dry friction plays a vital role in the performance of automobile tires. The
friction between the tire and the road surface provides the necessary grip for accelerating,
decelerating, and maintaining control while driving. It allows vehicles to navigate turns safely
and ensures effective braking.
2. Brakes and clutches: Dry friction is extensively used in brake systems and clutches. Frictional
force between the brake pads and the rotors or between the clutch plates allows for the controlled
transfer of rotational energy to heat, resulting in deceleration or engagement of the clutch.
MIDTERM DISCUSSIONS
Friction
MODULE V
3. Woodworking and carpentry: Dry friction is essential in woodworking applications. It allows
saws, sandpapers, and other tools to grip and cut through wood effectively, shaping and
smoothing the surfaces as desired.
Here are some real-life examples of slipping:
1. Vehicle skidding: slippery road conditions, such as rain, snow, or ice, can cause vehicles to lose
traction with the road surface. This can result in skidding or sliding, reducing the driver's control
over the vehicle and potentially leading to accidents.
2. Spills in the kitchen or workplace: Spilled liquids or substances on floors in kitchens, restaurants,
or workplaces can create slipping hazards. The lack of friction between the footwear and the
slippery surface increases the chances of accidents and injuries.
3. Slippery hand tools: Certain hand tools or objects can become slippery due to grease, oil, or other
substances. This can reduce the grip and control over the tool, increasing the risk of accidents or
injuries.
Here are a few real-life examples of tipping in mechanics:
1. Cranes and heavy machinery: Tipping is a significant concern when operating cranes, excavators,
or other heavy machinery. If the load being lifted is unbalanced, exceeds the machine's capacity,
or if the machine is on unstable ground, there is a risk of tipping over.
2. Forklifts: Forklifts are prone to tipping if they are not operated within their specified load
capacity or if the load is not properly positioned. Inaccurate weight distribution or sudden
maneuvers can cause the forklift to tip over.
3. Construction equipment: Tipping is a concern with construction equipment such as bulldozers,
loaders, and cranes. Uneven terrain, exceeding load limits, or sudden movements can lead to
tipping incidents, jeopardizing the operator's safety, and causing potential damage.
MIDTERM DISCUSSIONS
MODULE V
Friction
SAMPLE PROBLEM 1
The uniform crate shown in the figure has a mass of 20 kg. If a force P = 80 N is applied to the
crate, determine if it remains in equilibrium. The coefficient of static friction is μs = 0.3.
SOLUTION:
Step 1: Draw the Free-Body Diagram
Step 2: Summation of Forces and Moments to get the xdistance.
∑ 𝐹𝑥 = 0;
80 cos 3 0°𝑁 − 𝐹 = 0
𝐹 = 69.3 𝑁
∑ 𝐹𝑦 = 0;
−80 sin 3 0°𝑁 + 𝑁𝑐 − 196.2𝑁
𝑁𝑐 = 236.2 𝑁
∑ + ↺ 𝑀𝑜 = 0;
80 sin 3 0°𝑁(0.4𝑚) − 80 cos 3 0°𝑁(0.2𝑚) + 𝑁𝑐 (𝑥)
𝒙 = −𝟎. 𝟎𝟎𝟗𝟎𝟖𝒎 = − 𝟗. 𝟎𝟖𝟎𝟎𝒎𝒎
Note that: Since x is negative, it indicates the resultant normal force acts slightly to the left of
the crate's center line. NO tipping will occur since x is less than 0.4 meters. Also, the maximum
frictional force which can be developed at the surface of contact.
We can say that 𝑓𝑚𝑎𝑥 = 𝜇𝑠 𝑁𝑐 = 0.3(236.2) = 70.9𝑁. Since 𝐹 = 69.3 < 70.9 , the crate
will not slip, although it is very close to doing so.
MIDTERM DISCUSSIONS
Friction
MODULE V
Note: If the x distance of the normal force from the center of the body is greater than the half
distance of the body, then it must lie on the edge or outside the body. Therefore, TIPPING will occur.
SAMPLE PROBLEM 2
Determine the maximum value of P that can be applied without
causing movement of the 250lb crate that has a center of gravity at G.
𝜇𝑠 = 0.4
Note: When we look at cases where either slipping or tipping
may occur, we are usually interested in finding which of the options
will occur first or maximum force needed. Whichever option requires
less force is the option that will occur first and its force will be
considered.
SOLUTION:
Solution when considering Tipping:
∑ + ↺ 𝑀 = 0;
250(1.5) − 𝑃(4.5) = 0
𝑷 = 𝟖𝟑. 𝟑𝟑𝟑𝟑𝒍𝒃
Solution when considering Slipping:
∑ 𝐹𝑦 = 0;
−250 + 𝑁 = 0
𝑵 = 𝟐𝟓𝟎
∑ 𝐹𝑥 = 0;
𝑃 − 𝐹𝑓 = 0
𝑃 − 𝜇𝑠 𝑁 = 0
𝑃 − 0.4(𝑁) = 0
𝑷 = 𝟏𝟎𝟎𝒍𝒃
Note that tipping occurs when the normal force is at the edge of the contact surface.
Note: At P = 83.3333lb, the crate will tip and at P = 100lb, the crate will slip.
Tipping governs since P at tipping is smaller. Therefore, the maximum allowable P is 83.3333lb.
MIDTERM DISCUSSIONS
MODULE V
Friction
PRACTICE PROBLEM 1
The uniform 10-kg ladder rests against the smooth wall
at B, and the end A rests on the rough horizontal plane for which
the coefficient of static friction is μs = 0.3. Determine the angle
of inclination θ of the ladder and the normal reaction at B if the
ladder is on the verge of slipping.
SOLUTION:
Step 1: Draw the Free-Body Diagram
Step 2: Summation of Forces of x and y, and moments at point A.
Consider the ladder is on the verge of slipping based on the diagram.
∑ 𝐹𝑦 = 0;
𝑁𝐴 − 10(9.81) = 0
𝑁𝐴 = 98.1𝑁
𝐹𝐴 = 𝜇𝑠 𝑁𝐴 0 = .3(98.1𝑁) = 29.43𝑁
∑ 𝐹𝑥 = 0;
29.43𝑁 − 𝑁𝐵 = 0
𝑁𝐵 = 29.43𝑁
∑ + ↻ 𝑀𝐴 = 0;
29.43(4)𝑠𝑖𝑛𝜃 − 98.1𝑁(2)𝑐𝑜𝑠𝜃 = 0
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
= 𝑡𝑎𝑛𝜃 = 1.6667
𝜽 = 𝟓𝟗. 𝟎𝟒°
MIDTERM DISCUSSIONS
Friction
MODULE V
Wedges
Wedges, simple yet powerful mechanical devices, hold
significant importance in engineering mechanics. Their unique
design provides leverage and finds widespread applications
across various industries. In this chapter, we explore the
principles and practical uses of wedges in engineering. By
understanding their mechanics, we can harness the mechanical
advantage they offer to improve performance and safety in
diverse engineering systems. Discover the remarkable potential
of wedges and their ability to optimize mechanical designs in fields ranging from construction to
manufacturing.
A wedge is a simple machine that is often used to transform an applied force into much larger forces,
directed at approximately right angles to the applied force. Wedges also can be used to slightly move or
adjust heavy loads. Wedges are often used to adjust the elevation of structural or mechanical parts.
Also, they provide stability for objects.
The above figure shows a wedge under a block that is supported by a wall. If the force P is
large enough to push the wedge forward, then the block will rise. Here we have excluded the weight of
the wedge since it is usually small compared to the weight W of the block. Also, note that the frictional
forces F1 and F2 must oppose the motion of the wedge. Likewise, the frictional force F3 of the wall on
the block must act downward to oppose the block’s upward motion.
Provided the coefficient of friction is very small or the wedge angle θ is large, then the applied force P
must act to the right to hold the block. Otherwise, P may have a reverse sense of direction to pull on the
wedge to remove it. If P is not applied and friction forces hold the block in place, then the wedge is
referred to as self- locking.
MIDTERM DISCUSSIONS
Friction
MODULE V
Real life applications of Wedges
Wedges, with their simple yet effective design, find numerous practical applications in
engineering mechanics. Some real-life examples of the applications of wedges are:
1. Construction and Architecture: Wedges are extensively used in construction and architecture
for tasks such as splitting rocks, securing structural elements, and lifting heavy objects. They
are employed in masonry, stone carving, and roof truss assembly, where they provide stability
and support.
2. Manufacturing and Assembly: In manufacturing industries, wedges are used for tasks such as
assembling components, aligning parts, and securing objects in place. They play a crucial role
in automotive assembly, woodworking, and metalworking processes, ensuring precise and
secure positioning.
3. Heavy Equipment and Machinery: Wedges are utilized in heavy equipment and machinery for
tasks such as leveling, lifting, and stabilization. They are employed in hydraulic jacks, cranes,
and bulldozers to provide mechanical advantage and enhance safety during lifting and moving
operations.
4. Aerospace Industry: In the aerospace industry, wedges are used in various applications,
including aircraft assembly, maintenance, and repair. They assist in aligning and securing
components, adjusting clearances, and maintaining structural integrity.
5. Mining and Quarrying: Wedges are employed in mining and quarrying operations for rock
splitting, extraction, and excavation. They are utilized to create cracks and separate large rock
formations, facilitating the extraction of valuable minerals or creating space for further
excavation.
SAMPLE PROBLEM 1
The uniform stone has a mass of 500 kg
and is held in the horizontal position using a
wedge at B. If the coefficient of static friction
is μs = 0.3 at the surfaces of contact, determine
the minimum force P needed to remove the
wedge. Assume that the stone does not slip at
A.
MIDTERM DISCUSSIONS
Friction
SOLUTION:
Step 1: Draw the FBD of the Object and the Wedge Separately.
Step 2: Solve missing variables using the givens and moments at point A.
𝑊 = 500(9.81) = 4905𝑁
∑ + ↺ 𝑀𝐴 = 0;
−4905(0.5) + 𝑁𝐵 cos 7 °(1) + 0.3𝑁𝐵 sin 7 °(1) = 0
𝑁𝐵 = 2383.1𝑁
Step 3: Use the result from the object to the wedge through summation of forces.
∑ 𝐹𝑦 = 0;
𝑁𝑐 − 2383.1 cos 7 ° − 0.3(2383.1 sin 7 °)
𝑁𝐶 = 2452.5𝑁
∑ 𝐹𝑥 = 0;
2383.1 sin 7 ° − 0.3(2383.1 cos 7 °) + 𝑃 − 0.3(2452.5)
𝑷 = 𝟏𝟏𝟓𝟒. 𝟗𝑵 = 𝟏. 𝟏𝟓𝟒𝟗 𝒌𝑵
MODULE V
MIDTERM DISCUSSIONS
MODULE V
Friction
Friction Force on Flat Belts
Friction force on flat belts plays a crucial role in
engineering mechanics, particularly in power
transmission systems. When a flat belt is in motion, the
friction between the belt and the pulleys enables the
transfer of power and torque. Understanding the
principles and characteristics of this friction force is
essential for designing and optimizing belt-driven
systems. Consider the flat belt shown in Fig. a, which
passes over a fixed curved surface. The total angle of
belt-to-surface contact in radians is β, and the coefficient
of friction between the two surfaces is μ. We wish to
determine the tension T2 in the belt, which is needed to
pull the belt counterclockwise over the surface, and thereby overcome both the frictional forces at the
surface of contact and the tension T1 in the other end of the belt. Obviously, T2 > T1
In solving for T2, we can say that,
𝑇2 = 𝑇1 𝑒 𝜇𝛽
Where:
T2 , T1 = belt tensions; T1 opposes the direction of motion of the belt measured relative to the surface,
while T2 acts in the direction of the relative belt motion because of friction, T2 > T1
μ = coefficient of static or kinetic friction between the belt and the surface of contact
Β = angle of belt-to-surface contact, measured in radians
e = 2.718....., base of the natural logarithm
Note that T2 is independent of the radius of the drum, and instead it is a function of the angle of
belt to surface contact, β. As a result, this equation is valid for flat belts passing over any curved
contacting surface.
Real Life Applications of Friction Force on Flat Belts:
The friction force on flat belts finds practical applications in various industries and everyday
scenarios. Some real-life examples include:
MIDTERM DISCUSSIONS
Friction
MODULE V
1. Industrial Machinery: Friction force on flat belts is extensively used in industrial machinery for
power transmission. Flat belts connect pulleys on machines such as conveyor systems, milling
machines, and industrial drives. The friction between the belt and pulleys allows for the transfer
of rotational motion and power, enabling the smooth operation of these machines.
2. Automotive Industry: In automobiles, friction force on flat belts is crucial for driving auxiliary
components such as alternators, water pumps, and power steering systems. Serpentine belts, a
type of flat belt, transmit power from the engine crankshaft to these accessories, ensuring their
proper functioning.
3. Agricultural Machinery: Friction force on flat belts is commonly used in agricultural machinery,
such as combine harvesters and tractors. Belts transmit power from the engine to various
components, including grain threshers, conveyor systems, and cutting mechanisms, facilitating
agricultural operations.
4. Exercise Equipment: Many exercise machines, such as treadmills and stationary bikes, employ
friction force on flat belts for resistance and power transmission. The tension and friction
between the belt and pulleys provide the required resistance levels, allowing users to engage in
effective workouts.
5. Printing Industry: In printing presses, friction force on flat belts is utilized to transfer rotational
motion from the drive motor to the printing cylinders and other components. This ensures
accurate paper feed, precise registration, and consistent printing quality.
PRACTICE PROBLEM 1
The maximum tension that can be
developed in the cord is 600 N. If the
pulley at A is free to rotate and the
coefficient of static friction at the fixed
drums B and C is μs = 0.25, determine the
largest mass of the cylinder that can be
lifted by the cord.
MIDTERM DISCUSSIONS
MODULE V
Friction
SOLUTION:
Step 1: List the Given
𝑇𝑚𝑎𝑥 = 600𝑁
𝜇𝑠 = 0.25
Step 2: Solve 𝛽 by considering its angle of contact.
𝛽 = 180 − 45
𝛽 = 135° =
Solution Step 3: Consider Drum D.
𝑇 = 𝑇1 𝑒 𝜇𝛽
600 = 𝑇1 𝑒
𝜋
2
(0.25)( )
𝑇1 = 332.9129𝑁
Solution step 4. Consider pulley A which is frictionless.
𝑇1 = 𝑇2 𝑒 𝜇𝛽
332.9129 = 𝑇2 𝑒
𝜋
2
(0)( )
𝑇2 = 332.9129 𝑁
Solution Step 5: Consider Drum C.
𝑇2 = 𝑊𝑒 𝜇𝛽
332.9129 = 𝑊𝑒
3𝜋
)
4
(0.25)(
𝑊 = 184.718
𝑊 = 𝑚𝑔
183.7184 = 𝑚(9.81)
𝒎 = 𝟏𝟖. 𝟖𝟐𝟗𝟔 𝒌𝒈
3𝜋
𝑟𝑎𝑑
4
MIDTERM DISCUSSIONS
Friction
MODULE V
ACTIVITY 5
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Draw the FBD
on each problem, if necessary. Box and express your final answer in four decimal places.
1.
The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required
to move the car if the back brakes are locked, and the front wheels are free to roll. Also, solve for
the other unknown values such as the normal forces and friction force acting on the automobile.
Take us = 0.3.
2.
The man having a weight of 200 lb pushes horizontally on the crate. If the coefficient of static
friction between the 450-lb crate and the floor is us = 0.3 and between his shoes and the floor is us
= 0.6, determine if he can move the crate.
MIDTERM DISCUSSIONS
Friction
3.
MODULE V
Determine the smallest force P needed to lift the 3000-lb load. The coefficient of static friction
between A and C and between B and D is us = 0.3, and between A and B is us = 0.4. Neglect the
weight of each wedge.
4. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the
pipe. Determine the largest vertical force F that can be applied to the cord without moving the
cylinder. The cord passes (a) once over the pipe, β = 180oand (b) two times over the pipe, β = 540o.
Take us = 0.2. (10 points)
FINALS DISCUSSIONS
MODULE I
Centroids
I
Centroids
I
Objectives:
•
To discuss the concept of the center of gravity, center of mass, and the centroid.
•
To show how to determine the location of the center of gravity and centroid for a
body of arbitrary shape and one composed of composite parts.
•
To use the theorems of Pappus and Guldinus for finding the surface area and
volume for a body having axial symmetry.
•
To present a method for finding the resultant of a general distributed loading and
to show how it applies to finding the resultant force of a pressure loading caused
by a fluid.
Outline:
•
Centroid of a Body
▪
Centroid of a line
▪
Centroid of an Area
▪
Centroid of a Volume
•
Centroid of Composite Bodies
•
Theorems of Pappus and Guldinus
FINALS DISCUSSIONS
Centroids
MODULE I
I
Chapter Introduction
Understanding the combined weight of an object and its specific position holds significance in assessing
the impact exerted by this force on the object. The designated position is referred to as the center of
gravity, and within this portion, we will demonstrate the procedure to determine it for a body with an
irregular shape. Furthermore, we will expand this approach to illustrate the determination of the object's
center of mass and its geometric center or centroid.
Throughout this chapter, we will employ illustrative examples, diagrams, and equations to enhance our
understanding of centroids and their applications. By the end of this journey, readers will have acquired
a comprehensive understanding of centroids, enabling them to apply this knowledge to solve real-world
problems and make informed decisions in various disciplines.
Centroids
A centroid is a point that represents the geometric center or average position of a shape or object. It is
determined by calculating the average coordinates of all the points within the object. The centroid serves
as a reference point for analyzing the properties and behavior of the object, such as its balance, stability,
and moments. It is commonly used in various fields, including mathematics, physics, engineering, and
computer graphics. Looking at the figures shown, a circle and a rectangle, their respective centroids
which are denoted by the dots at the intersection of their centroidal axes.
Centroid refers to the geometrical center of a plane figure: a curve, area or volume. It is the average
position of all the points of an object.
Centroidal Axes are the lines passing
through the centroid of the figure. The
vertical centroidal axis is represented by line
Y and the horizontal centroidal axis is
denoted as line X.
Since the centroid is the average position of the points in a figure, integration is the process used to sum
up the infinite number of points a figure has. Integration is also equivalent to summing up finite elements.
