ENGINEERING MECHANICS STATICS SLIDES

this classic text continues to provide the same high quality material seen in previous editions. The text is extensively rewritten with updated prose for content clarity, superb new problems in new application areas, outstanding instruction on drawing free body diagrams, and new electronic supplements to assist readers. Furthermore, this edition offers more Webbased problem solving to practice solving problems, with immediate feedback; computational mechanics booklets offer flexibility in introducing

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HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

- Overall Educational Objectives/Learning Outcomes

At the end of the course a successful student should be able to

• utilize standard engineering approaches to model

mechanical systems

• utilize a variety of techniques to analyze rigid-body

equilibrium problems, select an appropriate technique for a

particular analysis, and evaluate the quality of results

- Student responsibility

• Student is expected that you will spend at least 8 hours/week

studying this course (4 lecture + 4 assignment)

• University regulations indicate that if students attend less

than 80% of scheduled classes they may be refused final

assessment

Engineering Mechanics – Statics 0.03 Introduction

- Course Assessment Policy

• One comprehensive final exam: 50%

• In-class quizzes, class participation and learning attitude: 20%

- Textbooks R.C Hibbeler,

Engineering Mechanics-Statics and Dynamics,

12Ed., Prentice Hall, 2010

- Grade scale: 100

- Course Outline General Principles - Force Vectors - Equilibrium of a Particle - Force System Resultants - Equilibrium of a Rigid Body - Structural Analysis - Internal Forces - Friction - Center of Gravity and Centroid - Moments of Inertia - Virtual Work HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 0.04 Introduction

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• To review the principles for applying the SI system of units

• To examine the standard procedures for performing numerical calculations

• To present a general guide for solving problems

§1 Mechanics

- Mechanics: the study how body react to the forces acting on

them

- Branches of mechanics

• Statics deals with the equilibrium of bodies, that is, those that

are either at rest or move with a constant velocity

• Dynamics is concerned with the accelerated motion of bodies

Engineering Mechanics – Statics 1.03 General Principles

§2 Fundamental Concepts

- Basic Quantities

• Length, 𝑚: used to locate the position of a point in space and

thereby describe the size of a physical system

• Time, 𝑠: although the principles of statics are time

independent, this quantity plays an important role

in the study of dynamics

• Mass, 𝑘𝑔: mass is a measure of a quantity of matter

• Force, 𝑁: considered as a push or pull exerted by one body

- Idealizations

• Particle: particle has a mass, but it size can be

neglected

• Rigid body: a rigid body can be considered as a

combination of a large number of particles

• Concentrated force: represents the effect of a loading which is

assumed to act at a point on a body

Three forces act

on the hook at 𝐴

Since these

forces all meet at

any force analysis,

the hook to be

particle

Steel is a common that does not deform load We can consider this railroad wheel to be a rigid the concentrated force of the rail

• Second Law A particle acted upon by an unbalanced

force experiences an acceleration that has the same direction as the force and a magnitude that is directly proportional to the force: 𝐹 = 𝑚𝑎

• Third Law The mutual forces of action and reaction

between two particles are equal, opposite and collinear

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 1.06 General Principles

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- Newton’s Law of Gravitational Attraction

Gravitational attraction between any two particles

𝐹 = 𝐺𝑚1𝑚2

𝑟2𝐹: force of gravitational between the two particles, 𝑁

𝐺: universal constant of gravitation, 𝑚3/𝑘𝑔𝑠2

From experiment: 𝐺 = 66.73 × 10−12𝑚3/𝑘𝑔𝑠2

𝑚𝑖: mass of each of the two particles, 𝑘𝑔

𝑟: distance between the two particles, 𝑚

- Mass

Mass is an absolute property of a body

The mass provides a measure of the resistance of a body to a

change in velocity, as defined by Newton's second law of

motion 𝑚 = 𝐹/𝑎

- WeighWeight refers to the gravitational attraction of the earth on a

body or quantity of mass

𝑊 = 𝐺𝑚𝑀𝑒

𝑟2 ≡ 𝑚𝑔, 𝑔 = 9.8066𝑚/𝑠2

The astronaut is weightless, for all practical purposes, since she is far removed from the

gravitational field of the earth

§3 Units of Measurement

- SI units: The International System of units abbreviated SI after

the French “Système International d’Unités”

- U.S Customary (FPS)

- Rules for Use: when performing calculations, represent the

numbers in terms of their base or derived units by converting

§5 Numerical Calculations

- Dimensional Homogeneity: dimensions have to be the same

on both sides of the equal sign, (e.g 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑠𝑝𝑒𝑒𝑑 × 𝑡𝑖𝑚𝑒)

- Significant Figures: the number of significant figures contained

in any number determines the accuracy of the number ⟹ use

an appropriate number of significant figures E.g 23400 might have three (234), four (2340), or five (23400) significant figures To avoid these ambiguities, rewrite

23 400 ⟶ 23.400 × 103 ⟶ 23.4 × 103

- Rounding Off Numbers: be consistent when rounding off

• greater than 5, round up: 3528 ⟶ 3530

• smaller than 5, round down: 0.03521 ⟶ 0.0352

• equal to 5 + the digit preceding the 5 is an even number: 75.25 ⟶ 75.2 Engineering Mechanics – Statics 1.12 General Principles

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§6 General Procedure for Analysis

Problem Solving Strategy

- Interpret: read carefully and determine what is given and what

is to be found/delivered Ask if not clear If necessary,

make assumptions and indicate them

- Plan: think about major steps (or road map) that you will

take to solve a given problem Think of

alternative/creative solutions and choose the best

one

- Execute: carry out your steps Use appropriate diagrams and

equations Estimate your answers, avoid simple

calculation mistakes, Reflect on/revise your work

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02 Force Vectors

Engineering Mechanics – Statics 2.01 Force Vectors

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.02 Force Vectors

§1 Scalars and Vectors

- Scalar: a scalar is any positive or negative physical quantity

that can be completely specified by its magnitude

Examples: length, mass, and time

- Vector: a vector is any physical quantity that requires both a

magnitude and a direction for its complete description

Examples: force, position, and moment

A vector is shown graphically by an arrow

• the length of the arrow: the magnitude

§3 Vector Addition Forces

- Experimental evidence has shown that a force is a vector quantity since it has a specified magnitude, direction, and sense and it adds according to the parallelogram law

