11

ME 162 BASIC MECHANICSME 162 BASIC MECHANICS

Course LecturerCourse Lecturer: Dr. Joshua Ampofo: Dr. Joshua Ampofo

Email: Email: josh.ampofo@gmail.com/auspi@hotmail.comjosh.ampofo@gmail.com/auspi@hotmail.com

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Course ContentCourse Content

Fundamental ConceptFundamental Concept Newton’s Laws of MotionNewton’s Laws of Motion Force Systems and Force Systems and

Characteristics of Forces Characteristics of Forces & Moments& Moments

Vector Representation of Vector Representation of Forces and MomentsForces and Moments

Basic StaticsBasic Statics Equilibrium of Rigid Equilibrium of Rigid

Bodies in 2D and 3DBodies in 2D and 3D

Structural AnalysisStructural Analysis Methods of JointsMethods of Joints Method of SectionsMethod of Sections

FrictionFriction Simple MachinesSimple Machines Basic Dynamics of Basic Dynamics of

ParticlesParticles Basic Dynamics of Rigid Basic Dynamics of Rigid

BodiesBodies Simple Harmonic Motion Simple Harmonic Motion

(SHM)(SHM)

33

Course ObjectivesCourse Objectives

Upon successful completion of this course, students should

be able to:

 

(i) Understand and apply Newton’s laws of motion

(ii) Understand the basics of force and moments, and draw free

body diagrams

(iii) Analyze a 2D and 3D determinant structures for support

reactions and internal forces

(iv) Solve static and dynamic problems that involve friction

(v) Evaluate Advantage and Velocity Ratio of Simple Machines

(vi) Solve simple one degree-of-freedom vibration problems

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NoteNote

1.1. Assignments are due one week after they Assignments are due one week after they are given.are given.

2.2. Late turn-ins will not be accepted.Late turn-ins will not be accepted.

3.3. Cell phones must be turned off in class-no Cell phones must be turned off in class-no flashing, texting, or any use of cell phone.flashing, texting, or any use of cell phone.

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MechanicsMechanics

Mechanics is the science which deals the conditions of rest or motion of Mechanics is the science which deals the conditions of rest or motion of bodies under the action of forces.bodies under the action of forces.

Categories of Mechanics:o Rigid bodies

o Staticso Dynamics

o Kinematicso Kinetics

o Deformable bodieso Mechanics of Materials /Strength of Materialso Theory of elasticityo Theory of plasticity

o Fluid Mechanicso Mechanics of Compressible fluidso Mechanics of incompressible fluids

Mechanics is an applied science - it is not an abstract or pure science but does not have the empiricism found in other engineering sciences.

Mechanics is the foundation of most engineering sciences and is an indispensable prerequisite to their study.

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A particle has a mass but it is so small that it may be regarded as geometric point.

When a body is idealised as a particle, the principles of mechanics reduce to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem.

All the forces acting on a body will be assumed to be applied at the same point, that is the forces are assumed concurrent.

ParticleParticle

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Rigid BodyRigid Body

A rigid body is a collection of particles. The size A rigid body is a collection of particles. The size and the shape of rigid bodies remain constant and the shape of rigid bodies remain constant at all times. at all times.

This is an ideal situation since real bodies will This is an ideal situation since real bodies will change their shape to a certain extent under a change their shape to a certain extent under a system of forcessystem of forces

88

Newton’s Laws of MotionNewton’s Laws of Motion

The entire subject of rigid-body mechanics is formulated onThe entire subject of rigid-body mechanics is formulated onthe basis of the Newton’s three laws of motion.the basis of the Newton’s three laws of motion.

1.1. A particle at rest or moving in a straight line with constant A particle at rest or moving in a straight line with constant velocity will remain in this state except compelled by an velocity will remain in this state except compelled by an external force to act otherwiseexternal force to act otherwise

2.2. Change of motion is proportional to the applied force Change of motion is proportional to the applied force impressed on it and it occurs in the direction of the forceimpressed on it and it occurs in the direction of the force

3. For every force acting on a particle, the particle exerts an equal, opposite and collinear reactive force.

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Newton’s Laws of MotionNewton’s Laws of Motion

From the 1From the 1stst Law Law For a particle to accelerated, it must be subjected to an external force.For a particle to accelerated, it must be subjected to an external force.

However, if the body is at rest or is moving in a straight line with constant velocity, the external forces However, if the body is at rest or is moving in a straight line with constant velocity, the external forces acting, if any, must be balanced.acting, if any, must be balanced.