FINALS DISCUSSIONS
MODULE I
Centroids
Centroid of a line
I
If a line segment (or rod) lies within the x–y plane and it
can be described by a thin curve y = f (x), then its centroid
is determined from the formula presented.
Here, the length of the differential element is given by the
Pythagorean theorem, dL = √(dx)2 + (dy)2 , which can
also be written in the form of
2
𝑑𝑦
dL = √1 + ( ) 𝑑𝑥, when expanded in terms of dx
𝑑𝑥
or
2
𝑑𝑥
dL = √( ) + 1 𝑑𝑦, when expanded in terms of dy.
𝑑𝑦
Either one of these expressions can be used; however, for application, the one that will result in a simpler
integration should be selected.
Note: The use of which dl formula to be used will depend on the given line function.
For example, consider the rod in the figure, defined by y =
2
𝑑𝑦
2x . The length of the element is dL = √1 + ( ) 𝑑𝑥,
2
𝑑𝑥
𝑑𝑦
and since
= 4𝑥, then dL = √1 + (4𝑥)2 𝑑𝑥. The
𝑑𝑥
centroid for this element is located at x = x and y = y.
FINALS DISCUSSIONS
MODULE I
Centroids
I
EXAMPLE 1
Locate the centroid of the rod bent into the shape of a
parabolic arc as shown in the figure.
SOLUTION:
Differential Element. The differential element is shown in the figure. It is located on the curve at the
arbitrary point (x, y).
Area and Moment Arms. The differential element of length dL can be expressed in terms of the
differentials dx and dy using the Pythagorean theorem.
𝑑𝐿 =
√(𝑑𝑥)2
+
(𝑑𝑦)2
= √(
𝑑𝑥 2
) + 1𝑑𝑦
𝑑𝑦
Since x = y2, then dx/dy = 2y. Therefore, expressing dL in terms of y and dy, we have
𝑑𝐿 = √(2𝑦)2 + 1𝑑𝑦
As shown in the figure, the centroid of the element is located at 𝑥̃
𝑥̅ =
∫𝐿 𝑥̃ 𝑑𝐿
∫𝐿 𝑑𝐿
1
𝑥̅ =
∫0 𝑥√4𝑦 2 + 1 𝑑𝑦
1
∫0 √4𝑦 2 + 1𝑑𝑦
1
𝑥̅ =
∫0 𝑦 2 √4𝑦 2 + 1 𝑑𝑦
1
∫0 √4𝑦 2 + 1𝑑𝑦
= 𝑥, ỹ = 𝑦
FINALS DISCUSSIONS
MODULE I
Centroids
1
∫0 𝑦 2 √4𝑦 2 + 1 𝑑𝑦
𝑥̅ =
1
∫0 √4𝑦 2
I
+ 1𝑑𝑦
𝑥̅ = 0.4099802175 𝑚
𝑥̅ = 0.4100 𝑚
𝑦̅ =
∫𝐿 𝑦̃ 𝑑𝐿
∫𝐿 𝑑𝐿
1
𝑦̅ =
∫0 𝑦 √4𝑦 2 + 1𝑑𝑦
1
∫0 √4𝑦 2 + 1𝑑𝑦
𝑦̅ = 0.5736270695 𝑚
𝑦̅ = 0.5736 𝑚
Centroid of an Area
Centroid of an Area. If an area lies in the x–y plane and is
bounded by the curve y = f (x), then its centroid will be in this
plane and can be determined from the formula shown.
These integrals can be evaluated by
performing a single integration if we use a
rectangular strip for the differential area
element. For example, if a vertical strip is
used, the area of the element is dA =
y dx, and its centroid is located at 𝑥̃ = x
and 𝑦̃ = y2. If we consider a horizontal
strip, then dA = x dy, and its centroid is
located at 𝑥̃ = 2x and 𝑦̃ = y.
FINALS DISCUSSIONS
MODULE I
Centroids
I
EXAMPLE 2
Determine the centroid (x, y) of the shaded area.
SOLUTION:
𝑑𝑦
𝑦 = 𝑥2
𝑑𝑥
= 2𝑥
Use the following formulas.
𝑑𝐴 = 𝑦 𝑑𝑥
𝑥̃ = x
y
𝑦̃ =
𝑥̅ =
2
∫ 𝑥̃ 𝑑𝐴
∫ 𝑑𝐴
𝑦̅ =
1𝑦
1
𝑥̅ =
∫0 𝑥 (𝑦 𝑑𝑥)
1
∫0 (𝑦
𝑑𝑥)
where, 𝑦 = 𝑥
2
𝑦̅ =
𝑥̅ =
1
∫0 (𝑥 2 𝑑𝑥)
𝑦̅ =
1
𝑥̅ =
∫0 𝑥 3 𝑑𝑥
1
∫0 𝑥 2
𝟑
𝑑𝑥
̅ = = 𝟎. 𝟕𝟓 𝒎
𝒙
𝟒
∫0 2 (𝑦 𝑑𝑥)
1
∫0 (𝑦
𝑑𝑥)
1𝑥2
1
∫0 𝑥 (𝑥 2 𝑑𝑥)
∫ 𝑦̃ 𝑑𝐴
∫ 𝑑𝐴
̅=
𝒚
∫0 2 (𝑥 2 𝑑𝑥)
1
∫0 (𝑥 2 𝑑𝑥)
𝟑
𝟏𝟎
= 𝟎. 𝟑 𝒎
where, 𝑦 = 𝑥 2
FINALS DISCUSSIONS
Centroids
Centroid of a Volume
MODULE I
I
The centroid of a volume refers to the geometric center or balance point of a three-dimensional object. It
is the point where the object would perfectly balance if supported at that location.
These equations represent a balance of the
moments of the volume of the body. Therefore, if
the volume possesses two planes of symmetry,
then its centroid must lie along the line of
intersection of these two planes.
For example, the cone in the figure has a centroid that lies
on the y axis so that x = z = 0. The location y can be found
using a single integration by choosing a differential
element represented by a thin disk having a thickness dy
and radius r = z. Its volume is dV = πr 2 dy = πz 2 dy and
its centroid is at 𝑥̃ = 0, 𝑦̃ = y, 𝑧̃ = 0.
EXAMPLE 3
Locate the y centroid for the paraboloid of revolution.
FINALS DISCUSSIONS
MODULE I
Centroids
I
SOLUTION:
𝑧 = 100 𝑦
Use the following formulas.
𝑑𝑉 = πz 2 dy
𝑦̅ =
∫ 𝑦̃ 𝑑𝑉
∫ 𝑑𝑉
100
𝑦̅ =
∫0
100
𝑦̅ =
∫0
𝑦 (πz2 dy)
1
∫0 (πz2
dy)
where 𝑧 2 = 100𝑦
𝑦 (π(100 y) dy)
1
∫0 (π(100 y) dy)
̅ = 𝟔𝟔. 𝟔𝟔𝟔𝟕 𝒎𝒎
𝒚
Centroid of Composite Bodies
A composite body consists of a series of
connected “simpler” shaped bodies,
which may be rectangular, triangular,
semicircular, etc. Such a body can often be
sectioned or divided into its composite
parts and, provided the weight and location
of the center of gravity of each of these parts are known, we can then eliminate the need for integration
to determine the center of gravity for the entire body. Therefore,
FINALS DISCUSSIONS
Centroids
MODULE I
PROCEDURE FOR ANALYSIS
I
The location of the center of gravity of a body or the centroid of a composite geometrical object
represented by a line, area, or volume can be determined using the following procedure.
Composite Parts.
• Using a sketch, divide the body or object into a finite number of composite parts that have simpler
shapes.
• If a composite body has a hole, or a geometric region having no material, then consider the composite
body without the hole and consider the hole as an additional composite part having negative weight or
size.
Moment Arms.
• Establish the coordinate axes on the sketch and determine the coordinates x , y , z of the center of
gravity or centroid of each part.
Summations.
• Determine x, y, z by applying the center of gravity equations or the analogous centroid equations.
• If an object is symmetrical about an axis, the centroid of the object lies on this axis.
***If desired, the calculations can be arranged in tabular form.***
FINALS DISCUSSIONS
Centroids
MODULE I
CENTROIDS OF COMMON SHAPES OF AREAS AND LINES I
FINALS DISCUSSIONS
Centroids
MODULE I
I
EXAMPLE 1
Locate the centroid of the plate area shown:
SOLUTION:
Note that this plate is divided into three segments. Here are the segments:
Note that: The rectangle is considered negatives since it is subtracted from the larger one.
Moment Arms. The centroid of each segment is located as indicated in the figure. Note that the x
coordinates of 2 and 3 are negative since they are located to the left of y axis.
FINALS DISCUSSIONS
MODULE I
Centroids
I
Segment
1
2
3
A ( 𝒇𝒕𝟐 )
4.5
9
-2
̃
𝒙
1
-1.5
-2.5
̃
𝒚
1
1.5
2
∑ 𝐴 = 11.5
̃A
𝒙
4.5
-13.5
5
̃𝑨 = − 4
∑𝒙
̃A
𝒚
4.5
13.5
-4
̃ 𝑨 = 14
∑𝒚
Take note that for triangle, x = 1/3 b and y = 1/3h if the distance of the centroid is from the
side opposite to vertex of the triangle.
NOTE: For triangle, x = 2/3 b and y = 2/3h if the distance of the centroid is from the vertex
of a triangle.
Therefore,
𝑥̅ =
∑ 𝑥̃𝐴
−4
=
= −𝟎. 𝟑𝟒𝟖𝒇𝒕
∑𝐴
11.5
𝑦̅ =
∑ 𝑦̃𝐴
14
=
= −𝟏. 𝟐𝟐𝒇𝒕
∑𝐴
11.5
PRACTICE PROBLEM
Locate the centroid (x,y) of the cross-sectional area.
SOLUTION:
𝑥̅ =
∑ 𝑥̃𝐴 0.25[4(0.5)] + 1.75[0.5(2.5)]
=
= 𝟎. 𝟖𝟐𝟕 𝒎
∑𝐴
4(0.5) + 0.5(2.5)
𝑦̅ =
∑ 𝑦̃𝐴 2[4(0.5)] + 0.25[0.5(2.5)]
=
= 𝟏. 𝟑𝟑𝒎
∑𝐴
4(0.5) + 0.5(2.5)
FINALS DISCUSSIONS
MODULE I
Centroids
Theorems of Pappus and Guldinus
I
The two theorems of Pappus and Guldinus are used to find the surface area and volume of any body of
revolution. They were first developed by Pappus of Alexandria during the fourth century a.d. and then
restated later by the Swiss mathematician Paul Guldin or Guldinus.
The first theorem of Pappus and Guldinus provides a way for computing the surface area of a
surface of revolution. It states that the .area of a surface of revolution equals the product of the
length of the generating curve and the distance traveled by the centroid of the curve in generating
the surface area.
Surface Area. If we revolve a plane curve about an
axis that does not intersect the curve, we will
generate a surface area of revolution.
The formula shown below is the surface area of a
curvethat is revolved only through an angle 𝜃
Where
A – surface area of revolution
𝜃 -angle of revolution measured in radians, 𝜃 ≤ 2𝜋
𝑟̅ - perpendicular distance from the axis of revolution
to the centroid of the generating curve
𝑙 – length of the generating curve
FINALS DISCUSSIONS
MODULE I
Centroids
Likewise, the second theorem of Pappus and Guldinus is for computing the volumeI of a body of
revolution: it states that the volume of a body of revolution equals the product of the generating area
and the distance traveled by the centroid of the area in generating the volume.
Volume. A volume can be generated by revolving a
plane area about an axis that does not intersect an area.
If the area is only revolved through an angle 𝜃
(radians), then,
Where
V – volume of revolution
𝜃 -angle of revolution measured in radians, 𝜃 ≤ 2𝜋
𝑟̅ - perpendicular distance from the axis of revolution
to the centroid of the generating curve
𝐴 – generating area
Composite Shapes. We may also apply the above two
theorems to lines or areas that are composed of a series of
composite parts. In this case the total surface area or
volume generated is the addition of the surface areas or
volumes generated by each of the composite parts. If the
perpendicular distance from the axis of revolution to the
centroid of each composite part is 𝑟̃ , then,
FINALS DISCUSSIONS
MODULE I
Centroids
I
EXAMPLE 1
Determine the surface area and volume of the full solid below.
SOLUTION:
AREA:
A = 2π∑𝑟̅ 𝐿
A = 2π[(25mm)(20mm) + (30mm)(√(10𝑚𝑚)2 + (10𝑚𝑚)2
+ (35𝑚𝑚)(30𝑚𝑚) + (30𝑚𝑚)(20𝑚𝑚)]
𝑨 = 𝟏𝟒𝟐𝟗𝟎 𝒎𝒎𝟐
VOLUME:
V = 2π∑𝑟̅ 𝐴
1
V = 2π[(31.667 𝑚𝑚) [ (10𝑚𝑚)(10𝑚𝑚)]
2
+ (30 mm)(20mm)(10mm)]
𝑽 = 𝟒𝟕𝟔𝟒𝟖 𝒎𝒎𝟑
FINALS DISCUSSIONS
MODULE I
Centroids
I
PRACTICE PROBLEM
Determine the surface area and volume of the solid formed by revolving the shaded area 360° about the z-axis.
SOLUTION:
AREA
A = 2π∑𝑟̅ 𝐿
A = 2π[1.95√(0.9)2 + (1.2)2 + 2.4(1.5) + 1.95(0.9) + 1.5(2.7)]
𝑨 = 𝟕𝟕. 𝟓 𝒎𝟐
VOLUME:
V = 2π∑𝑟̅ 𝐴
1
V = 2π[1.8 ( ) (0.9)(1.2) + 1.95(0.9)(1.5)]
2
𝑽 = 𝟐𝟐. 𝟔 𝒎𝟑
FINALS DISCUSSIONS
Centroids
ACTIVITY 6
MODULE I
I
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places. For problems nos. 3 and 4, kindly refer to the table
of centroids of common shapes.
1. Determine the distance ȳ to the center of gravity of the homogeneous rod.
2. Locate the centroid x̄ and ȳ of the shaded area.
FINALS DISCUSSIONS
Centroids
3. Determine the location (x̄, ȳ) of the centroid C of the area.
MODULE I
I
4. Determine the volume of the silo which consists of a cylinder and hemispherical cap. Neglect
the thickness of the plates.
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Moment of Inertia
II
Objectives:
• To establish a method for calculating the moment of inertia for an area.
• To introduce the product of inertia and demonstrate how to calculate
the maximum and minimum moments of inertia for a given area.
• To discuss the mass moment of inertia.
Outline:
•
•
•
•
•
Chapter Introduction
Definition of Moments of Inertia for Areas
Parallel-Axis Theorem for an Area
Radius of Gyration of an Area
Moments of Inertia for Composite Areas
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Chapter Introduction
When forces are continuously distributed over a surface on which they act, it is frequently necessary to
compute the moment of these forces about some axis that is either in or perpendicular to the area's plane.
The intensity of the force (pressure or stress) is frequently proportional to the distance of the force's line
of action from the moment axis. The elemental force acting on an area element is therefore proportional
to distance times differential area, and the elemental moment is proportional to distance squared times
differential area. Therefore, the total moment involves an integral of form (distance)² d (area). This
integral is referred to as the moment of inertia or the second moment of area. It is defined as a
geometric property of an area that is used to determine the strength of structural members and is also
defined as the capacity of a cross-section to resist bending.
This integral's numerical value is used to show how the area is distributed around a given axis. If the
axis is within the plane of a given area, that area’s second moment of the axis is called the rectangular
moment of inertia and is represented by the letter I. In any case an axis perpendicular to an area's plane,
the area’s second moment is known as the polar moment of inertia and is indicated as J.
Moments of Inertia for Areas
Fig. 7-1
Consider the area A in the x-y plane, Fig. 7-1. The moments of inertia of the element dA about
the x- and y-axes are, by definition, 𝑑𝑙𝑦 = 𝑥 2 𝑑𝐴 and 𝑑𝑙𝑥 = 𝑦 2 𝑑𝐴 , respectively. The moments of
inertia of A about the same axes are therefore
𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴
𝐼𝑦 = ∫ 𝑥 2 𝑑𝐴
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
where we carry out the integration over the entire area. We can also formulate this quantity for dA about
the “pole” O or z axis. The polar moment of inertia is defined as 𝑑𝐽𝑜 = 𝑟 2 𝑑𝐴, where r is the
perpendicular distance from the pole (z axis) to the element dA. For the entire area the polar moment of
inertia is
𝐽𝑜 = ∫ 𝑟 2 𝑑𝐴 = 𝐼𝑥 + 𝐼𝑦
This relation between Jo and Ix, Iy is possible since r2 = x2 + y2.
The moment of inertia of an element involves the square of the distance from the inertia axis to
the element. As a result, an element whose coordinate is negative contributes as much to the moment of
inertia as does an equal element with a positive coordinate of the same magnitude. Consequently, the
area moment of inertia about any axis is always a positive quantity. In contrast, the first moment of the
area, which was involved in the computations of centroids, could be either positive, negative, or zero.
The dimensions of moments of inertia of areas are clearly 𝐿4 , where L stands for the dimension
of length. Thus, the SI units for area moments of inertia are expressed as quartic meters (𝑚4 ) or quartic
rnillimeters (𝑚𝑚4 ). The U.S. customary units for area moments of inertia are quartic feet (𝑓𝑡 4 ) or
quartic inches (𝑖𝑛4 ).
The choice of the coordinates to use for the calculation of moments of inertia is important. Rectangular
coordinates should be used for shapes whose boundaries are most easily expressed in these coordinates.
Polar coordinates will usually simplify problems involving boundaries which are easily described in r
and 𝜃 . The choice of an element of area which simplifies the integration as much as possible is also
important.
Parallel-Axis Theorem
The parallel-axis theorem, also known as the transfer formula, establishes a relationship between the
moment of inertia with respect to any axis, as well as the moment of inertia in relation to a parallel axis
through the centroid. To develop this theorem, we will consider finding the moment of inertia of the
shaded area shown in Fig. 7-2 about the x axis.