- Finding a resultant force The two component forces 𝐹 1 and 𝐹 2 acting on the pin can be added together to form the resultant force 𝐹 𝑅= 𝐹 1+ 𝐹 2Engineering Mechanics – Statics 2.06 Force Vectors

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- Finding the components of a force

To resolve a force into two components in order to study its

pulling or pushing effect in two specific directions

- Addition of several forces

To resolve a force into two components in order to study its

pulling or pushing effect in two specific directions 𝑂

- Example 2.1 The screw eye is subjected to two forces, 𝐹 1 and

𝐹 2 Determine the magnitude and direction of the resultant force

- Example 2.2 Resolve the horizontal 600𝑁 force into components acting along the 𝑢 and axes and determine the

magnitudes of these components Solution

𝐹𝑢𝑠𝑖𝑛1200= 600

𝑠𝑖𝑛300 ⟹ 𝐹𝑢= 600𝑠𝑖𝑛120

0𝑠𝑖𝑛300 = 1039 𝑁

𝐹𝑣𝑠𝑖𝑛300= 600𝑠𝑖𝑛300 ⟹ 𝐹𝑣= 600𝑠𝑖𝑛30

0𝑠𝑖𝑛300= 600 𝑁 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.10 Force Vectors

- Example 2.3 Determine the magnitude of the component

force 𝐹 and the magnitude of the resultant force 𝐹 𝑅 if 𝐹 𝑅 is

directed along the positive 𝑦 axis

𝐹𝑅

𝑠𝑖𝑛750= 200

𝑠𝑖𝑛450⟹ 𝐹 = 200𝑠𝑖𝑛75

0𝑠𝑖𝑛450= 273 𝑁 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

- Example 2.4 It is required that the resultant force acting on the eyebolt be directed along the positive 𝑥 axis and that

𝐹 2 have a minimum magnitude Determine this magnitude, the angle 𝜃, and the corresponding resultant force

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Fundamental Problems

- F2.1: Determine the magnitude of the resultant force acting on the

screw eye and its direction measured clockwise from the 𝑥 axis

- F2.3: Determine the magnitude of the resultant force and its

direction measured counterclockwise from the positive 𝑥 axis

- F2.4: Resolve the 30𝑁 force into components along the 𝑢 and axes, and determine the magnitude of each of these components

- F2.5: The force acts on the frame Resolve this force into

components acting along members 𝐴𝐵 and 𝐴𝐶, and determine

the magnitude of each component

- F2.6: If force 𝑭 is to have a component along the 𝑢 axis of

𝑭𝑢= 6𝑘𝑁, determine the magnitude of 𝑭 and the magnitude of

its component 𝑭𝑣 along the 𝑣 axis Engineering Mechanics – Statics 2.18 Force Vectors

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§4 Addition of a System of Coplanar Forces

𝐹𝑥

𝐹=

𝑎𝑐

𝐹𝑦

𝐹= −

𝑏𝑐

- Coplanar Force Resultants

𝐹 = 𝐹 1+ 𝐹 2+ 𝐹 3

𝐹 = 𝐹1𝑥𝑖 + 𝐹1𝑦𝑗 + (−𝐹2𝑥𝑖 + 𝐹2𝑦𝑗 ) + (𝐹3𝑥𝑖 − 𝐹3𝑦𝑗 )

= 𝐹1𝑥− 𝐹2𝑥+ 𝐹3𝑥 𝑖 + (𝐹1𝑦+ 𝐹2𝑦− 𝐹3𝑦)𝑗

+→ 𝐹𝑅𝑥=𝐹𝑥 + ↑ 𝐹𝑅𝑦=𝐹𝑦

𝐹𝑅= 𝐹𝑅𝑥2 + 𝐹𝑅𝑦2 , 𝜃 = 𝑡𝑎𝑛−1 𝐹𝑅𝑥

𝐹𝑅𝑦HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.20 Force Vectors

- Example 2.5 Determine the 𝑥 and 𝑦 components of 𝐹 1 and

𝐹 2 acting on the boom Express each force as a Cartesian

vector

Solution

§4 Addition of a System of Coplanar Forces Scalar notation

𝐹1𝑥= −200𝑠𝑖𝑛300𝑁 = −100𝑁 = 100𝑁 ←

𝐹1𝑦= 200𝑐𝑜𝑠300𝑁 = 173𝑁 = 173𝑁 ↑

𝐹2𝑥= (12/13) × 260𝑁 = 240𝑁 = 240N → 𝐹2𝑦= −(5/13) × 260𝑁 = −100𝑁 = 100𝑁 ↓

Cartesian vector notation

𝐹 1= −100𝑖 + 173𝑗

𝐹 2= 240𝑖 − 100𝑗

- Example 2.6 The link is subjected to two forces 𝐹 1 and 𝐹 2 Determine the magnitude and direction of the resultant force Solution 1

Scalar Notation

+→ 𝐹𝑅𝑥=𝐹𝑥: 𝐹𝑅𝑥= 600𝑐𝑜𝑠300− 400𝑠𝑖𝑛450= 236.8𝑁 → + ↑ 𝐹𝑅𝑦 =𝐹𝑦: 𝐹𝑅𝑦= 600𝑠𝑖𝑛300+ 400𝑐𝑜𝑠450= 582.8𝑁 ↑

𝐹𝑅= (236.8)2+(582.8)2= 629𝑁

𝜃 = 𝑡𝑎𝑛−1(582.8/236.8) = 67.90 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.24 Force Vectors

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- Example 2.7 The end of the boom 𝑂 is subjected to three

concurrent and coplanar forces Determine the magnitude and direction of the resultant force

Solution

+→ 𝐹𝑅𝑥=𝐹𝑥: 𝐹𝑅𝑥= −400 + 250𝑠𝑖𝑛450− 2004

5= 383.2𝑁 ← + ↑ 𝐹𝑅𝑦 =𝐹𝑦: 𝐹𝑅𝑦= 250𝑐𝑜𝑠450+ 2003

5= 296.8𝑁 ↑

𝐹𝑅= (−383.2)2+(296.8)2= 485𝑁

𝜃 = 𝑡𝑎𝑛−1(296.8/383.2) = 37.80 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.26 Force Vectors