From the 2From the 2ndnd Law Law

From 3From 3rdrd Law Law Force is due to interaction between two different bodies.Force is due to interaction between two different bodies.

mvdt

dkF

kmavdt

dkmF

maF

If mass m = constant,

If k = unity,

1010

Basic DefinitionsBasic Definitions

Basic Definitions: Space-Is a geometric region in which the physical events of interest in

mechanics occur and it is given in terms of three coordinates measured from a reference point or origin.

Length-is used to specify the position of a point in space or size of a body

Time-Interval between two events

Matter-Anything that occupies space

Inertia- A property that cause a body to resist motion

Mass - Measure of inertia

Force-Action of a body upon another body

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Example 1Example 1

A body of mass 50 kg is acted upon by external force whose magnitude is 100N. What is the acceleration of the body?

2250

100

50100

:

?

100;50

sm

kg

Na

akgN

maFFrom

aonAccelerati

NFForcekgmMass

Solution

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Law of GravitationLaw of Gravitation

For any two bodies separated by For any two bodies separated by a distance r the force of a distance r the force of interaction between them is interaction between them is proportion to the product of their proportion to the product of their masses and inversely proportion masses and inversely proportion to square of their separation to square of their separation distancedistance

][.)10(673.6

:

311

221

unitSIskgmG

r

mmGF

allyMathematic

rm1 m2

1313

Mass and WeightMass and Weight

Mass (m)Mass (m) of a body is of a body is the quantity of matter in the quantity of matter in the body and it is the body and it is independent of location.independent of location.

Weight (W)Weight (W) is the is the product of mass and product of mass and gravitational gravitational acceleration thus acceleration thus depends on the locationdepends on the location

2

2

,

;

'

e

e

e

e

e

e

r

mGgTherefore

earthofradiusr

earthofmassm

mgr

mmGW

surfacesearthonThus

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Example 2Example 2

Calculate the weight W of a body at the surface Calculate the weight W of a body at the surface of the earth if it has a mass m of 675 kg.of the earth if it has a mass m of 675 kg.

Mass m = 675 kg; g = 9.81 m/s2

Solution

Weight W = mg = 675 kg x 9.81m/s2

= 6.62 x 103 N

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Basic UnitsBasic Units

QuantityQuantity SI UnitSI Unit SymbolSymbolLengthLength metermeter mm

MassMass kilogramkilogram kgkg

TimeTime secondsecond ss

Electric CurrentElectric Current AmpereAmpere AA

TemperatureTemperature KelvinKelvin KK

Amount of SubstanceAmount of Substance molemole molmol

Luminous intensityLuminous intensity candelacandela cdcd

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Derived UnitsDerived Units

QuantityQuantity Derived SI UnitDerived SI Unit SymbolSymbol Special NameSpecial Name

AreaArea square metersquare meter mm22

VolumeVolume cubic metercubic meter mm33

Linear VelocityLinear Velocity meter per secondmeter per second m/sm/s

Linear AccelerationLinear Acceleration meter per second meter per second squaredsquared

m/sm/s22

FrequencyFrequency (cycle) per second(cycle) per second HzHz HertzHertz

DensityDensity kilogram per cubic kilogram per cubic metermeter

Kg/mKg/m33

ForceForce Kilogram meter per Kilogram meter per second squaredsecond squared

NN NewtonNewton

Pressure /StressPressure /Stress Newton per meter Newton per meter squaredsquared

PaPa PascalPascal

Work /EnergyWork /Energy Newton.meterNewton.meter JJ JouleJoule

PowerPower Joule per secondJoule per second WW WattWatt

Moment of ForceMoment of Force Newton.meterNewton.meter N.mN.m

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Force Systems and Characteristics of Force Systems and Characteristics of ForcesForces

A force system comprises of several forces A force system comprises of several forces acting on a body of group of related bodiesacting on a body of group of related bodies

Force system are classified according to the Force system are classified according to the arraignment of constituent forces.arraignment of constituent forces.

– Collinear: act along the same lineCollinear: act along the same line– Parallel /Coplaner : lie in the same planeParallel /Coplaner : lie in the same plane– Concurrent: line of action of force interact at a Concurrent: line of action of force interact at a

common pointcommon point

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EquilibriumEquilibrium

• A body is in equilibrium if

1. it is at rest relative to an initial reference frame

2. the body moves with constant velocity along a straight line relative to an initial frame

Effects of a force on a rigid body depend on

(a) The magnitude of the force

(b) The direction of the force, and

(c) The line of action of the force

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Principle of TransmissibilityPrinciple of Transmissibility

• Principle of Transmissibility -The point of application of a force acting on a rigid body may be placed anywhere along its line of action without altering the conditions of equilibrium or motion of the rigid body.

• Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck.

• Principle of transmissibility may not always apply in determining internal forces and deformations.

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• Note : P, Q and R are magnitudes of forces and , respectively. A, B and C are interior angles of the force triangle.

Addition of VectorsAddition of Vectors

Q

P

R=Q+P

P

Q

Trapezoid rule for vector addition

Triangle rule for vector addition

BPQQPR cos2222

• Law of cosines

• Law of sines,

Q

C

R

B

P

A sinsinsin

• Vector addition is commutative

PQQP

A

B

Q

C

R=Q+P

P

,QP

R

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Addition of VectorsAddition of Vectors

• Addition of three or more vectors through repeated application of the triangle rule

• The polygon rule for the addition of three or more vectors.

• Vector addition is associative,

SQPSQPSQP

2222

Review of GeometryReview of Geometry

From Diagram (a),From Diagram (a),

D + A = 180D + A = 180oo

D + C = 180D + C = 180oo

A = C and B = DA = C and B = D

From Diagram (b),From Diagram (b),

G = E and H = FG = E and H = F

H =L and E = IH =L and E = I

From Diagram (c),From Diagram (c),

M + N + Q =180M + N + Q =180

N =180 - PN =180 - P

AB

CD

(a)

(b)

E FGH

I JKL

(c)

MN P

Q

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Example 3Example 3

Graphical SolutionGraphical Solution

The two forces act on a bolt at A. Determine their resultant.

Graphical solution - A half or full parallelogram with sides equal to P and Q is drawn to scale.

Q

P

Q’

P’

R

β

R = 98 N β = 35o

The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured,

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Example 3Example 3

Trigonometric solution

155cosN60N402N60N40

cos222

222 BPQQPR

Q=60 N

P =40 N

Rβ

20o

25o

β-20o

A

B

C

155o R = 97.73 N

From the Law of Cosines,

From the Law of Sines,

o

o

o

R

QB

R

B

Q

A

35

1520

N73.97

N60155sin

sin20sin

sinsin

2525

Rectangular Components of ForceRectangular Components of Force

yx FFF

x

y

θ

Fx

Fy

F

•A force vector may be resolved into perpendicular components so that the resulting parallelogram is a rectangle. The resulting x and y components are referred to as rectangular vector components and

Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components.

jFiFF yx

Fx and Fy are referred to as the scalar components

x

y

yx

F

F

FFF

1

22

tan

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Example 4 (mech 1)Example 4 (mech 1)

Solve Example 3 using Rectangular Components solution

RRxx = =ΣΣFFx x =Pcos 20=Pcos 20oo + Qcos (20 + Qcos (20oo +25 +25o)o)

= 40cos 20= 40cos 20oo + 60cos 45 + 60cos 45oo

= 80.014 N= 80.014 N

RRyy==ΣΣFFy y =Psin 20=Psin 20oo + Qsin (20 + Qsin (20oo +25 +25o)o)

= 40sin 20= 40sin 20oo + 60sin 45 + 60sin 45oo

= 57.107 N= 57.107 N

Q = 6

0 N

P = 40 N

R

β

20o

25o

o

x

y

yx

R

R

NR

R

RRR

35014.80

107.57tan

tan

72.97

107.57014.80

1

1

22

22

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Example 5Example 5

Four forces act on bolt A as shown. Determine the resultant of the force on the bolt.

SOLUTION:

• Resolve each force into rectangular components.

• Calculate the magnitude and direction of the resultant.

• Determine the components of the resultant by adding the corresponding force components.

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Example 5Example 5

SOLUTION:

• Resolve each force into rectangular components.

22 3.141.199 R N6.199R

• Calculate the magnitude and direction.

N1.199

N3.14tan 1.4

9.256.96100

0.1100110

2.754.2780

0.759.129150

4

3

2

1

F

F

F

F

compycompxmagforce

• Determine the components of the resultant by adding the corresponding force components.

1.199xR 3.14yR

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Example 5Example 5

Determine the magnitude of the Determine the magnitude of the resultant force and its direction resultant force and its direction measured from the positive x measured from the positive x axis.axis.

x

y

50 kN

40 kN20 kN

34

5

11

√260o

NR

CosR

CosFR

x

x

n

iiix

284.58

)2/1(406020)5/4(50

1

NR

SinR

SinFR

y

y

n

iiiy

604.15

)2/1(406020)5/3(50

1

oo

x

y

yx

R

R

NR

R

RRR

345or 15284.58

604.15tan

tan

34.60

604.15284.58

1

1

22

22

SolutionSolution