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Fig. 7-2
To start, we choose a differential element dA located at an arbitrary distance y' from the centroidal
x' axis. If the distance between the parallel x and x' axis is dy, then the moment of inertia of dA about the
2
x axis is 𝑑𝑙𝑥 = (𝑦 ′ + 𝑑𝑦 ) . For the entire area,
𝐼𝑥 = ∫ (𝑦 ′ + 𝑑𝑦)2 𝑑𝐴
= ∫ 𝑦 ′ 𝑑𝐴 + 2𝑑𝑦 ∫ 𝑦 ′ 𝑑𝐴 + 𝑑𝑦2 ∫ 𝑑𝐴
The first integral represents the area's moment of inertia about the centroidal axis, Ix. Because the x axis
passes through the area’s centroid C, the second integral is zero. Since the third integral represents the
total area A, the final result is therefore,
𝐼𝑥 = 𝐼𝑥 ′ + 𝐴𝑑𝑦2
For Iy, a similar expression can be written.
𝐼𝑦 = 𝐼𝑦 ′ + 𝐴𝑑𝑥2
And finally, for the polar moment of inertia, since 𝐽𝑐 = 𝐼𝑥 ′ + 𝐼𝑦 ′ and 𝑑2 = 𝑑𝑥2 + 𝑑𝑦2 ,, we have
𝐽𝑜 = 𝐽𝑐 + 𝐴𝑑2
Based on the form of each of these three equations, the moment of inertia for an area about an axis is
equal to its moment of inertia about a parallel axis passing through the area’s centroid plus the product
of the area and the square of the perpendicular distance between the axes.
Radius of Gyration
The radius of gyration of an area about an axis is measured in length units and is a quantity that
is frequently used in the design of structural columns mechanics. If the areas and moments of inertia are
known, the radii of gyration can be calculated using the formulas.
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
𝐼𝑥
𝑘𝑥 = √
𝐴
𝐼𝑦
𝑘𝑦 = √
𝐴
𝐽𝑜
𝑘𝑜 = √
𝐴
Because the form of these equations is similar to that of determining the moment of inertia for a
differential area about an axis, they are easily remembered. For example,𝐼𝑥 = 𝑘𝑥2 𝐴; whereas for a
differential area, 𝑑𝑙𝑥 = 𝑦 2 𝑑𝐴.
Procedure for Analysis
In most cases the moment of inertia can be determined using a single integration. The following
procedure shows two ways in which this can be done.
• If the curve defining the boundary of the area is expressed as y = f(x), then select a rectangular
differential element such that it has a finite length and differential width.
• The element should be located so that it intersects the curve at the arbitrary point (x, y).
Case 1.
Orient the element so that its length is parallel to the axis about which the moment of inertia is computed.
This situation occurs when the rectangular element shown in Fig. 7-3 is used to determine Ix for the area.
Here the entire element is at a distance y from the x axis since it has a thickness dy. Thus 𝐼𝑥 = 𝑦 2 𝑑𝐴.
Fig. 7-3
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Case 2.
The length of the element can be oriented perpendicular to the axis about which the moment of inertia
is computed; however, 𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴 𝑎𝑛𝑑 𝐼𝑦 = ∫ 𝑥 2 𝑑𝐴 does not apply since all points on the
element will not lie at the same moment-arm distance from the axis. For example, if the rectangular
element in Fig. 7-3 is used to determine Iy, it will first be necessary to calculate the moment of inertia of
the element about an axis parallel to the y axis that passes through the element’s centroid, and then
determine the moment of inertia of the element about the y axis using the parallel-axis theorem.
Fig. 7-4
EXAMPLE 1
Determine the moment of inertia for the rectangular area with
respect to (a) the centroidal x' axis, (b) the axis xb passing through the
base of the rectangle, and (c) the pole or z' axis perpendicular to the x’
- y’ plane and passing through the centroid C.
SOLUTION:
(CASE 1)
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Part (a). The differential element shown is chosen for integration. Because of its location and orientation,
the entire element is at a distance y’ from the x’ axis. Here it is necessary to integrate from 𝑦 ′ = −
′
to 𝑦 =
ℎ
2
ℎ
2
.. Since dA = b dy’ , then
ℎ ⁄2
′2
𝐼𝑥′ = ∫ 𝑦 𝑑𝐴 = ∫
ℎ ⁄2
𝑦
′2 (𝑏𝑑𝑦 ′ )
−ℎ ⁄2
𝟏
𝑰𝒙′ =
𝒃𝒉𝟑
𝟏𝟐
=𝑏∫
𝑦 ′2 (𝑑𝑦 ′ )
−ℎ⁄2
Part (b). The moment of inertia about an axis passing through the base of the rectangle can be obtained
by using the above result of part (a) and applying the parallel-axis theorem
𝐼𝑥𝑏 = 𝐼𝑥′ + 𝐴𝑑𝑦2
1
ℎ
𝟏
𝑏ℎ3 + 𝑏ℎ( )2 = 𝒃𝒉𝟑
12
2
𝟑
Part (c). To obtain the polar moment of inertia about point C, we must first obtain Iy, which may be
found by interchanging the dimensions b and h in the result of part (a), i.e.,
=
Using 𝐽𝑂 = ∫ 𝑟 2 𝑑𝐴 = 𝐼𝑥 + 𝐼𝑦 , 𝐼𝑦′ =
1
12
𝑏ℎ3 , the polar moment of inertia about C is therefore,
𝑱𝒄 = 𝑰𝒙′ + 𝑰𝒚′ =
𝟏
𝒃𝒉(𝒉𝟐 + 𝒃𝟐 )
𝟏𝟐
EXAMPLE 2
Determine the moment of inertia for the shaded
area about the x-axis.
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
SOLUTION:
Differential Element. Here 𝑥 = 2(1 − 𝑦 2 ) The area of the differential element parallel to
the x axis shown shaded in the figure is 𝑑𝐴 = 𝑥𝑑𝑦 = 2(1 − 𝑦 2 )𝑑𝑦
Moment of Inertia. Perform the integration,
1𝑚
𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴 = ∫0
𝑦 2 (2(1 − 𝑦 ′ ))
1𝑚
= 2∫
(𝑦 2 − 𝑦 4 )𝑑𝑦
0
𝑦 3 𝑦 5 1𝑚
= 2( − )|
3
5 0
4 4
=
𝑚 = 𝟎. 𝟐𝟔𝟕𝒎𝟒
15
EXAMPLE 3
Determine the moment of inertia for the shaded are shown about
the x axis.
SOLUTION (CASE 1).
A differential element of area that is parallel to the x axis, as shown in the fig a., is chosen for
integration. Since this element has a thickness dy and intersects the curve at the arbitrary point (x, y), its
area is dA = (100 - x) dy.
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
200𝑚𝑚
𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴 = ∫0
200𝑚𝑚
=∫
𝑦 2 (100 −
0
𝑦 2 (100 − 𝑥)𝑑𝑦
200𝑚𝑚
𝑦2
𝑦4
) 𝑑𝑦 = ∫
(100𝑦 2 −
) 𝑑𝑦
400
400
0
= 𝟏𝟎𝟕(𝟏𝟎)𝟔 𝒎𝒎𝟒
EXAMPLE 4
Determine the moment of inertia for the shaded area shown about the x axis.
SOLUTION II (CASE 2):
For the differential element shown in Fig. b, b = dx and h = y, and thus 𝑑𝐼𝑥 =
𝑦
1
12
𝑑𝑥 𝑦 3 ..
Since the centroid of the element is 𝑦 = from the x axis, the moment of inertia of the element about
2
this axis is
𝑑𝐼𝑥 = 𝑑𝐼𝑥 + 𝑑𝐴 𝑦 2 =
1
𝑑𝑥
12
𝑦
2
1
3
𝑦 3 + 𝑦 𝑑𝑥( )2 = 𝑦 3 𝑑𝑥
Integrating with respect to x, from x = 0 to x = 100mm, yields
100𝑚𝑚 1
𝐼𝑥 = ∫ 𝑑𝐼𝑥 = ∫0
3
100𝑚𝑚 1
𝑦 3 𝑑𝑥 = ∫0
= 𝟏𝟎𝟕(𝟏𝟎)𝟔 𝒎𝒎𝟒
3
(400𝑥)3⁄2 𝑑𝑥
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Moments of Inertia for Composite Areas
A composite area is made up of a series of "simpler" parts or shapes connected together, such as
rectangles, triangles, and circles. The moment of inertia of a composite area about a particular axis is
simply the sum of the moments of inertia of its component parts about the same axis. It is often convenient
to regard a composite
area as being composed of positive and negative parts. We may then treat the moment of inertia of a
negative area as a negative quantity.
Procedure for Analysis
The moment of inertia for a composite area about a reference axis can be determined using the
following procedure.
Composite Parts.
⚫ Using a sketch, divide the area into its composite parts and indicate the perpendicular
distance from the centroid of each part to the reference axis.
Parallel-Axis Theorem.
⚫ If the centroidal axis for each part does not coincide with the reference axis, the parallelaxis theorem, 𝐼 = 𝐼 ′ + 𝐴𝑑 2 , should be used to determine the moment of inertia of the
part about the reference axis.
Summation.
⚫ The moment of inertia of the entire area about the reference axis is determined by summing
the results of its composite parts about this axis.
If a composite part has an empty region (hole), its moment of inertia is found by subtracting the moment
of inertia of this region from the moment of inertia of the entire part including the region.
Moment of Inertia of Common Geometric Shapes
Rectangle
1
𝑏ℎ3
12
1
𝐼𝑦′ =
ℎ𝑏 3
12
1
𝐼𝑥 = 𝑏ℎ3
3
1
𝐼𝑦 = ℎ𝑏 3
3
𝐼𝑥′ =
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
𝐽𝐶 =
Triangle
1
𝑏ℎ(𝑏 2 + ℎ2 )
12
𝐼𝑥′ =
𝐼𝑥 =
1
36
1
12
𝑏ℎ3
𝑏ℎ3
Circle
1
𝐼𝑥′ = 𝐼𝑦 = 𝜋𝑟4
1
4
𝐽𝑂 = 𝜋𝑟4
2
Semicircle
1
𝐼𝑥 = 𝐼𝑦 = 𝜋𝑟4
1
8
𝐽𝑂 = 𝜋𝑟4
4
Quarter circle
𝐼𝑥 = 𝐼𝑦 =
1
1
16
𝜋𝑟4
𝐽𝑂 = 𝜋𝑟4
8
Ellipse
1
𝐼𝑥 = 𝜋𝑎𝑏4
4
1
𝐼𝑦 = 𝜋𝑏𝑎4
1
4
𝐽𝐶 = 𝜋𝑎𝑏(𝑎2 + 𝑏2 )
4
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
EXAMPLE 1
Determine the moments of inertia for the cross-sectional area of the
member shown in the figure about the x and y centroidal axes.
SOLUTION:
Composite Parts. The cross-section ca be subdivided into the three rectangular areas A, B and D shown in Fig. The crosssection ca be subdivided into the three rectangular areas A, B and D shown in Fig. B. For the calculation, the centroid of each of these
rectangles is in the figure.
1
Parallel-Axis Theorem. The moment of inertia of a rectangle about its centroidal axis is 𝐼 = 12 𝑏ℎ3 ..
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Rectangles A and D
1
𝐼𝑥 = 𝐼𝑥′ + 𝐴𝑑𝑦2 = 12 (100)(300)3 + (100)(300)(200)3
= 𝟏. 𝟒𝟐𝟓(𝟏𝟎)𝟗 𝒎𝒎𝟒
𝐼𝑦 = 𝐼𝑦′ + 𝐴𝑑𝑥2 =
1
(300)(100)3 + (100)(300)(250)3
12
= 𝟏. 𝟗𝟎(𝟏𝟎)𝟗 𝒎𝒎𝟒
Rectangle B
𝐼𝑥 =
1
(600)(100)3 = 0.05(10)9 𝑚𝑚4
12
𝐼𝑦 =
1
(100)(600)3 = 1.80(10)9 𝑚𝑚4
12
𝐼𝑥 = 2(1.425(10)9 + 0.005(10)9 )
= 𝟐. 𝟗𝟎(𝟏𝟎)𝟗 𝒎𝒎𝟒
𝐼𝑦 = 2(1.90(10)9 + 1.80(10)9 )
= 𝟓. 𝟔𝟎(𝟏𝟎)𝟗 𝒎𝒎𝟒
EXAMPLE 2
Determine the moment of inertia of the shaded are in the figure about
the x axis.
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
SOLUTION:
Segment
𝑨𝒊 (𝒎𝒎)𝟐
(𝒅𝒚)𝒊 (𝒎𝒎)
(𝑰𝒙′)𝒊 (𝒎𝒎)𝟒
(𝑨𝒅𝟐𝒚 )𝒊 (𝒎𝒎)𝟒
(𝑰𝒙 )𝒊 (𝒎𝒎)𝟒
1
200(300)
150
1
(200)(300)3
12
1.35(10)9
1.80(10)9
2
1
(500)(300)
2
100
1
(150)(300)3
36
0.225(10)9
0.3375(10)9
3
−𝜋(75)2
150
−0.3976(10)9
−0.4225(10)9
−
𝜋(75)4
4
Thus,
𝑰𝒙 = ∑(𝑰𝒙 )𝒊 = 𝟏. 𝟕𝟏𝟓(𝟏𝟎)𝟗 𝒎𝒎𝟒 = 𝟏. 𝟕𝟐(𝟏𝟎)𝟗 𝒎𝒎𝟒
PRACTICE PROBLEM
Determine the moment of inertia of the shaded area in the figure about
the y axis.
SOLUTION:
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
Segment
𝑨𝒊 (𝒎𝒎)𝟐
(𝒅𝒙)𝒊 (𝒎𝒎)
𝑰𝒚′ (𝒎𝒎)𝟒
(𝑨𝒅𝟐𝒙 )𝒊 (𝒎𝒎)𝟒
(𝑰𝒚)𝒊 (𝒎𝒎)𝟒
1
200(300)
100
1
(300)(200)3
12
0.6(10)9
0.800(10)9
2
1
(150)(300)
2
250
1
(300)(150)3
36
1.40625(10)9
1.434375(10)9
3
−𝜋(75)2
100
−0.1767(10)9
−0.20157(10)9
−
𝜋(75)4
4
Thus,
𝑰𝒚 = ∑(𝑰𝒚 )𝒊 = 𝟐. 𝟎𝟑𝟑(𝟏𝟎)𝟗 𝒎𝒎𝟒 = 𝟐. 𝟎𝟑(𝟏𝟎)𝟗 𝒎𝒎𝟒
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
ACTIVITY 7
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places.
1. Determine the moment of inertia for the shaded area about the x axis.
2. Determine the moment of inertia for the shaded area about the y axis.
FINALS DISCUSSIONS
MODULE II
Moment of Inertia
3. Determine the moment of inertia for the beam’s cross-sectional area about the x axis.
4. Determine the moment of inertia for the shaded area about the y axis.
FINALS DISCUSSIONS
MODULE IV
Kinematics of a Particle
Kinematics of a
III
Particle
Objectives:
• Discuss the fundamental kinematic relationships between position,
velocity, acceleration, and time.
• Solve problems using these fundamental kinematic relationships and
calculus or graphical methods
• Utilize a translating coordinate system to examine the relative motion of
multiple particles.
• Determine which coordinate system is best for solving a curvilinear
kinematics problem.
• Determine the position, velocity, and acceleration of a particle in
curvilinear motion.
• Analyze and solve problems of a projectile motion utilizing curvilinear
and rectilinear motion.
Outline:
• Chapter Introduction
• Kinematics of Particles
o Displacement
o Velocity in line of Particles
o Average Acceleration
• Rectilinear Motion of Particles
o Uniformly Accelerated Rectilinear Motion
• Curvilinear Motion of Particles
• Motion of Projectile
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Chapter Introduction
Dynamics, a fascinating field within mechanics, encompasses the study of objects in motion. In order to conduct
experiments that lay the foundation of dynamics, three essential units come into play: force, length, and time. This branch
of mechanics is further divided into two branches known as kinetics and kinematics, each serving a unique purpose.
1. Kinematics, often referred to as the geometry of motion, focuses on defining the movement of a particle or a
body. It delves into the intricacies of motion without considering the specific forces that drive it. Through
kinematics, we can comprehend fundamental concepts such as displacement, velocity, and acceleration, enabling
us to grasp the essence of an object's journey.
2. On the other hand, kinetics delves into the analysis of the forces responsible for initiating and sustaining motion.
It establishes a relationship between the force acting upon a body, its mass, and the resulting acceleration. By
studying kinetics, we gain insights into the underlying mechanisms that propel objects, allowing us to discern
the intricate interplay between forces and motion.
Dynamics serves as an extensive domain that investigates the movement of bodies. By employing three essential
units, namely force, length, and time, dynamics unfolds its two distinctive branches: kinematics, which focuses on the
geometry of motion without considering forces, and kinetics, which scrutinizes the forces that bring about motion,
intertwining them with mass and acceleration. This multifaceted framework enhances our understanding of the
mesmerizing world of motion and the fundamental principles governing it.
Kinematics of Particles
Kinematics of Particles
Position: In the realm of particle motion, the trajectory of an object
can be described by a single coordinate axis denoted as "s." This
axis represents a straight-line path along which the particle moves.
At the starting point, known as the origin "O," which remains
fixed, the position coordinate "s" is employed to determine the
precise location of the particle at any given moment.
Displacement: Displacement, on the other hand, pertains to the
vector quantity that signifies the distance and direction from the
origin to the particle's current position along its path. It
encapsulates the change in position experienced by the object as it undergoes motion.
Distance Traveled: The concept of distance traveled pertains to a scalar quantity that exclusively reflects
the total length of the path covered by the particle. It disregards the specific direction and only takes into
account the magnitude of the path traversed.
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Mathematically, displacement can be defined as:
Δs= sf−so
where:
Δs – Displacement
sf – final position
so – initial position
Displacement, a fundamental concept in physics, is characterized as a vector quantity. It signifies that
displacement possesses not only a magnitude but also a direction, thereby encompassing both essential
components of its nature. To visually represent this intriguing property, we envision displacement as an
arrow, extending from the initial position of an object to its final position.
To illustrate this visually, picture an arrow originating from the initial position of the object, reaching
out towards its final position. The length of the arrow denotes the magnitude of displacement, signifying
the extent of the object's travel. Meanwhile, the arrowhead, delicately pointing in a particular direction,
unravels the path the object follows throughout its motion. By comprehending displacement as a vector
quantity, we not only grasp the magnitude of an object's movement but also unlock the hidden aspect of
direction, providing a holistic understanding of its spatial transformation.
Velocity of a particle along a line
v =ds/dt
Velocity is a vector measurement of the rate and direction of motion. The rate at which something moves
in a single direction. In mathematics, The first derivative of position with respect to time is velocity.