- F2.7: Resolve each force acting on the post into its 𝑥 and 𝑦

components

- F2.8: Determine the magnitude and direction of the resultant force

- F2.9: Determine the magnitude of the resultant force acting

on the corbel and its direction 𝜃 measured counterclockwise

from the 𝑥 axis

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- F2.11: If the magnitude of the resultant force acting on the

bracket is to be 80𝑁 directed along the 𝑢 axis, determine the

magnitude of 𝐹 and its direction 𝜃

- Right-Handed Coordinate System

𝑢𝐴≡𝐴 𝐴

𝐴 = 𝐴𝑢𝐴= 𝐴𝑐𝑜𝑠𝛼𝑖 + 𝐴𝑐𝑜𝑠𝛽𝑗 + 𝐴𝑐𝑜𝑠𝛾𝑘 = 𝐴𝑥𝑖 + 𝐴𝑦𝑗 + 𝐴𝑧𝑘

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§6 Addition of Cartesian Vectors

𝐴 = 𝐴𝑥𝑖 + 𝐴𝑦𝑗 + 𝐴𝑧𝑘

𝐵 = 𝐵𝑥𝑖 + 𝐵𝑦𝑗 + 𝐵𝑧𝑘

𝑅 = 𝐴 + 𝐵

= (𝐴𝑥+ 𝐵𝑥)𝑖 + (𝐴𝑦+ 𝐵𝑦)𝑗 + (𝐴𝑧+ 𝐵𝑧)𝑘

- Example 2.8 Express the force 𝐹 as a Cartesian vector Solution

- Example 2.9 Determine the magnitude and the coordinate

direction angles of the resultant force acting on the ring

Solution

𝐹 𝑅= 𝐹 1+ 𝐹 2

= 50𝑖 − 100𝑗 + 100𝑘+ 60𝑖 + 80𝑘

= 50𝑖 − 40𝑗 + 180𝑘 The magnitude of the sum vector

𝐹 𝑅 = 502+ −402+1802

= 191.0𝑁

The coordinate direction angles

𝛼, 𝛽, 𝛾 are determined from the components of the unit vector acting in the direction of 𝐹 𝑅

𝑢𝐹𝑅=𝐹 𝑅𝐹𝑅

- Example 2.10 Express the force 𝐹 as a

𝐹 = 35.4𝑖 − 35.4𝑗 + 86.6𝑘

⟹ 𝐹 𝑅 = 35.42+ −35.42+86.62

= 100𝑁

|𝐹 | = 100𝑁, 𝐹𝑥= 35.4𝑁, 𝐹𝑦= 35.4𝑁, 𝐹𝑧= 86.6𝑁 The coordinate direction angles of 𝐹 can be determined from the components of the unit vector acting in the direction of 𝐹

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- Example 2.11 Two forces act on the hook Specify the

magnitude of and its coordinate direction angles of that the resultant force 𝐹 𝑅 acts along the positive 𝑦 axis and has a magnitude of 800𝑁 Solution

To satisfy this equation the 𝑖 , 𝑗 , 𝑘 components of 𝐹 𝑅 must be

equal to the corresponding 𝑖 , 𝑗 , 𝑘 components of 𝐹 1+ 𝐹 2

0 = 212.1 + 𝐹2𝑥 ⟹ 𝐹2𝑥= −212.1𝑁

800 = 150 + 𝐹2𝑦 ⟹ 𝐹2𝑦= 650𝑁

0 = −150 + 𝐹2𝑧 ⟹ 𝐹2𝑧= 150𝑁 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.44 Force Vectors

𝐹 𝑅= 800𝑗

𝐹 1= 212.1𝑖 + 150𝑗 −150𝑘

𝐹 2= −212.1𝑖 + 650𝑗 + 150𝑘 The magnitude of 𝐹 2

- F2.13: Determine its coordinate direction angles of the force

- F2.14: Express the force as a Cartesian vector

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- F2.18: Determine the resultant force acting on the hook

§7 Position Vectors

- 𝑥, 𝑦, 𝑧 Coordinates

A right handed coordinate system

to reference the location of points

An elastic rubber band is attached to points

𝐴 and 𝐵 Determine its length and its

direction measured from 𝐴 toward 𝐵 Solution

Establish a position vector from 𝐴 to 𝐵

𝑟 = (−2 − 1)𝑖 + (2 − 0)𝑗 + (3 − 3)𝑘

= −3𝑖 + 2𝑗 + 6𝑘 These components of 𝑟 can also be

determined directly by realizing that they

represent the direction and distance one must travel along each axis in order to move from 𝐴 to 𝐵

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§7 Position Vectors

The length of the rubber band

𝑟 = (−3)2+22+62= 7𝑚 Formulating a unit vector in the direction of 𝑟

𝑐𝑜𝑠𝛼 = −3/7 ⟹ 𝛼 = 1150𝑐𝑜𝑠𝛽 = 2/7 ⟹ 𝛽 = 73.40𝑐𝑜𝑠𝛾 = 6/7 ⟹ 𝛾 = 31.00

§8 Force Vector Directed Along A Line Quite often, the direction of a force is specified by two points through which its line of action passes

Formulate 𝐹 as a Cartesian vector by realizing that it has the

same direction and sense as the position vector 𝑟 directed from point 𝐴 to point 𝐵 on the cord

𝐹 = 𝐹𝑢 = 𝐹𝑟

𝑟= 𝐹

𝑥𝐵− 𝑥𝐴 𝑖 + 𝑦𝐵− 𝑦𝐴𝑗 + (𝑧𝐵− 𝑧𝐴)𝑘

𝑥𝐵− 𝑥𝐴2+ 𝑦𝐵− 𝑦𝐴2+ (𝑧𝐵− 𝑧𝐴)2 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.56 Force Vectors