Velocity can be calculated using a simple formula that includes rate, time and distance.
We represent the velocity v by an algebraic number that can be positive or negative. A positive value of
v indicates that x increases, i.e., that the particle moves in the positive direction. A negative value of v
indicates that x decreases, i.e., that the particle moves in the negative direction. The magnitude of v is
known as the speed of the particle. We define the average acceleration of the particle over the time
interval Dt as the quotient of Dv and Dt.
Average acceleration
a =dv/dt
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Acceleration is a vector quantity that indicates how quickly velocity changes over time. It involves the
dimensions of length and time. It is often called "speeding up," but it measures changes in velocity. If
an object alters its velocity, it is accelerating.
We represent the acceleration a by an algebraic number that can be positive or negative. A positive
value of a indicates that the velocity increases. This may mean that the particle is moving faster in the
positive direction or that it is moving more slowly in the negative direction. In both cases, Dv is
positive. A negative value of a indicates that the velocity decreases; either the particle is moving more
slowly in the positive direction, or it is moving faster in the negative direction. Sometimes we use the
term deceleration to refer to when the speed of the particle decreases; the particle is then moving more
slowly. For example, the particle of is decelerating in parts b and c; it is truly accelerating (i.e., moving
faster) in parts a and d.
Finally, an important differential relationship involving displacement, velocity, and acceleration along
the path can be obtained.
ads = vdv
If a relationship exists between any two of the four variables a, s, and t, then a one of the kinematic
equations can be used to calculate the third variable, v =ds/dt, a =dv/dt or ads = vdv, because each
equation relates all three variables.
Example:
The car in the figure moves in a straight line such that for a short time its velocity is defined by
v=(3t2+2t) ft/s , where t is in seconds. Also, s = 0 when t = 0. Determine its position and acceleration
when t = 3s.
𝐏𝐨𝐬𝐢𝐭𝐢𝐨𝐧 →+ 𝒗 =
𝒔
ds
dt
= (3𝑡 2 + 2𝑡)
𝒕
∫𝟎 𝒅𝑺 = ∫𝟎 (3𝑡 2 + 2𝑡)𝑑𝑡
⌈𝑆⌉0𝑆 = ⌈𝑡 3 + 𝑡 2 ⌉𝑡0
When t=3 s,
𝑺 = 𝟑𝟑 + 𝟑𝟐 = 𝟑𝟔 𝒎
Acceleration →+ 𝑎 =
dv
dt
d
= dt (3𝑡 2 + 2𝑡)
= 6𝑡 + 2
When t=3 s,
𝒂 = 𝟔(𝟑) + 𝟐 = 𝟐𝟎 𝒇𝒕/𝒔𝟐
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Rectilinear Motion
Rectilinear Motion
Motion is a fundamental concept in physics that encompasses the study of objects and how their
positions change over time. It is an important aspect of our daily lives because everything around us is
constantly in motion. Understanding motion allows us to describe, analyze, and predict the behavior of
objects ranging from tiny particles to celestial bodies. Physics provides us with a framework for
systematically studying and comprehending motion. Physics is divided into two major branches that
deal with different types of motion: rectilinear motion and curvilinear motion.
The motion of an object along a straight line is referred to as rectilinear motion, also known as linear
motion. The path of the object in this type of motion is a one-dimensional line, and its position changes
only along this line. Rectilinear motion can occur in a variety of situations, such as a car traveling
along a straight road or an object falling vertically due to gravity. Physicists can accurately describe
and predict the behavior of objects in rectilinear motion by analyzing the forces acting on the object
and understanding the laws of motion.
Curvilinear motion, on the other hand, is the movement of an object along a curved path. Curvilinear
motion occurs when an object's path deviates from a straight line, as opposed to rectilinear motion.
Curvilinear motion can be seen in a satellite orbiting the Earth or a baseball thrown in a curved
trajectory. Curvilinear motion analysis necessitates a deeper understanding of concepts such as
acceleration, centripetal force, and vectors.
Following are the rectilinear motion examples:
• Use of elevators in public places
• Gravitational forces acting on objects resulting in free fall
• Kids sliding down from a slide
• Motion of planes in the sky
Uniformly Accelerated Rectilinear Motion
Constant acceleration is one of the most common types of straight-line motion. When a body
moves with constant acceleration motion, also known as uniformly accelerated rectilinear motion, its
trajectory is straight, and its acceleration is constant. This implies that the velocity's magnitude
increases or decreases uniformly.
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
A common example of constant accelerated motion is when a body falls freely toward the
earth. If air resistance is ignored and the distance of fall is short, the downward acceleration of the body
when it is close to the earth is constant.
Velocity as a function of time
Integrate =
𝑑𝑣
𝑑𝑡
, assuming that initially 𝑣 = 𝑣𝑜 when 𝑡 = 0
𝑣
𝑡
∫ 𝑑𝑣 = ∫ 𝑎𝑐 𝑑𝑡
𝑣𝑜
0
𝑣 = 𝑣𝑜 + 𝑎𝑐 𝑡
Constant Acceleration
Position as a function of time
Integrate =
ds
dt
= 𝑣𝑜 + 𝑎𝑐 𝑡 , assuming that initially 𝑠 = 𝑠𝑜 when 𝑡 = 0
𝑠
𝑡
∫ 𝑑𝑠 = ∫ (𝑣𝑜 + 𝑎𝑐 𝑡)𝑑𝑡
𝑠𝑜
0
𝑠 = 𝑠𝑜 + 𝑣𝑜𝑡 +
1
𝑎 𝑡2
2 𝑐
Constant Acceleration
Velocity as a function of time
Integrate 𝑣 = 𝑎𝑐 𝑑𝑠 , assuming that initially 𝑣 = 𝑣𝑜 when 𝑠 = 𝑠𝑜
𝑣
𝑡
∫ 𝑑𝑣 = ∫ 𝑎𝑐 𝑑𝑡
𝑣𝑜
0
𝑣 2 = 𝑣𝑜2 + 2𝑎𝑐 (𝑠 − 𝑠𝑜)
Constant Acceleration
When the acceleration is constant and the initial conditions are s = s0 and v = v0 when t = 0, these
equations apply.
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Uniformly Accelerated Rectilinear Motion Equations:
Example Problem:
During a test a rocket travels upward at 75 m/s, and when it is 40 m from the ground its engine fails.
Determine the maximum height SB reached by the rocket and its speed just before it hits the ground.
While in motion the rocket is subjected to a constant downward acceleration of 9.81 m/s^2 due to gravity.
Maximum Height. Since the rocket is traveling upward.
𝑉𝑎 = +75 𝑚/𝑠 when 𝑡 = 0. At the maximum height 𝑠 = 𝑠𝑏 the
velocity 𝑉𝑏 = 0. For the entire motion, the acceleration is 𝑎𝑐 =
−9.81 𝑚/𝑠 2 . Since 𝑎𝑐 is constant the rocket’s position may be
related to its velocity at the two point A and B on the path by using
the equation 𝑣 2 = 𝑣𝑜2 + 2𝑎𝑐 (𝑠 − 𝑠𝑜).
𝑣𝑏2 = 𝑣𝑎2 + 2𝑎𝑐 (𝑠𝑏 − 𝑠𝑎 )
0 = (75 𝑚 /𝑠)2 + 2(−9.81 𝑚/𝑠 2 )(𝑠𝑏 − 40 𝑚)
𝒔𝒃 = 𝟑𝟐𝟕 𝒎
Velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply 𝑣 2 =
𝑣𝑜 2 + 2𝑎𝑐 (𝑠 − 𝑠𝑜) point between B and C.
𝑣𝑐2 = 𝑣𝑏2 + 2𝑎𝑐 (𝑠𝑐 − 𝑠𝑏 )
𝑣𝑐2 = 02 + 2(−9.81 𝑚/𝑠 2 )(0 − 327 𝑚)
𝒗𝟐𝒄 = −𝟖𝟎. 𝟏
𝒎
𝒎
𝒐𝒓 𝟖𝟎. 𝟏 𝒅𝒐𝒘𝒏𝒘𝒂𝒓𝒅𝒔
𝒔
𝒔
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Curvilinear Motion
Curvilinear motion refers to the motion of a particle along a curved path. Unlike rectilinear
motion, where the path is a straight line, curvilinear motion involves changes in direction as the particle
moves. This type of motion can be observed in various natural phenomena and everyday scenarios, such
as the orbit of planets around the sun, the flight path of a bird, or the trajectory of a baseball being hit.
When analyzing curvilinear motion, several key concepts come into play. One of the fundamental
concepts is velocity, which describes the rate at which the particle's position changes with respect to
time. In curvilinear motion, velocity has two components: tangential velocity and radial velocity.
Tangential velocity represents the speed of the particle along the curve, while radial velocity accounts
for the component of velocity perpendicular to the curve.
Another important concept is acceleration. Acceleration in curvilinear motion consists of two
components: tangential acceleration and radial acceleration. Tangential acceleration measures how the
particle's speed changes along the curve, while radial acceleration quantifies the change in direction or
curvature of the path.
To understand curvilinear motion mathematically, vector calculus is often employed. Vectors are
used to represent quantities such as position, velocity, and acceleration. By decomposing these vectors
into their tangential and radial components, it becomes possible to analyze the motion more effectively.
In curvilinear motion, the particle's path can be described using parametric equations or in terms of a
curvilinear coordinate system. Parametric equations express the position of the particle as a function of
time, with separate equations for each coordinate. This allows for a more detailed representation of the
particle's movement along the curve. Curvilinear coordinate systems, such as polar coordinates or
spherical coordinates, provide an alternative approach to describing the particle's position and motion.
Analyzing curvilinear motion also requires understanding centripetal force. Centripetal force is
the force that acts towards the center of the curved path, allowing the particle to stay on its trajectory. It
is essential for maintaining circular or curved motion and can be calculated using the mass of the particle,
its velocity, and the radius of the curve.
Engineers use the principles of curvilinear motion to design efficient and safe transportation
systems, such as roads, railways, and aircraft flight paths. Astronomers rely on the understanding of
curvilinear motion to describe the complex orbits of celestial bodies. Additionally, sports enthusiasts and
game developers utilize the principles of curvilinear motion to accurately simulate the trajectories of
projectiles in games like golf, basketball, and racing simulations. In this section we will integrate the
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
equation of motion with respect to time and thereby obtain the principle of impulse and momentum. The
resulting equation will be useful for solving problems involving force, velocity, and time.
Following are the curvilinear motion examples:
• Cyclist racing on curved tracks of velodrome
• Earth moving around the sun
• A car taking a turn on a road
• A ball thrown upwards at an angle
• Throwing of a javelin
• Motion of a snake
Procedure for analysis
• Since rectilinear motion occurs along each coordinate axis, the motion occurs along each
coordinate axis is found using 𝑎 =
𝑑𝑣
𝑑𝑡
and 𝑣 =
ds
dt
or in cases where the motion is not expressed
as a function of time, the equation 𝑣𝑑𝑣 = 𝑎𝑑𝑠 can be used.
𝑣𝑥 = 𝑥̇
𝑣𝑦 = 𝑦̇
𝑣𝑧 = 𝑧̇
𝑎𝑥 = 𝑣𝑥̇ = 𝑥̈
𝑎𝑦 = 𝑣𝑦̇ = 𝑦̈
𝑎𝑧 = 𝑣𝑧̇ = 𝑧̈
• In two dimensions, the equation of the path 𝑦 = 𝑓(𝑥) can be used to relate the x and y
components of velocity and acceleration by applying the chain rule of calculus.
𝑦̇ =
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑥
= 𝑑𝑥 × 𝑑𝑡
𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒓𝒖𝒍𝒆 𝑑(𝑢𝑣) = 𝑑𝑢𝑣 + 𝑢𝑑𝑣
• Once the x, y , z components of v and a have been determined, the magtinudes of these vectors
found from the Pythagorean theorem and coordinate direction angles from the components of
their unit vectors.
𝑣 = √𝑣𝑥2 + 𝑣𝑦2 + 𝑣𝑧2
𝑎 = √𝑎𝑥2 + 𝑎𝑦2 + 𝑎𝑧2
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Example Problem
For a short time, the path of the plane is described by 𝑦 = (𝑜. 001𝑥 2 )𝑚. If the plane is rising
𝑚
with a constant velocity 10 𝑠 , determine the magnitudes of the velocity and acceleration of the plane
when it is at 𝑦 = 100 𝑚
𝑚
When 𝑦 = 100 𝑚, then 100 = 0.001𝑥 2 or 𝑥 = 316.2. Also, since 𝑉𝑦 = 10 𝑠 , then
𝑦 = 𝑣𝑦 𝑡:
𝑚
100 𝑚 = (10 𝑠 ) 𝑡
𝑡 = 10 𝑠
Velocity. Using the chain rule to find the relationship between the velocity components we have
𝑣𝑦 = 𝑦 =
𝑑
(0.001𝑥 2 ) = (0.002𝑥)𝑥 = 0.002𝑥𝑉𝑥
𝑑𝑡
Thus
10
𝑚
= 0.002(316.2 𝑚)(𝑣𝑥 )
𝑠
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
𝑣𝑥 = 15.81
𝑚
𝑠
The magnitude of the velocity is therefore
𝑚 2
𝑚 2
𝑣 = √𝑣𝑥2 + 𝑣𝑦2 = √(15.81 𝑠 ) + (10 𝑠 ) = 𝟏𝟖. 𝟕
𝒎
𝒔
Acceleration. Using the chain rule, the time derivative of the equation gives the relation between the
acceleration components.
𝑎𝑦 = 𝑣𝑦 = 0.002𝑥̇ 𝑉𝑥 + 0.002𝑥𝑣𝑥̇ = 0.002(𝑣𝑥2 + 𝑥𝑎𝑥 )
𝑚
When 𝑥 = 316.2 𝑚, 𝑣𝑥 = 15.81 𝑠 , 𝑣𝑦 = 𝑎̇ 𝑦 = 0,
15.81𝑚 2
) + 316.2 𝑚(𝑎𝑥 ))
0 = 0.002 (
𝑠
𝑚
𝑎𝑥 = −0.791 2
𝑠
The magnitude of the plane’s acceleration is therefore;
𝑚 2
𝑚 2
𝒎
a= √𝑎𝑥2 + 𝑎𝑦2 = √(−0.791 𝑠2 ) + (0 𝑠2 ) = 𝟎. 𝟕𝟗𝟏 𝒔𝟐
Motion of a Projectile
Motion of Projectile
The trajectory followed by an object that is launched into the air and moves under the influence
of gravity, with no additional propulsion after the initial launch, is referred to as projectile motion. The
study of projectile motion is classified as classical mechanics and is distinguished by its parabolic path.
A projectile's free-flight motion is frequently studied in terms of its rectangular components.
Consider a projectile launched at point (𝑥𝑜 , 𝑦𝑜 ) with an initial velocity of 𝑣𝑜 and components 𝑣𝑜 𝑥 and
𝑣𝑜 𝑦 to demonstrate the kinematic analysis. When air resistance is ignored, the only force acting on the
projectile is gravity, resulting in a constant downward acceleration of approximately 𝑎𝑐 = 𝑔 =
𝑚
𝑓𝑡
9.81 𝑠𝑠 𝑜𝑟 𝑔 = 32.2 𝑠2 .
A projectile's motion includes the parabolic trajectory followed by an object under gravity with
no additional propulsion. Various parameters related to the trajectory can be analyzed and calculated by
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
considering the independent horizontal and vertical components of motion. The study of projectile
motion provides insights into classical mechanics principles and has practical applications in a variety of
scientific, engineering, and sporting endeavors.
HORIZONTAL MOTION
Since ax = 0, application of the constant acceleration equations yields:
(→+)
𝑣𝑥 = (𝑣𝑜 )𝑥
𝑣 = 𝑣𝑂 + 𝑎𝑐 𝑡 ;
1
(→+)
𝑥 = 𝑥𝑂 + 𝑣𝑂 𝑡 + 2 𝑎𝑐 𝑡 2 ;
(→+)
𝑣 2 = 𝑣𝑂2 + 2𝑎𝑐 (𝑥 − 𝑥𝑜) ; ;
𝑥 = 𝑥𝑂 + (𝑣𝑜 )𝑥𝑡
𝑣𝑥 = (𝑣𝑜 )𝑥
VERTICAL MOTION
Since the positive y axis is directed upward, then ay = -g.
𝑣𝑥 = (𝑣𝑜 )𝑦 − 𝑔𝑡
(↑+)
𝑣 = 𝑣𝑂 + 𝑎𝑐 𝑡 ;
(↑+)
𝑦 = 𝑦𝑂 + 𝑣𝑂 𝑡 + 2 𝑎𝑐 𝑡 2 ;
(↑+)
𝑣 2 = 𝑣𝑂2 + 2𝑎𝑐 (𝑦 − 𝑦𝑜) ;
2𝑔(𝑦 − 𝑦𝑜)
1
1
𝑦 = 𝑦𝑂 + (𝑣𝑜 )𝑦𝑡 − 2 𝑔𝑡 2
𝑣𝑦2 = (𝑣𝑜 )2𝑦 −
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
Example Problem
A sack slides off the ramp, with a horizontal velocity of 12 m/s. If the height of
the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the
range R where sacks begin to pile up.
Coordinate System. The origin of coordinates is
established at the beginning, point A. The initial
𝑚
velocity of a sack has components (𝑣𝑎 )𝑥 = 12 𝑠 and
𝑚
(𝑣𝑎 )𝑦 = 0 𝑠 . Also between points A and B the
𝑚
acceleration is 𝑎𝑦 = −9.81 𝑠2. Since (𝑣𝑏 )𝑥 =
(𝑣𝑎 )𝑥 = 12
𝑚
𝑠
, the three unknowns are (𝑣𝑏 )y, R,
and the time of flight 𝑡𝑎𝑏 .
Vertical Motion. The vertical distance from A to B
is known, and therefore we can obtain a direct solution for 𝑡𝑎𝑏 using the equation
1
𝑦 = 𝑦𝑂 + 𝑣𝑂 𝑡 + 2 𝑎𝑐 𝑡 2 .
1
𝑦 = 𝑦𝑂 + (𝑣𝑜 )𝑦𝑡 − 2 𝑔𝑡 2
(↑+)
1
𝑚
6 𝑚 = 0 + 0 − 2 (9.81 𝑠2 ) 𝑡𝑎𝑏 2
𝑡𝑎𝑏 = 𝟏. 𝟏𝟏 𝒔
Horizontal Motion. Since 𝑡𝑎𝑏 has been calculated. R is determined as follows:
(→+)
𝑥 = 𝑥𝑂 + (𝑣𝑜 )𝑥𝑡
𝑥𝑏 = 𝑥𝑎 + (𝑣𝑎 )𝑥𝑡𝑎𝑏
𝑅 = 0 + 12
𝑚
(1.11 𝑠)
𝑠
𝑅 = 𝟏𝟑. 𝟑 𝒎
FINALS DISCUSSIONS
MODULE III
Kinematics of a Particle
ACTIVITY 8
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places.
1. A particle travels along a straight line with a velocity v = (12 - 3t2) m/s, where t is in
seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the
acceleration when t = 4 s and the displacement from t = 0 to t = 10.
2. A ball A is thrown vertically upward from the top of a 30-m-high building with an initial
velocity of 5 m/s. At the same instant, another ball B is thrown upward from the ground
with an initial velocity of 20 m/s. Determine the height from the ground and the time at
which they pass each other.
3. At any instant, the horizontal position of the weather balloon is defined by x = (8t) ft, where
t is in seconds. If the equation of the path is y = x2/10, determine the magnitude and
direction of the velocity and the acceleration when t = 2 s.
4. The track for this racing event was designed so that riders jump off the slope at 30°, from
a height of 1 m. During a race it was observed that the rider remained in mid-air for 1.5 s.
Determine the speed at which he was traveling off the ramp, the horizontal distance he
travels before striking the ground, and the maximum height he attains. Neglect the size of
the bike and rider.
FINALS DISCUSSIONS
MODULE IVIV
MODULE
Kinetics of a Particle
Kinetics of a
IV
Particle
Objectives:
•
To state Newton’s Second Law of Motion and to define mass and weight.
•
To analyze the accelerated motion of a particle using the equation of motion with
different coordinate systems.
•
To develop the principle of work and energy and apply it to solve problems that
involve force, velocity, and displacement.
•
To study problems that involve power and efficiency.
•
To develop the principle of linear impulse and momentum for a particle and apply
it to solve problems that involve force, velocity, and time.
•
To analyze the mechanics of impact.
Outline:
•
Kinetics of a Particle: Force and Acceleration
o Newton’s Second Law of Motion
o Equation of Motion for a System of Particles
o Equations of Motion: Rectangular Components
•
Kinetics of a Particle: Work and Energy
o The Work of a Force
o Principle of Work and Energy
o Power and Efficiency
•
Kinetics of a Particle: Impulse and Momentum
o Principle of Linear Impulse and Momentum
o Conservation of Linear Momentum for a System of Particles
o Impact
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
Chapter Introduction
The study of kinetics examines the connections between forces and motion. if the motion creates the forces
(causality) or if the forces cause the motion is not a relevant problem in this situation. The particles are the main topic of
this chapter. We will also analyze motion of a particle using the concepts of work and energy. Lastly, we
will integrate the equation of motion with respect to time and thereby obtain the principle of impulse and
momentum. The resulting equation will be useful for solving problems involving force, velocity, and
time.
Kinetic problems may be solved using at least three different methods: (a) Newton's second law; (b) the work
and energy technique; and (c) the impulse and momentum method.
Kinetics of a Particle: Force and Acceleration
Newton’s Second Law of Motion
Kinetics is a branch of dynamics that deals with the relationship between the change in motion
of a body and the forces that cause this change. The basis for kinetics is Newton’s second law, which
states that when an unbalanced force acts on a particle, the particle will accelerate in the direction of
the force with a magnitude that is proportional to the force.
This law can be verified experimentally by applying a known unbalanced force F to a particle,
and then measuring the acceleration a. Since the force and acceleration are directly proportional, the
constant of proportionality, m, may be determined from the ratio m = F/a. This positive scalar m is called
the mass of the particle. Being constant during any acceleration, m provides a quantitative measure of
the resistance of the particle to a change in its velocity, that is its inertia.
The basis for kinetics is Newton’s second law, which states that when an unbalanced force acts
on a particle, the particle will accelerate in the direction of the force with a magnitude that is proportional
to the force.
𝑭 = 𝒎𝒂
The above equation, which is referred to as the equation of motion, is one of the most important
formulations in mechanics. As previously stated, its validity is based solely on experimental evidence.
In 1905, however, Albert Einstein developed the theory of relativity and placed limitations on the use of
Newton’s second law for describing general particle motion.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
Equation of Motion for a System of Particles
The equation of motion will now be extended to include a system of particles isolated within an
enclosed region in space, as shown in Fig. 9.1-1. In particular, there is no restriction in the way the
particles are connected, so the following analysis applies equally well to the motion of a solid, liquid, or
gas system.
At the instant considered, the arbitrary i-th particle, having a mass mi, is subjected to a system of
internal forces and a resultant external force. The internal force, represented symbolically as fi, is the
resultant of all the forces the other particles exert on the i-th particle. The resultant external force Fi
represents, for example, the effect of gravitational, electrical, magnetic, or contact forces between the ith particle and adjacent bodies or particles not included within the system.
When the equation of motion is applied to each of the other particles of the system, similar
equations will result. And, if all these equations are added together vectorially, we obtain
𝚺𝑭𝒊 + 𝚺𝒇𝒊 = 𝚺𝒎𝒊 𝒂𝒊
Fig. 9.1-1
If rG is a position vector which locates the center of mass G of the particles, then by definition of
the center of mass, mrG = miri, where m = Σmi is the total mass of all the particles. Differentiating this
equation twice with respect to time, assuming that no mass is entering or leaving the system, yields
𝑚𝒂𝒊 = 𝚺𝒎𝒊 𝒂𝒊
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
Hence, the sum of the external forces acting on the system of particles is equal to the total mass
of the particles times the acceleration of its center of mass G. Since in reality, all particles must have a
finite size to possess mass. The equation below justifies application of the equation of motion to a body
that is represented as a single particle.
𝚺𝑭 = 𝑚𝒂𝑮
Equations of Motion: Rectangular Components
When a particle moves relative to an inertial x, y, z frame of reference, the forces acting on the
particle, as well as its acceleration, can be expressed in terms of their x, y, z components. We may write
the following three scalar equations:
𝚺𝑭𝒙 = 𝑚𝒂𝒙
𝚺𝑭𝒚 = 𝑚𝒂𝒚
𝚺𝑭𝒛 = 𝑚𝒂𝒛
In particular, if the particle is constrained to move only in the x–y plane, then the first two of
these equations are used to specify the motion.
Procedure for Analysis
The equations of motion are used to solve problems which require a relationship between the
forces acting on a particle and the accelerated motion they cause.
Free-Body Diagram
● Select the inertial coordinate system. Most often, rectangular or x, y, z coordinates are chosen to
analyze problems for which the particle has rectilinear motion.
● Once the coordinates are established, draw the particle’s free body diagram. Drawing this
diagram is very important since it provides a graphical representation that accounts for all the
forces (ΣF) which act on the particle, and thereby makes it possible to resolve these forces into
their x, y, z components.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
● The direction and sense of the particle’s acceleration (a) should also be established. If the sense
is unknown, for mathematical convenience assume that the sense of each acceleration component
acts in the same direction as its positive inertial coordinate axis.
● The acceleration may be represented as the ma vector on the kinetic diagram.
● Identify the unknowns in the problem.
Equation of Motion
● Friction. If a moving particle contacts a rough surface, it may be necessary to use the frictional
equation, which relates the frictional and normal forces Ff and N acting at the surface of contact
by using the coefficient of kinetic friction, Ff = μkN. If the particle is on the verge of relative
motion, then the coefficient of static friction should be used.
● Spring. If the particle is connected to an elastic spring having negligible mass, the spring force
Fs can be related to the deformation of the spring by the equation Fs = ks. Here k is the spring's
stiffness measured as a force per unit length, and s is the stretch or compression defined as the
difference between the deformed length, I and the undeformed length lo, s = l - lo.
Kinematics
● If the velocity or position of the particle is to be found, it will be necessary to apply the necessary
kinematic equations once the particle's acceleration is determined from ΣF = ma.
● If acceleration is a function of time, use a = dv/dt and v = ds/dt which, when integrated, yield
the particle's velocity and position, respectively.
● If acceleration is a function of displacement, integrate ads = vdv to obtain the velocity as a
function of position.
● If acceleration is constant, use v = vo + at, s = so + vot + 1/2at2, v2 = vo2 + 2a(s - so) to determine
the velocity or position of the particle.
● If the solution for an unknown vector component yields a negative scalar, it indicates that the
component acts in the direction opposite to that which was assumed.
EXAMPLE 1
The 50-kg crate rests on a horizontal surface for which the coefficient of kinetic friction is µk =
0.3. If the crate is subjected to a 400-N towing force as shown, determine the velocity of the crate in 3 s
starting from rest.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
Σ𝐹𝑥 = 𝑚𝑎𝑥 ;
Σ𝐹𝑦 = 𝑚𝑎𝑦 ;
400𝑐𝑜𝑠30° − 0.3𝑁𝑐 = 50𝑎
𝑁𝑐 − 490.5 + 400𝑠𝑖𝑛30° = 0
𝑁𝑐 = 290.5
𝑎 = 5.185
𝑚
𝑠2
𝑣 = 𝑣𝑜 + 𝑎𝑐 𝑡 = 0 + 5.185(3)
𝒗 = 𝟏𝟓. 𝟔
𝒎
𝒔
EXAMPLE 2
A 10-kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m/s,
Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected.
𝑚
𝑊 = 𝑚𝑔 = 10𝑘𝑔 (9.81 𝑠2 ) = 98.1𝑁
Σ𝐹𝑦 = 𝑚𝑎𝑦
−98.1 = (10)𝑎𝑦
𝑎𝑦 = −9.81
By using the formula: 𝑣 2 = 𝑣𝑜2 + 2𝑎𝑆
02 = 502 + 2(−9.81)(ℎ)
𝒉 = 𝟏𝟐𝟕𝒎
𝑚
𝑠2
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
Kinetics of a Particle: Work and Energy
The Work of a Force
The resulting equation will be useful for solving problems that involve force, velocity, and
displacement. Before we do this, however, we must first define the work of a force. Specifically, a force
F will do work on a particle only when the particle undergoes a displacement in the direction of the force.
For example, if the force F in Fig. below causes the particle to move along the path s from position r to
a new position r', the displacement is then dr = r’- r. The magnitude of dr is ds, the length of the
differential segment along the path. If the angle between the tails of dr and F is θ, Fig. 9.2–1, then the
work done by F is a scalar quantity, defined by
𝑑𝑈 = 𝐹 𝑑𝑠 𝑐𝑜𝑠𝜃
Fig. 9.2–1
This equation may alternatively be expressed as, according to the definition of the dot product:
𝒅𝑼 = 𝑭 ⋅ 𝒅𝒓
This result may be interpreted in one of two ways: either as the product of F and the component
of displacement ds cos θ in the direction of the force, or as the product of ds and the component of force,
F cos u, in the direction of displacement. Note that if 0° ≤ θ < 90°, then the force component and the
displacement have the same sense so that the work is positive; whereas if 90° < θ ≤ 180°, these vectors
will have opposite sense, and therefore the work is negative. Also, dU = 0 if the force is perpendicular
to displacement, since cos 90° = 0, or if the force is applied at a fixed point, in which case the
displacement is zero.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
The unit of work in SI units is the joule (J), which is the amount of work done by a one-newton
force when it moves through a distance of one meter in the direction of the force (1 J = 1 N・m). In the
FPS system, work is measured in units of foot-pounds (ft・lb), which is the work done by a one-pound
force acting through a distance of one foot in the direction of the force.
Work of a Variable Force
If the particle acted upon by the force F undergoes a finite displacement along its path from r1 to
r2 or s1 to s2, Fig. 9.2-2a, the work of force F is determined by integration. Provided F and θ can be
expressed as a function of position, then
𝑟2
𝑠2
𝑈1−2 ∫ 𝑭 ⋅ 𝑑𝒓 = ∫ 𝐹 𝑐𝑜𝑠𝜃 𝑑𝑠
𝑟1
𝑠1
Sometimes, this relation may be obtained by using experimental data to plot a graph of F cos θ
vs. s. Then the area under this graph bounded by s1 and s2 represents the total work, Fig. 9.2-2b.
Fig. 9.2–2
Work of a Constant Force Moving Along a Straight Line
If the force Fc has a constant magnitude and acts at a constant angle θ from its straight-line path,
Fig. 9.2-3a, then the component of Fc in the direction of displacement is always Fc cos u. The work done
by Fc when the particle is displaced from s1 to s2 is determined from Eq. 9.2–1, in which case
𝑟2
𝑈1−2 = 𝐹𝑐 𝑐𝑜𝑠𝜃 ∫ 𝑑𝑠
𝑟1
or
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
𝑈1−2 = 𝐹𝑐 𝑐𝑜𝑠𝜃(𝑠2 − 𝑠1 )
Here the work of Fc represents the area of the rectangle in Fig. 9.2–3b
.
Fig. 9.2–3
Work of a Weight
Consider a particle of weight W, which moves up along the path. The work done by the
gravitational force acting on a particle (or weight of an object) can be calculated by using
𝑦2
∫ −𝑊 𝑑𝑦 = −𝑊(𝑦2 − 𝑦1 )
𝑦1
𝑈1−2 = −𝑊 ∆𝑦
The work of a weight is the product of the magnitude of the particle’s weight and its vertical
displacement. The work is negative since the W is downwards and Δy is upward. However, if the particle
is displaced downward (-Δy), the work of the weight is positive.
Work of a Spring Force
Consider a spring with stiffness k that was elongated by a distance s. If a particle is attached to
the spring, the force Fs exerted on the particle is opposite to that exerted on the spring. Thus, the
work done on the particle by the spring force will be negative. If the particle displaces from s1 to s2
the work of Fs is then
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
𝐹 = −𝑘𝑠
𝑠2
𝑈1−2 = ∫ 𝐹 ⋅ 𝑑𝑠
𝑠1
𝑠2
𝑈1−2 = ∫ −𝑘𝑠𝑑𝑠
𝑠1
1
𝑈1−2 = − 𝑘[𝑠22 − 𝑠12 ]
2
Principle of Work and Energy
The work-energy principle states that an increase in the kinetic energy of a rigid body is caused
by an equal amount of positive work done on the body by the resultant force acting on that body. The
work-energy theorem relates work and energy using the basic differential equation from kinematics and
the law of acceleration.
The equation below represents the principle of work and energy for the particle. The term on the
left is the sum of the work done by all the forces acting on the particle as the particle moves from point
1 to point 2. The two terms on the right side, define the particle’s final and initial kinetic energy,
respectively.
Σ𝑈1−2 =
1
1
𝑚𝑣22 − 𝑚𝑣12
2
2
𝑇1 + Σ𝑈1−2 = 𝑇2
Procedure for Analysis
Work (Free-Body Diagram)
● Establish the inertial coordinate system and draw a free-body diagram of the particle in order to
account for all the forces that do work on the particle as it moves along its path.
Principle of Work and Energy
● Apply the principle of work and energy, T1 + U1-2 = T2.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
● The kinetic energy at the initial and final points is always positive, since it involves the speed
squared (T = 1/2 mv2).
● A force does work when it moves through a displacement in the direction of the force.
● Work is positive when the force component is in the same sense of direction as its displacement,
otherwise it is negative.
● Forces that are functions of displacement must be integrated to obtain the work. Graphically, the
work is equal to the area under the force-displacement curve.
● The work of a weight is the product of the weight magnitude and the vertical displacement, UW
= ±Wy. It is positive when the weight moves downwards.
● The work of a spring is of the form Us = 1/2ks2, where k is the spring stiffness and s is the stretch
or compression of the spring.
Power and Efficiency
Power. The term “power” provides a useful basis for choosing the type of motor or machine
which is required to do a certain amount of work in each time. For example, two pumps may each be
able to empty a reservoir if given enough time; however, the pump having the larger power will complete
the job sooner. The power generated by a machine or engine that performs an amount of work dU within
the time interval dt is therefore:
𝑃=
𝑑𝑈
𝑑𝑡
If the work dU is expressed as dU = F・dr, then
𝑃=
𝑑𝑈 𝐹 ⋅ 𝑑𝑟
𝑑𝑟
=
=𝐹⋅
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑷=𝑭⋅𝒗
Hence, power is a scalar, where in this formulation v represents the velocity of the particle
which is acted upon by the force F. The basic units of power used in the SI and FPS systems are the
watt (W) and horsepower (hp), respectively. These units are defined as
𝐽
𝑚
1𝑊 = 1 = 1𝑁 ⋅
𝑠
𝑠
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
1ℎ𝑝 = 550𝑓𝑡 ⋅
𝑙𝑏
𝑠
For conversion between the two systems of units, 1 hp = 746 W
Efficiency. The mechanical efficiency of a machine is defined as the ratio of the output of useful
power produced by the machine to the input of power supplied to the machine. Hence,
𝜀=
𝑝𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡
𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡
If energy supplied to the machine occurs during the same time interval at which it is drawn, then
the efficiency may also be expressed in terms of the ratio.
Since machines consist of a series of moving parts, frictional forces will always be developed
within the machine, and as a result, extra energy or power is needed to overcome these forces.
Consequently, power output will be less than power input and so the efficiency of a machine is always
less than 1. The power supplied to a body can be determined using the following procedure.
EXAMPLE 1
When s = 0.6 m, the spring is unstretched and the 10-kg block, which is subjected to a force of
100 N, has a speed of 5 m/s down the smooth plane. Determine the distance s when the block stops.
𝑇1 + ∫ 𝐹𝑑𝑡 = 𝑇2
1
1
1
𝑚𝑣12 + 𝐹(𝑠2 − 𝑠1 ) + 𝑊𝑠𝑖𝑛𝜃(𝑠2 − 𝑠1 ) − 𝑘(𝑠2 − 𝑠1 ) = 𝑚𝑣22
2
2
2
1
(10)(5)2 + 100[(𝑠 − 0.6) − (0.6 − 0.6)] + 10(9.81)𝑠𝑖𝑛30°
2
1
1
[(𝑠 − 0.6) − (0.6 − 0.6)] − (200)[(𝑠 − 0.6)2 − (0.6 − 0.6)2 ] = (10)(0)2
2
2
𝒔 = 𝟐. 𝟔𝟖𝟖𝟗 𝒎
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
EXAMPLE 2
The man pushes on the 50-kg crate with a force of F = 150 N. Determine the power supplied by
the man when t = 4 s. The coefficient of kinetic friction between the floor and the crate is μk = 0.2.