§8 Force Vector Directed Along A Line

- Example 2.13 The man pulls on the cord with

a force of 70𝑁 Represent this force acting on the support 𝐴 as

a Cartesian vector and determine its direction Solution

Force 𝐹 is shown in the figure

The direction of this vector, 𝑢, is

determined from the position vector 𝑟 (𝐴 → {−24𝑘} → {−8𝑗} →{12𝑖} → 𝐵)

𝑟 = 12𝑖 − 8𝑗 − 24𝑘 𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

The magnitude of 𝑟 , which

represents the length of cord 𝐴𝐵

|𝑟 | = 122+ (−8)2+(−24)2= 28𝑚 Forming the unit vector that defines the direction and sense

and a direction specified by 𝑢

𝑐𝑜𝑠𝛼 = 12/28 ⟹ 𝛼 = 64.60𝑐𝑜𝑠𝛽 = −8/28 ⟹ 𝛽 = 1070𝑐𝑜𝑠𝛾 = −24/28 ⟹ 𝛾 = 1490

- Example 2.14 Express the force acts on the hook as a Cartesian

vector Solution The coordinates for points 𝐴 and 𝐵

𝐹 𝐵= 𝐹𝐵𝑢𝐵= 750(−0.7428𝑖 + 0.6433𝑗 + 0.1857𝑘)

= −577𝑖 + 482𝑗 + 139𝑘 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 2.60 Force Vectors

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- Example 2.15

The cables exert forces 𝐹𝐴𝐵= 100𝑁, 𝐹𝐴𝐶= 120𝑁 on the wall hook

at 𝐴, determine the resultant force acting at 𝐴 as a Cartesian vector

𝑟 𝐴𝐵= 4𝑖 − 4𝑘 𝑟𝐴𝐵= 42+ 42= 5.66

- F2.20: Determine the length of the rod and the position vector

directed from 𝐴 to 𝐵 What is the angle 𝜃?

- F2.23: Determine the resultant force at 𝐴 Engineering Mechanics – Statics 2.66 Force Vectors

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- F2.24: Determine the magnitude of the resultant force at 𝐴

§9 Dot Product

- Example 2.17 The frame is subjected to a horizontal force

𝐹 = 300𝑗 𝑁 Determine the magnitude of the components of this force parallel and perpendicular to member 𝐴𝐵

§9 Dot Product

Solution

𝑢𝐵=𝑟 𝐵

𝑟𝐵=2𝑖 + 6𝑗 + 3𝑘

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§9 Dot Product

- Example 2.18 The pipe is subjected to the force of 𝐹 = 80𝑁

Determine the angle 𝜃 between 𝐹 and the pipe segment 𝐵𝐴 and the projection

of 𝐹 along this segment Solution

Angle 𝜃

𝑟 𝐵𝐴= −2𝑖 − 2𝑗 + 𝑘, 𝑟𝐵𝐴= 3𝑚

𝑟 𝐵𝐶= −3𝑗 + 𝑘, 𝑟𝐵𝐶= 10𝑚 𝑐𝑜𝑠𝜃 =𝑟 𝐵𝐴∙ 𝑟 𝐵𝐶

𝑟𝐵𝐴𝑟𝐵𝐶

= −2 0 + −2 −3 +(1)(1)

3 10 = 0.7379 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

⟹ 𝜃 = 42.50

§9 Dot Product Components of 𝐹

The component of 𝐹 along

𝐵𝐴 is shown in the figure

𝑢𝐵𝐴=𝑟 𝐵𝐴

𝑟𝐵𝐴=

−2𝑖 − 2𝑗 + 𝑘(−2)2+(−2)2+12

- F2.25: Determine the angle 𝜃 between the force and the line 𝐴𝑂

- F2.26: Determine the angle 𝜃 between the force and the line 𝐴𝐵

- F2.27: Determine the angle 𝜃 between the force and the line 𝑂𝐴

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- F2.29: Find the magnitude of the projected component of the

force along the pipe

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03 Equilibrium of a Particle

Engineering Mechanics – Statics 3.01 Equilibrium of a Particle

Chapter 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

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 3.02 Equilibrium of a Particle

§1 Condition for Equilibrium of a Particle

- Newton’s First Law of Physics: If the resultant force on a

particle is zero, the particle will remain at rest or will continue

at constant speed in a straight line

∑𝐹 𝑖= 0 ⟺ Equilibrium

- Equilibrium: a key concept in statics is that of equilibrium If an

object is at rest, we will assume that it is in equilibrium and that

the sum of the forces acting on the object equal zero

∑𝐹 𝑖= 𝐹 𝑅= 0

• Equations for 2D Equilibrium: 𝐹𝑥= 0, 𝐹𝑦= 0

• Equations for 3D Equilibrium : 𝐹𝑥= 0, 𝐹𝑦= 0, 𝐹𝑧= 0

§2 The Free-Body Diagram

- A free-body diagram (FBD): drawing that shows the particle with all the forces that act on it

- Two types of connections often encountered in particle equilibrium problems

• Springs: can be used to apply forces of tension/compression

• Cables and Pulleys

+ Ideal pulleys simply change the direction of a force

+ The tension on each side of an ideal pulley is the same

+ The tension is the same everywhere in a given rope or cable if ideal pulleys are used

- A Procedure for Drawing a Free-body Diagram

• Imagine the particle to be isolated or cut free from its

surroundings

• Show all the forces that act on the particle

+ Active forces: they want to move the particle

+ Reactive forces: they tend to resist the motion

• Identify each force and show all known magnitudes and

directions Show all unknown magnitudes and/or directions

as variables

- Example 3.1 The sphere has a mass of 6𝑘𝑔 and is supported as shown Draw a free-body diagram of the sphere, the cord 𝐶𝐸, and the knot at 𝐶

Solution

- Knot 𝐶

- Cord 𝐶𝐸

- Sphere Engineering Mechanics – Statics 3.06 Equilibrium of a Particle

Trang 20

§3 Coplanar Force Systems

- 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 𝑖

- Procedure for Analysis + Free-Body Diagram

• Establish the 𝑥, 𝑦 axes in any suitable orientation

• Label all the known/unknown force magnitudes and directions

• The sense of an unknown magnitude force can be assumed + Equations of Equilibrium