Initially the crate is at rest.
Σ𝐹𝑦 = 𝑚𝑎𝑦 ;
Σ𝐹𝑥 = 𝑚𝑎𝑥 ;
3
𝑁 − 150𝑁 − 50(9.81)𝑁 = 0
5
4
150𝑁 − 0.2(580.5)𝑁 = (50𝑘𝑔)𝑎
5
𝑁=0
𝑎 = 0.078
𝑚
𝑠2
The velocity of the crate when t = 4s is therefore,
𝑣 = 𝑣0 + 𝑎𝑐 𝑡
𝑣 = 0 + (0.078
𝑚
𝑚
) (4𝑠) = 0.321
2
𝑠
𝑠
The power supplied to the crate by the man when t = 4s is therefore
4
𝑚
𝑃 = 𝐹 ⋅ 𝑣 = 𝐹𝑥 𝑣 = ( )(150𝑁)(0.312 )
5
2
𝑷 = 𝟑𝟕. 𝟒𝑾
Kinetics of a Particle: Impulse and Momentum
Principle of Linear Impulse and Momentum
Using kinematics, the equation of motion for a particle of mass m can be written as
𝑑𝑣
∑ 𝐹 = 𝑚𝑎 = 𝑚
𝑑𝑡
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
where a and v are both measured from an inertial frame of reference. Rearranging the terms and
integrating between the limits v = v1 at t = t1 and v = v2 at t = t2, we have
𝑡2
𝑣2
∑ ∫ 𝑭𝑑𝑡 = 𝑚 ∫ 𝑑𝑣
𝑡1
𝑡2
𝑜𝑟
∑ ∫ 𝑭𝑑𝑡 = 𝑚𝑣2 − 𝑚𝑣1
𝑣1
𝑡1
This equation is referred to as the principle of linear impulse and momentum. From the derivation
it is simply a time integration of the equation of motion. It provides a direct means of obtaining the
particle’s final velocity v2 after a specified time period when the particle’s initial velocity is known and
the forces acting on the particle are either constant or can be expressed as functions of time. By
comparison, if v2 was determined using the equation of motion, a two-step process would be necessary;
i.e., apply ΣF = ma to obtain a, then integrate a = dv/dt to obtain v2.
Impulse and momentum
The equation of principles of linear impulse and momentum were derived from the basic
differential equation for acceleration and the law of acceleration.
𝑡2
𝑚𝑣1 + ∑ ∫ 𝑭𝑑𝑡 = 𝑚𝑣2
𝑡1
This equation simply states that the initial momentum of the particle at time t1 plus the sum of
all the impulses applied to the particle from t1 to t2 is equivalent to the final momentum of the particle
at time t2.
Fig. 9.3–1
Similar to the free-body diagram, the impulse diagram is an outlined shape of the particle
showing all the impulses that act on the particle when it is located at some intermediate point along its
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
path. If each of the vectors in Fig. 9.3–1 is resolved into its x, y, z components, we can write the
following three scalar equations of linear impulse and momentum.
𝑡2
𝑚(𝑣𝑥 )1 + ∑ ∫ 𝑭𝒙 𝑑𝑡 = 𝑚(𝑣𝑥 )2
𝑡1
𝑡2
𝑚(𝑣𝑦 )1 + ∑ ∫ 𝑭𝒚 𝑑𝑡 = 𝑚(𝑣𝑦 )2
𝑡1
𝑡2
𝑚(𝑣𝑧 )1 + ∑ ∫ 𝑭𝒛 𝑑𝑡 = 𝑚(𝑣𝑧 )2
𝑡1
Conservation of Linear Momentum for a System of Particles
When the sum of the external impulses acting on a system of particles is zero. This equation is
referred to as the conservation of linear momentum. It states that the total linear momentum for a system
of particles remains constant during the time period t1 to t2.
∑ 𝑚𝑖 (𝑣𝑖 )1 = ∑ 𝑚𝑖 (𝑣𝑖 )2
Impulsive forces normally occur due to an explosion or the striking of one body against another,
whereas nonimpulsive forces may include the weight of a body, the force imparted by a slightly deformed
spring having a relatively small stiffness, or for that matter, any force that is very small compared to
other larger (impulsive) forces. When making this distinction between impulsive and nonimpulsive
forces, it is important to realize that this only applies during the time t1 to t2.
The impulsive force causes the change in momentum before and after
impact. Consider two colliding particles A and B, the impulsive force on A
due to B should be equal to the impulsive force on B due to A. These are
similar in magnitude but
are oppositely directed.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
Fig. 9.3–2
𝑡2
𝑡2
𝑚𝐴 𝑣𝐴1 − ∫ 𝐹𝑑𝑡 = 𝑚𝐴 𝑣𝐴2
𝑡1
𝑚𝐵 𝑣𝐵1 − ∫ 𝐹𝑑𝑡 = 𝑚𝐵 𝑣𝐵2
𝑡1
𝑚𝐴 𝑣𝐴1 + 𝑚𝐵 𝑣𝐵1 = 𝑚𝐴 𝑣𝐴2 + 𝑚𝑏 𝑣𝐵2
The internal impulses for the system will always be cancel out, since they occur in equal but
opposite collinear pairs.
Impact
Impact occurs when two bodies collide during a very short time period, causing large impulsive
forces to be exerted between the bodies.
The line of impact is a line through the centers of mass of the colliding particles. When two
particles A and B have a direct impact, the internal impulse between them is equal, opposite, and
collinear. Consequently, the conservation of momentum for this system applies along the line of
impact.
In general, there are two types of impact:
Central Impact
Central Impact occurs when the directions of motion
of the two colliding particles are along the line of impact.
During the collision the particles must be thought of as
deformable or nonrigid. The particles will undergo a period
of deformation such that they exert an equal but opposite
deformation impulse on each other. Only at the instant of
maximum deformation will both particles move with a common velocity v, since their relative motion is
zero.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
Oblique Impact
Oblique impact occurs when the direction of motion
of one or both particles is at an angle to the line of impact.
When oblique impact occurs between two smooth particles,
the particles move away from each other with velocities
having unknown directions as well as unknown magnitudes
Coefficient of Restitution, e
The ratio of the particles’ relative separation velocity after impact to the particles’ relative
approach velocity before impact is called the coefficient of restitution.
(𝑉𝐵 )2 − (𝑉𝐴 )2
𝑒=
(𝑉𝐴 )1 − (𝑉𝐵 )1
it is states that e is equal to the ratio of the relative velocity of the particles’ separation just after
impact, (vB)2 - (vA)2, to the relative velocity of the particles’ approach just before impact, (vA)1 - (vB)1. By
measuring these relative velocities experimentally, it has been found that e varies appreciably with impact
velocity as well as with the size and shape of the colliding bodies. For these reasons the coefficient of
restitution is reliable only when used with data which closely approximate the conditions which were
known to exist when measurements of it were made. In general e has a value between zero and one, and
one should be aware of the physical meaning of these two limits.
Elastic impact (e = 1)
In a perfectly elastic collision, no energy is lost, and the relative separation velocity equals
the relative approach velocity of the particles. In practical situations, this condition cannot be
achieved.
Plastic impact (e = 0)
In a plastic impact, the relative separation velocity is zero and the energy lost during the
collision is a maximum. The particles stick together and move with a common velocity after the impact.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
EXAMPLE 1
The 15-Mg boxcar A is coasting at 1.5 m/s on the horizontal track when it encounters a12-Mg
tank car B coasting at 0.75 m/s toward it as shown in the figure. If the cars collide and couple together,
determine (a) the speed of both cars just after the coupling, and (b) the average force between them if
the coupling takes place in 0.8 s
𝑚𝐴 (𝑣𝐴 )1 + 𝑚𝐵 (𝑣𝐵 )1 = (𝑚𝐴 + 𝑚𝐵 )𝑣2
(15000𝑘𝑔) (1.5
𝑚
𝑚
) − 12000𝑘𝑔 (0.75 ) = (27000)𝑣2
𝑠
𝑠
𝒗𝟐 = 𝟎. 𝟓
𝒎
𝒔
𝑚𝐴 (𝑣𝐴 )1 + Σ ∫ 𝑓 𝑑𝑡 = 𝑚𝐴 𝑣2
(15000𝑘𝑔) (1.5
𝑚
𝑚
) − 𝐹𝑎𝑣𝑔 (0.8𝑠) = (15000)(0.5 )
𝑠
𝑠
𝑭𝒂𝒗𝒈 = 𝟏𝟖. 𝟖 𝒌𝑵
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
EXAMPLE 2
The ball strikes the smooth wall with a velocity vB1 = 20 m/s. The coefficient of restitution
between the ball and the wall is e = 0.75. Calculate the velocity of the ball just after the impact.
𝑚[(𝑣𝐵 )1 ]𝑦 = 𝑚[(𝑣𝐵 )2 ]𝑦
𝑚
[(𝑣𝐵 )2 ]𝑦 = [(𝑣𝐵 )1 ]𝑦 = (20 2 ) 𝑠𝑖𝑛30° = 10
𝑒=
(𝑉𝑤 )2 − [(𝑉𝑏 )2 ]𝑥
[(𝑉𝑏 )1 ]𝑥 − (𝑉𝑤 )1
0 − [(𝑉𝑏 )2 ]𝑥
𝑚
(20 𝑠 ) 𝑐𝑜𝑠30° − 0
𝑚
𝑚
[(𝑉𝑏 )2 ]𝑥 = −12.99 = 12.99 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡
𝑠
𝑠
0.75 =
(𝑉𝑏 )2 = √[(𝑉𝑏 )2 ]𝑥 2 + [(𝑉𝑏 )2 ]𝑦 2
𝑚 2
𝑚 2
𝑠
𝑠
= √(12.99 ) + (10 )
= 𝟏𝟔. 𝟒
𝒎
𝒔
[(𝑣𝐵 )2 ]𝑦
𝜃 = 𝑡𝑎𝑛−1 ([(𝑣
𝐵 ) 2 ]𝑦
= 𝑡𝑎𝑛−1 (
= 𝟑𝟕. 𝟔°
𝑚
𝑠
𝑚
12.99
𝑠
10
)
)
𝑚
𝑠
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
ACTIVITY 9
GENERAL INSTRUCTIONS: Answer all the problems with full honesty and integrity. Box and
express your final answer in four decimal places.
1. The 10-lb block has a speed of 4 ft/s when the force of F = (8t2) lb is applied. Determine the time and
velocity of the block when it moves s = 30 ft. The coefficient of kinetic friction at the surface is μs = 0.2.
2. The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at A. If the plane is
smooth, determine the time needed for the block to reach the ground and the distance d, measured from
the wall, to where the block strikes the ground. Neglect the size of the block.
3. An automobile having a mass of 2 Mg travels up a 7° slope at a constant speed of v = 100 km/h. If
mechanical friction and wind resistance are neglected, determine the power developed by the engine if
the automobile has an efficiency ɛ = 0.65.
FINALS DISCUSSIONS
MODULE IV
Kinetics of a Particle
4. If the jet on the dragster supplies a constant thrust of T = 20 kN, determine the power generated by the
jet as a function of time. Neglect drag and rolling resistance, and the loss of fuel. The dragster has a mass
of 1 Mg and starts from rest.
5. The 20-g bullet is traveling at 400 m/s when it becomes embedded in the 2-kg stationary block.
Determine the distance the block will slide before it stops. The coefficient of kinetic friction between the
block and the plane is μk = 0.2.
6. Disk A has a mass of 2 kg and is sliding forward on the smooth surface with a velocity (vA)1 = 5 m/s
when it strikes the 4-kg disk B, which is sliding towards A at (vB)1 = 2 m/s with direct central impact.
If the coefficient of restitution between the disks is e = 0.4, compute the velocities of A and B just after
collision.
APPENDICES
SOLUTIONS
ACTIVITY 1
ANSWER KEY:
1.
𝐹𝑅 = √6002 + 8002 − 2(600(800) cos 6 0°)
𝐹𝑅 = 721.1102 𝑁 (answer)
𝑠𝑖𝑛 ∝ sin 60
∝= 30­° ;
=
800
721
800 sin 6 0
𝜃 = sin (
) = 73.93°
721
𝜃 = 43.8979° (answer)
2.
𝐹𝑅2 = (600)2 + (800)2 − 2(600)(800) cos 4 0.9 = 274 300
𝐹𝑅 = 523.7444 𝑙𝑏 (answer)
sin 4 0.8934
𝑠𝑖𝑛𝜃
𝜃=(
)=
523.7444
600
𝜃 = 48.5874° (answer)
3.
Σ𝑥 = 770 cos(36.87) − 725 cos(60) − 750 cos(45)
Σ𝑥 = −276.83𝑁
Σ𝑦 = −770 sin(36.87) − 725 sin(60) + 750 sin(45)
Σ𝑦 = −559.54𝑁
APPENDICES
SOLUTIONS
𝐹𝑅 = √𝑥 2 + 𝑦 2
𝐹𝑅 = √(−276.83)2 + (−559.54)2
𝐹𝑅 = 624.2755𝑁 (answer)
−559.54
)
−275.83
𝜃 = 63.6763 + 180
𝜃 = 243.6763° (answer)
𝜃 = tan−1 (
4.
𝑥 = 1 + 2 + 2.5 cos 4 5°
𝑥 = 4.77𝑚
𝑦 = 2.5 sin 4 5°
+↺ 𝑀𝑜 = −(600)(1) + (500)(4.77) − (300)(1.77)
+↺ 𝑀𝑜 = 1254𝑛𝑚 (answer)
ACTIVITY 2
Answer Key
1.
𝑀𝑜 = ∑ 𝑥𝐹
3
5
𝑀𝑜 = −500 ( ) (0.25) +
4
500 (5) (0.425) +
600 cos 60° (0.25) +
600 sin 60° (0.425)
𝑀𝑜 = 𝟐𝟔𝟖. 𝟑𝟑𝟔𝟓 𝑵(𝒎)
APPENDICES
SOLUTIONS
2.
3
5
𝐹𝑥 = 500 ( )
4
5
𝐹𝑥 = 500 ( )
3
𝑅𝑥 = 500 ( ) − 200 + 200 = 300𝑁(𝑚)
5
4
𝑅𝑦 = 500 ( ) − 250 = −350𝑁(𝑚)
5
𝑅 = √3002 + 3502 = 𝟒𝟔𝟎. 𝟕𝟕𝟕𝟐 𝑵(𝒎)
𝑀𝑜 = 𝑅𝑑
4
3
= 300 ( ) (2.50) − 500 ( ) (1) − 700(1.25) + 200
5
5
= −𝟑𝟕. 𝟓 𝑵(𝒎) 𝒐𝒓 𝟑𝟕. 𝟓 𝑵(𝒎) 𝒄𝒐𝒖𝒏𝒕𝒆𝒓 𝒄𝒍𝒐𝒄𝒘𝒊𝒔𝒆
3.
𝑅𝑑 = ∑ 𝐹𝑥
300(4) = −100(0) + 𝑃(2) − 𝐹(5) + 200(7)
1200 = 2𝑃 − 5𝐹 + 1400
1400 − 1200 = 2𝑃 − 5𝐹
200 = 2𝑃 − 5F
2𝑃 − 5𝐹 − 200 = 0
200 = 2𝑃 − 5F
2(200 + 𝐹) − 5𝐹 + 200 = 0
400 + 2𝐹 − 5𝐹 + 200 = 0
400 − 3𝐹 + 200 = 0
600 3𝐹
=
3
3
APPENDICES
SOLUTIONS
𝑭 = 𝟐𝟎𝟎𝒍𝒃
𝑃 = 200 + 200
𝑷 = 𝟒𝟎𝟎𝒍𝒃
4.
∑ 𝑀𝐶 = −6(1.5) + 8(1)
= −1𝑘𝑁. 𝑚
3
∑ 𝑅𝑥 = −8 + 5 ( ) = −5𝑘𝑁 = 5𝑘𝑁. 𝑚 ←
5
4
∑ 𝑅𝑦 = −6 − 5 ( ) = −10𝑘𝑁 = 10𝑘𝑁. 𝑚 ↓
5
𝑅 = √52 + 102 = 11.1803 𝑘𝑁. 𝑚
∑ 𝑀𝐶 = 𝑅𝑥 𝑑
−1 = −5𝑚
𝒅 = 𝟎. 𝟐𝒎
ACTIVITY 3
ANSWER KEY
1. FBD:
+ → ∑Fx = 0
;
TB − TA cos15° = 0
+ ↓ ∑Fy = 0
;
−𝑇𝐴 𝑠𝑖𝑛15° + 10 (9.81) = 0
98.1 = 𝑇𝐴 𝑠𝑖𝑛15°
98.1
𝑇𝐴 = 𝑠𝑖𝑛15
𝑻𝑨 = 𝟑𝟕𝟗. 𝟎𝟐𝟗𝟑 𝑵
TB = TA cos15°
TB = 379.0293cos15°
𝐓𝐁 = 𝟑𝟔𝟔. 𝟏𝟏𝟒𝟐 𝑵
APPENDICES
SOLUTIONS
+ → ∑Fx = 0
−TB + TD cosθ° = 0
;
TD cosθ° = 366.1142 → eq 1
+ ↓ ∑Fy = 0
−TD sinθ° + 15(9.81) = 0
;
𝑇𝐷 sinθ° = 147.15
𝑇𝐷 sinθ° = 147.15
147.15
𝑡𝑎𝑛θ° = 366.1142
147.15
147.15
𝑠𝑖𝑛121.8964°
θ° = tan−1 (366.1142)
𝑇𝐷 =
𝜽° = 𝟐𝟏. 𝟖𝟗𝟔𝟒
𝑻𝑫 = 𝟑𝟗𝟒. 𝟓𝟕𝟗𝟐 𝑵
2. FBD:
𝑐 = √1.12 + 32
𝑐 = 3.1953
+ → ∑Fx = 0
;
𝐵𝐴 (
+ ↓ ∑Fy = 0
➔ 𝑩𝑨 = 𝐁𝐂 = 𝑻; 𝑻 = 𝒌𝒔;
➔ 𝒔 = 𝒍 − 𝒍𝒐
lo = 3𝑚;
k = 500N/m
lo = √1.12 + 32 = 3.1953 𝑚
SINCE: 𝑩𝑨 = 𝑩𝑪 = 𝑻
BA (
1.1
1.1
) + 𝐵𝐶 (
)−𝐹 = 0
3.1953
3.1953
97.6545(0.3443) + 97.6545(0.3443) = 𝐹
𝑭 = 𝟔𝟕. 𝟐𝟑𝟔𝟏 𝑵
1.1
1.1
) + BC (
)−F = 0
3.1953
3.1953
3
)+
3.1953
; −𝐵𝐴 (
➔ 𝑇 = 𝑘𝑠 = 𝑘(𝑙 − 𝑙𝑜 )
𝑇 = 500(3.1953 − 3)𝑚
𝑻 = 𝟗𝟕. 𝟔𝟓𝟒𝟓 𝑵
3
)
3.1953
BC (
=0
APPENDICES
SOLUTIONS
3.