• Apply the equations of equilibrium, ∑𝐹𝑥= 0 and ∑𝐹𝑦= 0

• 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 (+), then if the solution for a force yields a (−) result, this indicates its sense is the reverse of that shown on the free-body diagram HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 3.08 Equilibrium of a Particle

- Example 3.2 Determine the tension in cables 𝐵𝐴 and 𝐵𝐶

necessary to support the 60𝑘𝑔 cylinder

Solution

Free-body diagram

§3 Coplanar Force Systems Equation of equilibrium +→ ∑𝐹𝑥= 0: 𝑇𝐶𝑐𝑜𝑠450−4

5𝑇𝐴= 0 + ↑ ∑𝐹𝑦= 0: 𝑇𝐶𝑠𝑖𝑛450+3

5𝑇𝐴− 60 × 9.81 = 0

The tension in cables

• 𝐵𝐴: 𝑇𝐴= 420𝑁

• 𝐵𝐶: 𝑇𝐶= 475.66𝑁 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 3.10 Equilibrium of a Particle

- Example 3.3 The 200𝑘𝑔 crate is suspended using the ropes

𝐴𝐵 and 𝐴𝐶 Each rope can withstand a maximum force of

10𝑘𝑁 before it breaks If 𝐴𝐵 always remains horizontal,

determine the smallest angle 𝜃 to which the crate can be

suspended before one of the ropes breaks

Solution

Equation of equilibrium +→ ∑𝐹𝑥= 0: −𝐹𝐶𝑐𝑜𝑠𝜃 + 𝐹𝐵= 0 + ↑ ∑𝐹𝑦= 0: 𝐹𝐶𝑠𝑖𝑛𝜃 − 200 × 9.81 = 0 The smallest angle 𝜃

𝜃 = 𝑠𝑖𝑛−1 1962

10 × 103 = 11.310≈ 11.30

𝐹𝐵= 10 × 103𝑐𝑜𝑠11.310= 9.81 × 103𝑁 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

- Example 3.4 Determine the required length of cord 𝐴𝐶 so

that the 8𝑘𝑔 lamp can be suspended in the position shown

The undeformed length of spring 𝐴𝐵 is 𝑙′𝐴𝐵= 0.4𝑚 Solution

𝑇𝐴𝐵= 𝑘𝐴𝐵𝑠𝐴𝐵⟹ 𝑠𝐴𝐵= 0.453𝑚 The stretched length 𝑙𝐴𝐵= 𝑙′𝐴𝐵+ 𝑠𝐴𝐵= 0.4 + 0.453 = 0.853𝑚 Distance from 𝐶 to 𝐵 2 = 𝑙𝐴𝐶𝑐𝑜𝑠300+ 0.853 ⟹𝑙𝐴𝐶= 1.32𝑚HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 3.12 Equilibrium of a Particle

Trang 21

- F3.1: The crate has a weight of 550𝑁 Determine the force in

each supporting cable

- F3.2: The beam has a weight of 700𝑁 Determine the shortest cable 𝐴𝐵𝐶 that can be used to lift it if the maximum force the

cable can sustain is 1500𝑁

- F3.3: If the 5𝑘𝑔 block is suspended from the pulley 𝐵 and the

sag of the cord is 𝑑 = 0.15𝑚, determine the force in cord 𝐴𝐵𝐶

Neglect the size of the pulley

- F3.5: If the mass of cylinder 𝐶 is 40𝑘𝑔, determine the mass of

cylinder 𝐴 in order to hold the assembly in the position shown

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§4 Three-Dimensional Force Systems

- The necessary and sufficient condition for particle equilibrium

- Procedure for Analysis + Free-Body Diagram

• Establish the 𝑥, 𝑦, 𝑧 axes in any suitable orientation

• Label all the known/unknown force magnitudes and directions

• The sense of an unknown magnitude force can be assumed + Equations of Equilibrium

• Apply the equations of equilibrium, ∑𝐹𝑥= 0, ∑𝐹𝑦= 0, ∑𝐹𝑧= 0

• If the 3D geometry appears difficult, express each force on the free-body diagram as a Cartesian vector, substitute these vectors into ∑𝐹 = 0, set the 𝑖 , 𝑗 , 𝑘 components equal to zero

• Since the magnitude of a force is always (+), then if the solution for a force yields a (−) result, this indicates its

sense is the reverse of that shown on the free-body diagram

- Example 3.5 A 90𝑁 load is suspended from the hook If the

load is supported by two cables and a spring having a stiffness 𝑘 = 500𝑁/𝑚, determine the force in the cables and the stretch of the spring for equilibrium

Cable 𝐴𝐷 lies in the 𝑥– 𝑦 plane and

cable 𝐴𝐶 lies in the 𝑥– 𝑧 plane Solution

Free-body diagram Equation of equilibrium

𝐹𝐶= 150𝑁

𝐹𝐷= 240𝑁

𝐹𝐵= 207.8𝑁 The stretch of the spring

- Example 3.6 The 10𝑘𝑔 lamp is suspended from the three

equal-length cords Determine its smallest vertical distance

𝑠from the ceiling if the force developed in any cord is not

allowed to exceed 50𝑁

Solution Free-body diagram Equation of equilibrium

𝑠 = 600𝑡𝑎𝑛49.160= 519𝑚𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

- Example 3.7 Determine the force in each cable used to

support the 40𝑁 crate Solution

Free-body diagram Equation of equilibrium

𝐹 𝐵= 𝐹𝐵 −3𝑖 − 4𝑗 + 8𝑘(−3)2+(−4)2+82

= −0.318𝐹𝐵𝑖 − 0.424𝐹𝐵𝑗 + 0.848𝐹𝐵𝑘

𝐹 𝐶= 𝐹𝐶 −3𝑖 + 4𝑗 + 8𝑘(−3)2+42+ 82

= −0.318𝐹𝐶𝑖 + 0.424𝐹𝐶𝑗 + 0.848𝐹𝐶𝑘

𝐹 𝐷= 𝐹𝐷𝑖 , 𝑊 = −40𝑘

Trang 23

Equilibrium required

𝐹 𝐵+ 𝐹 𝐶+ 𝐹 𝐷+ 𝑊 = 0

⟹ −0.318𝐹𝐵− 0.318𝐹𝐶+ 𝐹𝐷𝑖 + −0.424𝐹𝐵+ 0.424𝐹𝐶𝑗 + 0.848𝐹𝐵+ 0.848𝐹𝐶− 40 𝑘 = 0

Equating the respective 𝑖 , 𝑗 , 𝑘 components

- Example 3.7 Determine the tension in each cord used to support the 100𝑘𝑔 crate

Solution Free-body diagram Equation of equilibrium

𝐹 𝐵= 𝐹𝐵𝑖

𝐹 𝐶= 𝐹𝐶𝑐𝑜𝑠1200𝑖 + 𝐹𝐶𝑐𝑜𝑠1350𝑗 + 𝐹𝐶𝑐𝑜𝑠600𝑘

= −0.5𝐹𝐶𝑖 − 0.707𝐹 𝐶𝑗 + 0.5𝐹𝐶𝑘

𝐹 𝐷= 𝐹𝐷 −𝑖 + 2𝑗 + 2𝑘(−1)2+22+ 22

Equating the respective 𝑖 , 𝑗 , 𝑘 components

- F3.8: Determine the tension developed in cables 𝐴𝐵, 𝐴𝐶, 𝐴𝐷

- F3.9: Determine the tension developed in cables 𝐴𝐵, 𝐴𝐶, 𝐴𝐷 Engineering Mechanics – Statics 3.30 Equilibrium of a Particle

Trang 24

- F3.10: Determine the tension developed in cables 𝐴𝐵, 𝐴𝐶, 𝐴𝐷

Trang 25

04 Force System Resultants

Engineering Mechanics – Statics 4.01 Force System Resultants

Chapter 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 an axis

• To define the moment of a couple

• To present methods for determining the resultants of concurrent force systems

non-• To indicate how to reduce a simple distributed loading to a resultant force having a specified location

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 4.02 Force System Resultants

§1 Moment of a Force – Scalar Formulation

- The moment of a force about a point provides a measure of

the tendency for rotation (sometimes called a torque)

𝑀 = 𝐹𝑑 𝑀 = 𝐹𝑑′= 𝐹𝑑𝑠𝑖𝑛𝜃 𝑀 = 𝐹𝑑 = 0

- Example 4.1 For each case illustrated in the figure,

determine the moment of the force about point 𝑂

Solution

𝑀𝑂= 100𝑁 × 2𝑚 = 200𝑁𝑚 ↻

𝑀𝑂= 50𝑁 × 0.75𝑚 = 37.5𝑁𝑚 ↻

𝑀𝑂= 40𝑁 × (4𝑚 + 2𝑐𝑜𝑠300) = 229𝑁𝑚 ↻

- Example 4.1 For each case illustrated in the figure, determine the moment of the force about point 𝑂

Solution

𝑀𝑂= 60𝑁 × 1𝑠𝑖𝑛450𝑚 = 42.4𝑁𝑚 ↺

𝑀𝑂= 7𝑘𝑁 × (4𝑚 − 1𝑚) = 21.0𝑁𝑚 ↺ Engineering Mechanics – Statics 4.06 Force System Resultants

Trang 26

Specified by the right-hand rule

curling the fingers of the right hand from vector 𝐴 (cross) to

vector 𝐵, the thumb points in the direction of 𝐶

§3 Moment of a Force-Vector Formulation

- The moment of a force 𝐹 about point 𝑂

𝑀𝑂= 𝑟 × 𝐹

- Magnitude

𝑀𝑂= 𝑟𝐹𝑠𝑖𝑛𝜃 = 𝐹 𝑟𝑠𝑖𝑛𝜃 = 𝐹𝑑

- Direction Determined by the right-hand rule

as it applies to the cross product

- Resultant Moment of a System of Forces

𝑀𝑅𝑂= ∑(𝑟 × 𝐹 )

Trang 27

- Example 4.3 Determine the moment produced by the force 𝐹

about point 𝑂 Express the result as a Cartesian vector

= −16.5𝑖 + 5.5𝑗 𝑘𝑁𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Solution

⟹ 𝑢𝐴𝐵= 𝑟 𝐵− 𝑟 𝐴

- Example 4.4 Two forces act on the rod, determine the resultant moment they create about the flange at 𝑂 Express the result as a Cartesian vector

§4 Principle of Moments (Varignon’s Theorem)

- The moment of a force about a point is equal to the sum of the

moments of the components of the force about the point

𝑀𝑂= 𝑟 × 𝐹 = 𝑟 × 𝐹 1+ 𝐹 2 = 𝑟 × 𝐹 1+ 𝑟 × 𝐹 2

For two-dimensional problems

𝑀𝑂= 𝐹𝑑 = 𝐹𝑥𝑦 − 𝐹 𝑦𝑥

- Example 4.6 Determine the moment of the force about point 𝑂

- Example 4.6 Force 𝐹 acts at the end of the angle bracket, determine the moment of the force about point 𝑂

Solution 1

↺ + 𝑀𝑂= 400𝑠𝑖𝑛300× 0.2 − 400𝑐𝑜𝑠300× 0.4

= −98.6𝑁

= 98.6𝑁 ↻ Engineering Mechanics – Statics 4.18 Force System Resultants

Trang 28

- Example 4.6 Force 𝐹 acts at the end of the angle bracket,

determine the moment of the force about point 𝑂

- F4.1 Determine the moment of the force about point 𝑂

Trang 29

§5 Moment of a Force about a Specified Axis

- The moment of 𝐹 about a axis

𝑢𝑎𝑥,𝑢𝑎𝑥,𝑢𝑎𝑥 : 𝑥,𝑦,𝑧 components of the unit vector 𝑢𝑎 of 𝑎 axis

𝑟𝑥,𝑟𝑦,𝑟𝑧 : 𝑥 , 𝑦 , 𝑧 components of the position vector

extended from any point 𝑂 on the 𝑎 axis to any

point 𝐴 on the line of action of the force 𝐹

𝐹𝑥,𝐹𝑦,𝐹𝑧 : 𝑥,𝑦,𝑧 components of the force vector

𝑀𝑎= 𝑢𝑎(𝑟 × 𝐹 )