+ ∑MA = 0
;
12
52 (13) (0.3) + 30𝑠𝑖𝑛60(0.7)
∑𝑴𝑨 = 𝟑𝟐. 𝟓𝟖𝟔𝟓 𝑵𝒎, P
+→ ∑Fx = 0
;
5
13
AX − 52 ( ) + 30𝑐𝑜𝑠60° = 0
AX = 5 𝑁
R = √𝐴𝑋2 + 𝐴2𝑦
R = √52 + 73.98082
𝑹 = 𝟕𝟒. 𝟏𝟒𝟗𝟓 𝑵
+ ↓ ∑Fy = 0
;
12
−𝐴𝑦 + 52 (13) + 30𝑠𝑖𝑛60 = 0
𝐀 𝐲 = 𝟕𝟑. 𝟗𝟖𝟎𝟖 𝐍
𝑡𝑎𝑛θ° =
Ay
Ax
73.9808
−1(
)
5
θ = tan
𝜽 = 𝟖𝟔. 𝟏𝟑𝟑𝟓°
APPENDICES
SOLUTIONS
ACTIVITY 4
ANSWER KEY
1.
∑ 𝐹𝑥 = 0 = 𝐻𝑥
Q + ∑ 𝑀𝐻 = 𝐴𝑦 (12) + 30(9) + 30(6) + 90(3) = 0
𝐴𝑦 = 60 𝑘𝑁
∑ 𝐹𝑥 = 0 = 60 + 𝐻𝑦 − 30 − 30 − 90
𝐻𝑦 = 90 𝑘𝑁
@joint A
3
∑ 𝐹𝑦 = 0 = 60 + 𝐹𝐴𝐵 (
)
3√2
𝐹𝐴𝐵 = −84.8528𝑘𝑁
𝑭𝑨𝑩 = 𝟖𝟒. 𝟖𝟓𝟐𝟖 𝒌𝑵, 𝑪
3
∑ 𝐹𝑥 = 0 = 𝐹𝐴𝐵 (
) + 𝐹𝐴𝐶
3√2
3
−84.8528(
) + 𝐹𝐴𝐶 = 0
3√2
𝑭𝑨𝑪 = 𝟔𝟎 𝒌𝑵, 𝑻
@joint B
3
3
∑ 𝐹𝑥 = −𝐹𝐴𝐵 (
) + 𝐹𝐵𝐷 (
)=0
3√2
√10
3
3
= 84.8528(
) + 𝐹𝐵𝐷 (
)=0
3√2
√10
APPENDICES
SOLUTIONS
𝐹𝐵𝐷 = −63.2455𝑘𝑁
𝑭𝑩𝑫 = −𝟔𝟑. 𝟐𝟒𝟓𝟓 𝒌𝑵, 𝑪
1
3
∑ 𝐹𝑦 = 0 = 𝐹𝐵𝐷 (
) − 𝐹𝐴𝐵 (
) − 𝐹𝐵𝐶
3√2
√10
𝑭𝑩𝑪 = 𝟒𝟎 𝒌𝑵, 𝑻
@joint C
4
∑ 𝐹𝑥 = 0 = 𝐹𝐵𝐶 − 30 + 𝐹𝐶𝐷 ( )
5
4
40 − 30 + 𝐹𝐶𝐷 ( ) = 0
5
𝐹𝐶𝐷 = −12.5𝑘𝑁
𝑭𝑪𝑫 = 𝟏𝟐. 𝟓 𝒌𝑵, 𝑪
3
∑ 𝐹𝑋 = 0 = −𝐹𝐴𝐶 + 𝐹𝐶𝐸 + 𝐹𝐶𝐷 ( )
5
3
∑ 𝐹𝑋 = −60 + 𝐹𝐶𝐸 + (−12.5)( )
5
𝑭𝑪𝑬 = 𝟔𝟕. 𝟓 𝒌𝑵, 𝑻
@joint H
3
∑ 𝐹𝑦 = 0 = 𝐻𝑦 + 𝐹𝐻𝐹 (
)=0
3√2
3
90 + 𝐹𝐻𝐹 (
)=0
3√2
𝐹𝐻𝐹 = −127.2792𝑘𝑁
𝑭𝑯𝑭 = 𝟏𝟐𝟕. 𝟐𝟕𝟗𝟐 𝒌𝑵, 𝑪
3
∑ 𝐹𝑥 = 0 − 𝐹𝐻𝐹 (
) − 𝐹𝐺𝐻 = 0
3√2
3
−(−127.2792)(
) − 𝐹𝐺𝐻 = 0
3√2
𝑭𝑮𝑯 = 𝟗𝟎 𝒌𝑵, 𝑻
@joint F
∑ 𝐹𝑥 = 0
APPENDICES
SOLUTIONS
3
3
−𝐹𝐷𝐹 (
) + 𝐹𝐹𝐻 (
)=0
3√2
√10
3
3
−𝐹𝐷𝐹 (
) + (−127.2792)(
)=0
3√2
√10
𝑭𝑫𝑭 = −𝟗𝟒. 𝟖𝟔𝟖𝟑 𝒌 𝑵, 𝑻
𝑭𝑫𝑭 = 𝟗𝟒. 𝟖𝟔𝟖𝟑 𝒌𝑵, 𝑪
∑ 𝐹𝑦 = 0
3
1
−𝐹𝐹𝐺 − 𝐹𝐹𝐻 (
) + 𝐹𝐷𝐹 (
)=0
3√2
√10
3
1
−𝐹𝐹𝐺 − (−127.2792)(
) + (−94.8683)(
)=0
3√2
√10
𝑭𝑭𝑮 = 𝟔𝟎 𝒌𝑵, 𝑻
@joint G
∑ 𝐹𝑦 = 0
4
𝐹𝐷𝐺 ( ) + 𝐹𝐹𝐺 − 90 = 0
5
4
𝐹𝐷𝐺 ( ) + 60 − 90 = 0
5
𝑭𝑫𝑮 = 𝟑𝟕. 𝟓𝒌𝑵, 𝑻
𝐹𝑥 = 0
3
−𝐹𝐷𝐺 ( ) − 𝐹𝐸𝐺 + 𝐹𝐺𝐻 = 0
5
3
−(37.5)( ) − 𝐹𝐸𝐺 + (90) = 0
5
𝑭𝑬𝑮 = 𝟔𝟕. 𝟓𝒌𝑵, 𝑻
@joint E
∑ 𝐹𝑦 = 0
−30 + 𝐹𝐷𝐸 = 0
𝑭𝑫𝑬 = 𝟑𝟎𝒌𝑵, 𝑻
APPENDICES
SOLUTIONS
∑ 𝐹𝑥 = 0
𝐴𝑥 = 0
2.
𝑃 + 𝑀𝐺 = −5000(9) − 8000(36) − 4000(45) − 𝐴𝑥 + 𝐴𝑦 (54) = 0
𝑨𝒚 = 𝟗𝟓𝟎𝟎𝒍𝒃
∑ 𝐹𝑥 = 0
9
9
𝐴𝑥 + 𝐶𝐷 + 𝐶𝐽(15) + 𝐾𝑇 = 0 → 𝐶𝐷 = −𝐶𝐽(15) − 𝐾𝑇
∑ 𝐹𝑦 = 0
12
𝐴𝑦 − 4000 − 8000 − 𝐶𝐽(15) = 0
12
9500 − 4000 − 8000 + 𝐶𝐽(15) = 0
𝐶𝐽 = −3125𝑙𝑏
𝑪𝑱 = 𝟑𝟏𝟐𝟓𝒍𝒃, 𝑪
𝑃 + 𝑀𝐷 = 0
12
𝐴𝑦 (24) − 4000(18) − 8000(9) − 𝐶𝐽(15)(9) − 𝐾𝑇(12) = 0
12
9500(27) − 4000(18) − 8000(9) + (3125)(15)(9) − 𝐾𝑇(12) = 0
𝑲𝑻 = 𝟏𝟏𝟐𝟓𝟎𝒍𝒃, 𝑻
9
𝐶𝐷 = −𝐶𝐽( ) − 𝐾𝑇
15
9
𝐶𝐷 = −(3125)(15) − (11250)
𝐶𝐷 = −9375𝑙𝑏
𝑪𝑫 = 𝟗𝟑𝟕𝟓𝒍𝒃, 𝑪
3.
W = 75(9.31)
W = 735.75N = T
∑ 𝐹𝑥 = 0
𝐷𝑥 − 𝑇 = 0
𝐷𝑥 − 735.75 = 0
𝑫𝒙 = 𝟕𝟑𝟓. 𝟕𝟓𝑵
APPENDICES
SOLUTIONS
∑ 𝐹𝑦 = 0
𝐷𝑦 − 𝑇 = 0
𝐷𝑦 − 735.75 = 0
𝐷𝑦 = 735.75𝑁
∑ 𝐹𝑥 = 0
4
5
𝐶𝑥 − 𝐴𝐵( ) + 735.75 = 0 → 𝒆𝒒. 𝟏
∑ 𝐹𝑦 = 0
3
𝐶𝑦 − 𝐴𝐵(5) + 𝐷𝑦 = 0
3
𝐶𝑦 − 𝐴𝐵(5) + 735.75 = 0 → 𝒆𝒒. 𝟐
𝑷 + 𝑀𝑐 = 0
4
3
−𝐴𝐵( )(0) + ( )(2) + 𝐷𝑥 (0) − 𝐷𝑦 (0.5 +
5
5
3
𝐴𝐵(5)(2) − 735.75(0.5 + 2) = 0
2) = 0
𝑨𝑩 = 𝟏𝟓𝟑𝟐. 𝟖𝟏𝟐𝟓𝑵
Substitute
4
𝐶𝑥 − (1532.8125)( ) + 735.75 = 0
5
𝑪𝒙 = 𝟒𝟗𝟎. 𝟓𝑵
3
𝐶𝑦 − (1532.8125)( ) + 735.75 = 0
5
𝑪𝒚 = 𝟏𝟖𝟑. 𝟗𝟑𝟕𝟓𝑵
4
4
𝐵𝑥 = 𝐴𝐵( ) = 1532.8125( )
5
5
𝑩𝒙 = 𝟏𝟐𝟐𝟔. 𝟐𝟓𝑵
3
3
𝐵𝑦 = 𝐴𝐵( ) = 1532.8125( )
5
5
𝑩𝒚 = 𝟗𝟏𝟗. 𝟔𝟖𝟕𝟓𝑵
APPENDICES
SOLUTIONS
ACTIVITY 5
ANSWER KEY:
1.
Step 1: Draw the FBD and list the Given
𝑚 = 2000𝑘𝑔
𝑤 = 𝑚𝑔 = (2000)(9.81)
𝜇 = 0.3
Step 2: Get the Summation of Forces in X and Y, and Moments
∑ 𝐹𝑥 = 0 = 𝐹𝐵 − 𝐹 cos(30) (eq. 1)
∑ 𝐹𝑦 = 𝑁𝐴 + 𝑁𝐵 + 𝐹 sin(30) − 2000(9.81) (eq. 2)
∑ 𝑀𝐴 = 0 = 𝐹 cos 3 0(0.3) − 𝐹 sin 3 0(0.75) + 𝑁𝐵 (2.6) − 19620 (eq. 3)
Friction Force
𝐹𝐵 = 𝜇𝐵 𝑁𝐵 = (0.3)𝑁𝐵 (eq. 4)
From equations 1 and 4
0.3𝑁𝐵 − 𝐹 cos 3 0 = 0
𝑁𝐵 =
𝐹 cos 30
0.3
(eq. 5)
Substitute eq. 5 to eq. 1
𝐹 cos 3 0(0.3) − 𝐹 sin 3 0(0.75) +
𝐹 cos 3 0
(2.6) = 19620 𝑘𝑁
0.3
𝐹(0.2598 − 0.375 + 7.5055) = 19260
APPENDICES
SOLUTIONS
𝐹=
19260
7.3903
𝐹 = 2654.8313 𝑘𝑁
𝐹 = 2.6548 𝑘𝑁
2.
Given:
200𝑙𝑏 = 𝜇𝑁 = 0.3(450)
𝐹=𝜇
𝐹 = 120𝑙𝑏
𝑃 = 0.6(200)
Solution and Analysis:
200 < 135 + 120; 200 < 235
Therefore, he cannot move the crate.
3.
Step 1: Draw the FBD for the object and the wedge.
APPENDICES
SOLUTIONS
For the wedge:
Step 2: Summation of Fx and Fy
∑ 𝐹𝑥 = 0
𝐹 cos 1 5 + 𝑁 sin 1 5 − 𝑁𝐷
0.4 cos 1 5 + 𝑁 sin 1 5 = 𝑁𝐷
0.3863𝑁 + 0.259𝑁 = 𝑁𝐷
𝑁𝐷 = 0.6453𝑁
∑ 𝐹𝑦 = 0
𝑁 cos 1 5 − 𝐹 sin 1 5 − 𝐹𝐷 − 3000 = 0
𝑁 cos 1 5 − 𝜇𝑠 𝑁(sin 1 5) − 𝜇𝑠 𝑁𝐷 − 3000 = 0
𝑁 cos 1 5 − 0.4𝑁 sin 1 5 − 0.3𝑁𝐷 − 3000 = 0
0.8625 − 0.3𝑁𝐷 − 3000 = 0
𝑁 = 4485𝑙𝑏
𝑁𝐷 = 0.6453(4485)
𝑁𝐷 = 2894.17𝑙𝑏
For the object:
APPENDICES
SOLUTIONS
∑ 𝐹𝑦 = 0
𝑁𝑐 + 𝐹 sin 1 5 − 𝑁 cos 1 5 = 0
𝑁𝑐 + 0.4(4485) sin 1 5 − (4485) cos 1 5 = 0
𝑁𝑐 + 464.32 − 4332.18 = 0
𝑁𝑐 = 2867.86 𝑙𝑏
∑ 𝐹𝑥 = 0
𝑃 − 𝐹𝑐 − 𝑁 sin 1 5 − 𝐹 cos 1 5 = 0
𝑃 − [(0.3)(3867.85)] − 4.485 sin 1 5 − [(0.4)(4485)] cos 1 5 = 0
𝑃 − 1160.36 − 1160.80 − 1732.87 = 0
𝑃 = 4054.03𝑙𝑏
APPENDICES
SOLUTIONS
4.
𝜇 = 0.2
2452.5 = 𝑇1 𝑒(0.2𝑥)
𝛽 = 130°
2𝑥 − 360
𝑇2 = 2452.5𝑁
𝑇1 = 𝐹
𝛽 = 540°
𝛽=
2𝑥(540)
= 3𝑥
360
𝑇2 = 2452.5𝑁
APPENDICES
SOLUTIONS
𝑇1 = 𝐹
𝜇 = 0.2
𝑇2 = 𝑇1 𝑒 𝜇𝛽
2452.5 = 𝑇1 𝑒 (0.2)(3𝑥)
𝑇1 = 372.38𝑁
ACTIVITY 6
ANSWER KEY:
1.
𝑦 = 2𝑥 3
𝑑𝑦
= 6𝑥 2
𝑑𝑥
𝑑𝐿 = √1 +
𝑑𝑦 2
𝑑𝑦
𝑑𝑥
𝑑𝐿 = √1 + (6𝑥 2 )2 𝑑𝑥
ȳ=
ȳ=
∫ ỹ𝑑𝐿
∫ 𝑑𝐿
∫ (√1 + (6𝑥 2 )2 𝑑𝑥) (2𝑥 3 )
∫ √1 + (6𝑥 2 )2 𝑑𝑥
ȳ = 0.8568𝑚 (Answer)
2.
1
𝑦1 = 100 𝑥 2
𝑦2 = 𝑥
𝑑𝐴 = (𝑦2 − 𝑦1 ) dx
1
𝑑𝐴 = (𝑥 − 100 𝑥 2 ) dx
𝑥̃ = 𝑥
𝑥̅ =
∫ 𝑥̃𝑑𝐴
∫ 𝑑𝐴
𝑦̃ =
𝑦2 −𝑦1
2
𝑥̅ =
∫ 𝑦̃𝑑𝐴
∫ 𝑑𝐴
APPENDICES
100
𝑥̅ =
SOLUTIONS
1 2
𝑥 )𝑑𝑥
100
100
1 2
∫0 (𝑥−100𝑥 )𝑑𝑥
∫0
𝑥(𝑥−
𝑦̅ =
𝑦̅ =
̅ = 𝟓𝟎 𝒎𝒎
𝒙
100 𝑦2 −𝑦1
1
(
)(𝑥− 𝑥 2 )𝑑𝑥
2
100
100
1
∫0 (𝑥−100𝑥 2 )𝑑𝑥
1 2
𝑥
100 𝑥+
1
)(𝑥− 𝑥 2 )𝑑𝑥
∫0 ( 100
2
100
100
1
∫0 (𝑥−100𝑥 2 )𝑑𝑥
∫0
̅ = 𝟒𝟎 𝒎𝒎
𝒙
3.
2
2
𝑥̃ = 𝑏 = (1.5) + 1.5 = 2.5 𝑖𝑛
3
3
2
2
𝑦̃ = ℎ = (1.5) + 1.5 = 2.5 𝑖𝑛
3
3
SOLUTION:
Segment
1
2
3
∑
Thus,
𝑨(𝒊𝒏)𝟐
̃(𝒊𝒏)
𝒙
̃(𝒊𝒏)
𝒚
̃𝑨(𝒊𝒏)𝟑
𝒙
̃𝑨(𝒊𝒏)𝟑
𝒚
3(3)] = 9
1.5
1.5
13.5
13.5
2
𝜋
2
𝜋
−1.125
−1.125
2.5
2.5
−2.8125
−2.8125
9.5625
9.5625
𝜋
− (1.5)2
4
9
=− 𝜋
16
1
− (1.5)(1.5)
2
9
=− 𝜋
8
6.1079
APPENDICES
SOLUTIONS
𝟗.𝟓𝟔𝟐𝟓 𝒊𝒏𝟑
̅=
𝒙
̃𝑨
∑𝒙
∑𝑨
= 𝟔.𝟏𝟎𝟕𝟗 𝒊𝒏𝟐 = 𝟏. 𝟓𝟔𝟓𝟔 𝐢𝐧
̅=
𝒚
̃𝑨
∑𝒚
∑𝑨
= 𝟔.𝟏𝟎𝟕𝟗 𝒊𝒏𝟐 = 𝟏. 𝟓𝟔𝟓𝟔 𝐢𝐧
𝟗.𝟓𝟔𝟐𝟓 𝒊𝒏𝟑
4.
𝑉 = 𝜋𝛿 2 ℎ + 2𝜋𝛿 3
𝑉 = 𝜋 𝑥 102 𝑥 86 +
2
𝜋(103 )
3
𝑉 = 27227.1363 𝑓𝑡 3 (Answer)
ACTIVITY 7
Answer key
5.
SOLUTION:
𝑑𝐴 = 2𝑥𝑑𝑦
𝑦=
1 2
𝑥
50
50𝑦 = 𝑥 2
√50𝑦 = 𝑥
𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴
200
𝑦 2 (2𝑥𝑑𝑦)
=∫
0
200
𝑦 2 (2)(√50𝑦)𝑑𝑦
=∫
0
200
= 2∫
𝑦 2 (√50𝑦)𝑑𝑦
0
𝑰𝒙 = 𝟒𝟓𝟕, 𝟏𝟒𝟐, 𝟖𝟓𝟕𝒎𝒎𝟒
APPENDICES
SOLUTIONS
6.
SOLUTION:
𝑑𝐴 = (𝑦2 − 𝑦1 )𝑑𝑥
𝑦2 = √2𝑥
𝑦1 = 𝑥
𝐼𝑦 = ∫ 𝑥 2 𝑑𝐴
2
= ∫ 𝑥 2 (𝑦2 − 𝑦1 )𝑑𝑥
0
2
= ∫ 𝑥 2 (√2𝑥 − 𝑥)𝑑𝑥
0
𝑰𝒚 = 𝟎. 𝟓𝟕𝟏𝟒𝒎𝟒
7.
SOLUTION:
Segment
𝑨𝒊 (𝒊𝒏)𝟐
1
1(8)
4
(𝒅𝒚)𝒊 (𝒊𝒏)
2
8(1)
0.5
3
1(3)
1.5
Thus,
𝑰𝒙 = 𝟏𝟖𝟐. 𝟑𝟑𝟑𝟑 𝒊𝒏𝟒
4.
SOLUTION:
𝑰𝒙′ (𝒊𝒏)𝟒
(𝑨𝒅𝟐𝒚 )𝒊 (𝒊𝒏)𝟒
1
(1)(8)3
12
128
=
3
1
(8)(1)3
12
2
=
3
1
(1)(3)3
12
9
=
4
1(8)(4)2
= 128
(𝑰𝒙)𝒊 (𝒊𝒏)𝟒
512
3
8(1)(0.5)2
=2
8
3
1(3)(1.5)2
= 6.75
9
APPENDICES
Segment
1
SOLUTIONS
𝑨𝒊 (𝒊𝒏)𝟐
6(6)
3
(𝒅𝒚)𝒊 (𝒊𝒏)
2
1
(6)(3)
2
1
(6) = 2
3
3
1
(6)(9)
2
1
(6) = 2
3
4
−𝜋(2)2
3
Thus,
𝑰𝒚 = 𝟓𝟐𝟐. 𝟑𝟑𝟔𝟑 𝒊𝒏𝟒 s
𝑰𝒙′ (𝒊𝒏)𝟒
(𝑨𝒅𝟐𝒚 )𝒊 (𝒊𝒏)𝟒
1
(6)(6)3
12
= 108
1
(3)(6)3
36
= 18
1
(9)(6)3
36
= 54
1
− 𝜋(2)4
4
= −4𝜋
6(6)(3)2
= 324
1
(6)(3)(2)2
2
= 36
1
(6)(9)(2)2
2
= 108
−𝜋(2)2 (3)2
= −36𝜋
(𝑰𝒙)𝒊 (𝒊𝒏)𝟒
432
54
162
−40𝜋
APPENDICES
SOLUTIONS
ACTIVITY 8
Answer Key
1.
𝑎=
𝑑𝑣
𝑑𝑠
𝑠 = 12𝑡 − 𝑡 3 − 21
𝑣 = 𝑑𝑡
𝑑𝑡
𝑑
𝑆1
𝑡
𝑎 = 𝑑𝑡 (12 − 3𝑡 2 )
∫𝑆𝑜 𝑑𝑠 = ∫𝑡𝑜 𝑣𝑑𝑡
𝑎 = −6𝑡
∫−10 𝑑𝑠 = ∫𝑡𝑜 𝑣𝑑𝑡
𝑆
when t=0s
𝑡
𝑠 = 12(0) − (0)3 − 21
𝑠 + 10 = 12𝑡 − 𝑡 3 − (12(1) − (1)3 )
𝑎 = −6(4)
𝒎
𝑠 = 12𝑡 − 𝑡 3 − 21
𝑎 = −𝟐𝟒 𝒔𝟐
𝑠 = −𝟐𝟏𝒎
when t = 10s
𝑠 = 12(10) − (10)3 − 21
𝑠 = −𝟗𝟎𝟏 𝒎
∆𝑠 = −901 𝑚 − (−21 𝑚)
∆𝑠 = −𝟖𝟖𝟎 𝒎
2.
𝑠𝑎 = 𝑠𝑏
𝑠𝑜 + 𝑣𝑜𝑡 +
30 + 5𝑡 +
1
1
𝑎𝑐 𝑡 2 = 𝑠𝑜 + 𝑣𝑜𝑡 + 𝑎𝑐 𝑡 2
2
2
1
1
(−9.81)𝑡 2 = 0 + 20𝑡 + (9.81)22
2
2
𝑡 = 2𝑠
1
𝑠𝑎 = 𝑠𝑜 + 𝑣𝑜𝑡 + 2 𝑎𝑐 𝑡 2
1
= 30 + 5(2) + 2 (−9.81)(2)2
𝑠𝑎 = 𝟐𝟎. 𝟑𝟖 𝒎
1
𝑠𝑏 = 𝑠𝑜 + 𝑣𝑜𝑡 + 2 𝑎𝑐 𝑡 2
1
= 0 + 20(2) + 2 (−9.81)(2)2
𝑠𝑏 = 𝟐𝟎. 𝟑𝟖 𝒎
APPENDICES
SOLUTIONS
3.
Product rule
Velocity
𝑣𝑥 =
Acceleration
𝑑𝑥
; 𝑥 = 8𝑡
𝑑𝑡
𝑑
; 𝑣𝑥 = 8
= 𝑑𝑡 (8)
8𝑓𝑡
𝑓𝑡
𝑎𝑥 = 0 𝑠 2
𝑠
𝑎𝑦 =
𝑑
𝑑𝑡
𝑑
= 𝑑𝑡 (8𝑡)
𝑣𝑥 =
𝑑𝑣𝑥
𝑎𝑥 =
𝑑𝑣𝑦
1
; 𝑣𝑦 = 5 𝑥𝑣𝑥
𝑑𝑡
1
= 𝑑𝑡 (5 𝑥𝑣𝑥
Chain rule :
𝑣𝑦 =
𝑑
𝑑𝑦
𝑑𝑡
𝑥2
; 𝑦 = 10
𝑥2
𝑑𝑡
𝑑𝑦
𝑑𝑥
𝑑𝑥
𝑓𝑡
1
𝑣𝑦 = 5 × 𝑣𝑥
1
𝑓𝑡
𝑣 = 𝟐𝟔. 𝟖𝟐𝟎𝟗
𝒇𝒕
𝒔
1
= (16)(8)
𝑓𝑡
𝑠
𝑣𝑦
∅ = tan−1(𝑣 )
1
25.6
∅ = 𝟕𝟐. 𝟔𝟒𝟔𝟎°
8
)
𝑑𝑡
𝑑𝑣𝑥
𝑑𝑡
1
1
1
= 5 (8)2 + 5 (16)(0)
𝑓𝑡
𝑠2
𝑎 = √(𝑎𝑥 )2 + (𝑎𝑦 )2
𝑥
∅ = tan−1(
1
𝑑𝑣𝑥
𝑎𝑦 = 5 (𝑣𝑥)2 + 5 (16)(0)
𝑎𝑦 = 12.8
5
1
= 5 (𝑣𝑥)(𝑣𝑥) + 5 ×
𝑣 = √(8 𝑠 )2 + (25.6 𝑠 )2
= 10 ( 𝑑𝑡 )
𝒗𝒚 = 25.6
1 𝑑𝑥
= 5 ( 𝑑𝑡 ) 𝑣𝑥 + 5 (𝑥)
= 𝑑𝑥 × 𝑑𝑡
𝑣 = √(𝑣𝑥 )2 + (𝑣𝑦 )2
= 𝑑𝑡 (10)
2
𝑑𝑦
𝑎 = √0)2 + (12.8)2
𝒇𝒕
𝑎 = 𝟏𝟐. 𝟖 𝒔𝟐
𝑎𝑦
∅ = tan−1(𝑎 )
𝑥
∅ = tan−1(
∅ = 𝟗𝟎°
12.8
0
)
APPENDICES
SOLUTIONS
4.
Vertical Motion:
1
𝑦𝑏 = 𝑦𝑎 + (𝑣𝑜 )𝑦𝑡 − 𝑔𝑡 2
2
−1𝑚 = 0 + (𝑣𝑎 sin 30°)(1.5𝑠)
1
− (9.81)(1.5𝑠)2
2
𝒗𝒂 = 𝟏𝟑. 𝟑𝟖𝟏𝟕
𝒎
𝒔
Horizontal Motion:
Consider maximum height:
𝑥𝑏 = 𝑥𝑎 + (𝑣𝑎 )𝑥𝑡
𝑣𝑐2 = (𝑣𝑎 )2𝑦 − 2𝑔(𝑦𝑐 − 𝑦𝑎)
𝑚
𝑅 = 0 + (13.3817 cos 30° 𝑠 ) (1.5 𝑠)
02 = (13.3817 𝑠𝑖𝑛30°)2 − 2(9.81)((ℎ −
1) − 0)
𝑅 = 𝟏𝟕. 𝟑𝟖𝟑𝟑 𝒎
ℎ = 𝟑. 𝟐𝟖𝟏𝟕 𝒎
APPENDICES
SOLUTIONS
ACTIVITY 9
ANSWER KEY
1.
Σ𝐹𝑦 = 𝑚𝑎𝑦
𝑁 − 10 = 0
𝑁 = 10𝑙𝑏
𝑎=
𝑑𝑣
Σ𝐹𝑥 = 𝑚𝑎𝑥
8𝑡 2 − 0.2(10) =
10
(𝑎)
32.2
𝑎 = 3.22(8𝑡 2 − 2)
𝑑𝑠
𝑣 = 𝑑𝑡
𝑑𝑡
𝑑𝑣 = 𝑎𝑑𝑡
𝑑𝑠 = 𝑑𝑡
𝑣
∫0 𝑑𝑠 = ∫0 ( 75 𝑡 3 − 6.44𝑡 + 4) 𝑑𝑡
𝑡
∫4 𝑑𝑣 = ∫0 3.22(8𝑡 2 ) 𝑑𝑡
8
𝑣 − 4 = 3,22 (3 𝑡 3 − 2𝑡)
𝑣=
644 3
𝑡
75
𝑠
𝑠=
𝑡 644
161 4
𝑡
75
− 6.44𝑡 + 4
At s = 30ft,
𝑠=
30 =
161 4
𝑡
75
161 4
𝑡
75
− 3.22𝑡 2 + 4𝑡
At t = 2.0089s,
− 3.22𝑡 2 + 4𝑡
− 3.22𝑡 2 + 4𝑡
𝒕 = 𝟐. 𝟎𝟎𝟖𝟗𝒔
2.
𝑣=
𝑣=
644
75
644 3
𝑡
75
− 6.44𝑡 + 4
(2.0089)3 − 6.44(2.0089) + 4
𝒗 = 𝟔𝟎. 𝟔𝟕𝟐𝟔 𝒇𝒕/𝒔
Horizontal Motion
𝑥𝑐 = 𝑥𝐵 + (𝑣𝐵 )𝑥 𝑡
4
𝑑 = 0 + 33.0874(5)(1,36925)
𝒅 = 𝟑𝟔. 𝟐𝟒𝟐𝟒 𝒇𝒕
Consider the principle of work and energy:
𝑇1 + ΣU1−2 = 𝑇2
Vertical Motion
1
𝑣𝑐 = 𝑣𝐵 + (𝑉𝐵 )𝑦 𝑡 − 2 𝑔𝑡 2
APPENDICES
1
2
1
SOLUTIONS
1
3
𝑚𝑣12 + 2 𝑘𝑠 2 − 𝑊𝑠 = 2 𝑚𝑣22
1
0 = 3 + 33.0878 (5) 𝑡 − 2 (32.2)𝑡 2
𝒕=
𝟏. 𝟑𝟔𝟗𝟐 𝒔
1 10
1
10
1 10 2
(
)(0)2 + (100)(2)2 − (
)(32.2)(5 − 2) = (
)𝑣
2 32,2
2
32,2
2 32,2 𝐵
𝑉𝐵 = 33.0878 𝑓𝑡/𝑠
3.
Σ𝐹𝑥 = 𝑚𝑎𝑥
𝑃𝑜 = 𝐹 ⋅ 𝑣
𝐹 = 2000(9.81)𝑠𝑖𝑛7° = 2000(0)
= 2391.0765𝑁(
𝐹 = 2391.0765 𝑁
𝑣 = 100
𝑃𝑖𝑛𝑝𝑢𝑡 =
𝑘𝑚 1000
ℎ
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝜀
4.
9
𝑚/𝑠)
𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 66,418.7922𝑊
1ℎ
( 1𝑘𝑚 ) (3600𝑠) =
=
250
25
9
𝑚/𝑠
66,418.7922𝑊
0.65
𝑷𝒊𝒏𝒑𝒖𝒕 = 𝟏𝟎𝟐, 𝟏𝟖𝟐. 𝟕𝟓𝟕𝟐 𝑾
Σ𝐹𝑥 = 𝑚𝑎𝑥
𝑣 = 𝑣𝑜 + 𝑎𝑡
20000 = 1000(𝑎)
𝑣 = 0 + 200𝑚/𝑠 2 (𝑡)
𝑎 = 20𝑚/𝑠 2
𝑣=0
𝑃 =𝐹⋅𝑣
= 20000(200𝑡)
𝑷 = (𝟒𝟎𝟎, 𝟎𝟎𝟎𝒕) 𝑾
5.
Conservation of linear momentum
𝑚𝑏 𝑣𝑏 + 𝑚𝐵 𝑣𝐵 = (𝑚𝑏 + 𝑚𝐵) 𝑣
0.02(400) + 2(0) = (0.02 + 2)𝑣
𝑣 = 3.9604𝑚/𝑠
Principle of impulse and momentum along vertical
Principle of impulse and momentum along horizontal
APPENDICES
SOLUTIONS
𝑡
𝑡
𝑚(𝑣𝑦 )1 + Σ ∫𝑡 2 𝐹𝑦𝑑𝑡 = 𝑚(𝑣𝑦 )2
𝑚(𝑣𝑥 )1 + Σ ∫𝑡 2 𝐹𝑥𝑑𝑡 = 𝑚(𝑣𝑥 )2
2(0) + 𝑁(𝑡) − 2.02(9.81)𝑡 = 2,02(0)
2.02(3.9604) + −0.2(19.8162𝑡) = 2.02𝑉2
𝑁 = 19.8162𝑡 𝑁
3.9604 − 1.962𝑡 = 𝑣2 = 0
1
1
𝑑𝑠
𝑡 = 2.0186𝑠
𝑠 = 3.9604𝑡 − 0.981𝑡 2
𝑣 = 𝑑𝑡
𝑠
𝑡
𝑠
𝑡
𝑠 = 3.9604(2.0186) − 0.981(2.0186)2
∫0 𝑑𝑠 = ∫0 𝑣𝑑𝑡
∫0 𝑑𝑠 = ∫0 (3.9604 − 1.962𝑡)𝑑𝑡
𝑠 = 3.9604𝑡 −
𝒔 = 𝟑. 𝟗𝟗𝟕𝟏 𝒎
1.962𝑡 2
2
6.
Conservation of linear momentum:
2(1−𝑉𝐴2 )
𝑚𝐴 𝑉𝐴 + 𝑚𝐵 𝑉𝐵1 = 𝑚𝐴 𝑉𝐴2 + 𝑚𝐵 𝑉𝐵2
2(5) + 4(−2) = 2𝑉𝐴2 + 4𝑉𝐵2
2 − 2𝑉𝐴2 = 4𝑉𝐵2
2(1 − 𝑉𝐴2 ) = 4𝑉𝐵2
Coefficient of restitution:
𝑣
−𝑣
𝜀 = 𝑣𝐵2−𝑣𝐴2
𝐴1
0.4 =
0.4 =
𝒗𝑨𝟐 = −𝟏. 𝟓𝟑𝟑𝟑
𝐵1
𝑣𝐵2 −𝑣𝐴2
5−(−2)
1
2
[ (1−𝑉𝐴2 )]−𝑉𝐴2
7
𝒎
𝒔
1
=
𝟏.𝟓𝟑𝟑𝟑𝒎
𝒔
𝑣𝐵2 = 2 (1 − (−1.5333))
𝒗𝑩𝟐 = 𝟏. 𝟐𝟔𝟔𝟕
𝒎
𝒔
𝒕𝒐 𝒕𝒉𝒆 𝒍𝒆𝒇𝒕
4
1
2
= 𝑉𝐵2
((1 − 𝑉𝐴2 ) = 𝑉𝐵2