𝑀𝑎= 𝑀𝑎𝑢𝑎

= 𝑢𝑎𝑥 𝑟𝑦𝐹𝑧− 𝑟𝑧𝐹 𝑦

- Example 4.7 Determine the resultant moment of the three forces 𝐹 160𝑁 , 𝐹 250𝑁 , 𝐹 340𝑁 about the 𝑥,𝑦,𝑧 axes

- Example 4.8 Determine the moment produced by the force

𝐹 , which tends to rotate the rod about the 𝐴𝐵 axis

Solution 𝑀𝐴𝐵= 𝑢𝐵 𝑟 × 𝐹 where

𝑢𝐵=𝑟 𝐵

𝑟𝐵=

0.4𝑖 + 0.2𝑗 0.42+ 0.22

𝑀𝐴𝐵= 𝑀𝐴𝐵𝑢𝐵

= 80.50 0.8944𝑖 + 0.4472𝑗

= 72.0𝑖 + 36.0𝑗 (𝑁𝑚) Engineering Mechanics – Statics 4.30 Force System Resultants

Trang 30

- Example 4.9 Determine the magnitude of the moment of force

𝐹 about segment 𝑂𝐴 of the pipe assembly

Solution

𝑀𝑂𝐴= 𝑢𝑂𝐴𝑟 × 𝐹 where

𝑢𝑂𝐴=𝑟 𝑂𝐴

𝑟𝑂𝐴=

0.3𝑖 + 0.4𝑗 0.32+ 0.42= 0.6𝑖 + 0.8𝑗

𝑟 = 𝑟 𝑂𝐷= 0.5𝑖 + 0.5𝑘

𝐹 = 𝐹𝑟 𝐶𝐷

𝑟𝐶𝐷

= 300 0.4𝑖 − 0.4𝑗 + 0.2𝑘0.42+ (−0.4)2+0.22

= 200𝑖 − 200𝑗 + 100𝑘 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

= 100 0.6𝑖 + 0.8𝑗

= 60.0𝑖 + 80.0𝑗 (𝑁𝑚) HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 4.32 Force System Resultants

- F4.13 Determine the magnitude of the moment of the force

𝐹 = 300𝑖 − 200𝑗 + 150𝑘 (𝑁) about the 𝑥 axis Express the

result as a Cartesian vector

𝐹 = 300𝑖 − 200𝑗 + 150𝑘 (𝑁) about the 𝑂𝐴 axis Express the result as a Cartesian vector

- F4.15 Determine the magnitude of the moment of the 200𝑁

force about the 𝑥 axis

Trang 31

𝐹 = 50𝑖 − 40𝑗 + 20𝑘 (𝑁) about the 𝐴𝐵 axis Express the result

as a Catersian vector

- A couple: two parallel forces that have the same magnitude but

opposite in direction separated by a perpendicular distance 𝑑

- A couple moment: the moment produced by a couple

𝑀 = 𝑟 𝐵× 𝐹 + 𝑟 𝐴× −𝐹

= 𝑟 𝐵− 𝑟 𝐴 × 𝐹

⟹𝑀 = 𝑟 × 𝐹

Trang 32

§6 Moment of a Couple

- Example 4.11 Determine the magnitude and direction of the

couple moment acting on the gear

§6 Moment of a Couple

- Example 4.12 Determine the couple moment acting on the

pipe Segment 𝐴𝐵 is directed 300 below the 𝑥– 𝑦 plane

Trang 33

- F4.19 Determine the resultant couple moment acting on the

beam

- F4.21 Determine the magnitude of 𝐹 so that the resultant

couple moment acting on the beam is 1.5𝑘𝑁𝑚 clockwise

- F4.22 Determine the couple moment acting on the beam

Trang 34

§7 Simplification of a Force and Couple System

- System of Forces and Couple Moments

𝐹 𝑅= ∑𝐹

𝑀𝑅𝑂= ∑𝑀𝑂+ ∑𝑀

If the force system lies in the 𝑥– 𝑦 plane and any couple

moments are perpendicular to this plane, then the above

equations reduce to the following three scalar equations

𝐹𝑅= ∑𝐹𝑥, 𝐹𝑅 = ∑𝐹𝑦, 𝑀𝑅𝑂= ∑𝑀𝑂+ ∑𝑀

- Example 4.14 Replace the force and couple system by an

equivalent resultant force and couple moment acting at point 𝑂

Solution Force summation +→ 𝐹𝑅= 3𝑐𝑜𝑠300+3

55 = 5.598𝑘𝑁 + ↑ 𝐹𝑅 = 3𝑠𝑖𝑛300−4

55 − 4

= −6.50𝑘𝑁 = 6.50𝑘𝑁 ↓ Moment summation

Summation +→ 𝐹𝑅= 3𝑐𝑜𝑠300+3

55 = 5.598𝑘𝑁 + ↑ 𝐹𝑅 = −6.50𝑘𝑁 = 6.50𝑘𝑁 ↓ +↺ 𝑀𝑅𝑂= 2.46𝑘𝑁𝑚 ↻ Using the Pythagorean theorem

𝐹𝑅= 𝐹𝑅𝑥 2+ 𝐹𝑅𝑦 2

= 5.5982+ 6.502

= 8.58𝑘𝑁 Its direction 𝜃

𝜃 = 𝑡𝑎𝑛−1𝐹𝑅

𝐹𝑅= 𝑡𝑎𝑛

−1 6.505.598= 49.3

0

- Example 4.15 Replace the force and couple system acting on

the member by an equivalent resultant force and couple moment acting at point 𝑂 Solution

Force summation +→ 𝐹𝑅=3

5500 = 300𝑁 = 300𝑁 →+ ↑ 𝐹𝑅 =45500 − 750 = −350𝑁 = 350 ↓ Moment summation

300= 49.40 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 4.58 Force System Resultants

- Example 4.16 The structural member is subjected to a couple

moment 𝑀 and forces 𝐹 1 and 𝐹 2 Replace this system by an

equivalent resultant force and couple moment acting at its

Force summation

𝐹 𝑅= ∑𝐹 = 𝐹 1+ 𝐹 2= −800𝑘 − 249.6𝑖 + 166𝐽

⟹ 𝐹 𝑅= −250𝑖 + 166𝑗 − 800𝑘 (𝑁)Moment summation

Trang 35

- F4.25 Replace the loading system by an equivalent resultant

force and couple moment acting at point 𝐴

force and couple moment acting at point 𝑂

Trang 36

§8 Further Simplification of a Force and Couple System

- Concurrent force system: the system in which the lines of

action of all the forces intersect at a common point 𝑂

𝐹 𝑅= ∑𝐹 𝑖

- Coplanar force system

- Parallel force system

𝑀𝑅𝑂= ∑𝑑 𝑖× 𝐹 𝑖

𝑑 = 𝑀𝑅𝑂/𝐹𝑅

- Reduction to a wrench

𝐹 𝑅𝑂

𝑀𝑅𝑂= 𝑀⊥+ 𝑀∥

𝐹 𝑅𝑃, 𝑑 = 𝑀⊥/𝐹𝑅𝑂𝑀∥

- Example 4.17 Replace the force and couple moment system acting on the beam by an equivalent resultant force, and find where its line of action intersects the beam, measured from point 𝑂 Solution

Force summation +→ 𝐹𝑅=3

58 = 4.8𝑘𝑁 = 4.8𝑘𝑁 →+ ↑ 𝐹𝑅 = −4 +4

58 = 2.4𝑘𝑁 = 2.4𝑘𝑁 ↑and 𝐹𝑅= 4.82+ 2.42= 5.37𝑘𝑁, 𝜃 = 𝑡𝑎𝑛−12.4

4.8= 26.60 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 4.72 Force System Resultants

Trang 37

- Example 4.19 The jib crane is subjected to three coplanar forces Replace this loading by an equivalent resultant force and specify where the resultant’s line of action intersects the column 𝐴𝐵 and boom 𝐵𝐶 Solution

Force summation +→ 𝐹𝑅= −35250 − 175 = −325𝑁+ ↑ 𝐹𝑅 = −4

5250 − 60 = −260𝑁and 𝐹𝑅= 3252+ 2602= 416𝑁

= 175 × 5 − 60 × 3 +35250 × 11 −45250 × 8

⟹ 𝑦 = 2.29𝑚

By the principle of transmissibility, can be placed at a distance 𝑥 where

it intersects 𝐵𝐶 +↺ 𝑀𝑅𝐴= 325 × 11 − 260 × 𝑥

= 175 × 5 − 60 × 3 +35250 × 11 −4

5250 × 8

⟹ 𝑦 = 10.9𝑚

- Example 4.18 The slab is subjected to four parallel forces Determine the magnitude and direction of a resultant force equivalent to the given force system and locate its point of application on the slab

Force summation + ↑ 𝐹𝑅𝑧= −600 + 100 − 400 − 500

= −1400𝑁 = 1400𝑁 ↓Moment summation +↺ 𝑀𝑅= −1400 × 𝑦

= 100 × 5 − 400 × 10

⟹ 𝑦 = 2.50𝑚 +↺ 𝑀𝑅= −1400 × 𝑥

= 600 × 8 − 100 × 6

⟹ 𝑥 = 3𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Solution

force and specify where the resultant’s line of action intersects

the beam measured from 𝑂

- F4.32 Replace the loading system by an equivalent resultant force and specify where the resultant’s line of action intersects the beam measured from 𝐴

Trang 38

force and specify where the resultant’s line of action intersects

the beam measured from 𝐴

- F4.34 Replace the loading system by an equivalent resultant force and specify where the resultant’s line of action intersects the member 𝐴𝐵 measured from 𝐴

- F4.35 Replace the loading system by an equivalent single

resultant force and specify the 𝑥 and 𝑦 coordinates of its line of

action

- F4.36 Replace the loading system by an equivalent single resultant force and specify the 𝑥 and 𝑦 coordinates of its line of action

§9 Reduction of a Simple Distribute Loading

- Surface forces are caused by direct contact

- Concentrated forces are applied to a point

- Linear Distributed forces 𝑤(𝑠) (force/length) are idealizations

The resultant force𝐹 𝑅is equivalent to the area under distributed

loading curve and acts through centroid or geometric center of

this area

- Body force is developed when one body exerts a force on

another body without direct physical contact

- Uniform Loading Along a Single Axis

𝑝 = 𝑝 𝑥 𝑁/𝑚2

⟹ 𝑤 = 𝑏𝑝 𝑥 = 𝑤(𝑥)

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- Magnitude of Resultant Force

+↓ 𝐹𝑅= 𝑤 𝑥 𝑑𝑥

𝐿

= 𝑑𝐴𝐴

- Example 4.21 Determine the magnitude and location of the equivalent resultant force acting on the shaft

Solution +↓ 𝐹𝑅= 𝑑𝐴𝐴

= 60𝑥2𝑑𝑥2 0

= 60 𝑥330

2

= 160𝑁 The location 𝑥 of 𝐹 𝑅 measured from 𝑂

𝑥 = 𝑥𝑑𝐴𝐴

𝑑𝐴𝐴 =

𝑥60𝑥02 2𝑑𝑥 60𝑥02 2𝑑𝑥=

60 𝑥4 40 2

160

⟹ 𝑥 = 1.5𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Engineering Mechanics – Statics 4.86 Force System Resultants

- Example 4.22 A distributed loading of Pa acts over the top

surface of the beam Determine the magnitude and location of

the equivalent resultant force

- Example 4.23 The granular material exerts the distributed loading on the beam Determine the magnitude and location of the equivalent resultant of this load

Solution

Divide the trapezoidal loading into a rectangular and triangular loading

a trapezoidal loading = a rectangular + a triangular loading

+

=

+

=

The magnitude of the force represented by each of these

loadings is equal to its associated area

The two parallel forces 𝐹 1 and 𝐹 2 can be reduced to a single resultant 𝐹 𝑅

+↓ 𝐹𝑅= ∑𝐹 𝐹𝑅= 225 + 450 = 675𝑁 +↺ 𝑀𝑅𝐴= ∑𝑀𝐴 𝑥 × 675 = 3 × 225 + 4.5 × 450 ⟹ 𝑥 = 4𝑚 Engineering Mechanics – Statics 4.90 Force System Resultants

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