Chapter Two Laith Batarseh - Philadelphia

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11/3/2015 1 Chapter Two Laith Batarseh Definition Engineering mechanics Deformable body mechanics Rigid body mechanics Dynamics Statics Fluid mechanics Constant Velocity Variable Velocity

Transcript of Chapter Two Laith Batarseh - Philadelphia

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Chapter Two

Laith Batarseh

Definition

Engineering mechanics

Deformable body

mechanics

Rigid body mechanics

Dynamics Statics

Fluid mechanics

Constant Velocity

Variable Velocity

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Definition

Rigid body is the body that has the same volume parameters before and after applying the load

Deformable body is the body changes its volume parameters when the load is applied on it.

Definition

Static Cases:

Dynamic Case:

Velocity = 0P

BA

PVelocity = Constant

BA

PVelocity is changeable

Acceleration or Deceleration

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Definition

P

At rest Acceleration Deceleration Constant velocity

Dynamics Statics

EndStart

Cartesian coordinate system

x

y

x

y

z

y-z plane

x-y plane

x-z

pla

ne

x-y plane

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Definition

x

y

y

P(x,y)

Newtown's Laws of Motion

Newtown's Laws of Motion

First Law: a particle at rest or moves in constant velocity will remain on its state unless it is subjected to unbalance force.

F1F4

F3

F2

0F

F

Second Law: a particle subjected to unbalance force will move at acceleration has the same direction of the force.

maF

Third Law: each acting force has a reaction equal in magnitude and opposite in direction

AAction: Force of A acting on B B Reaction: Force of B acting on A

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Newtown's Law of Gravity

Newtown proved his most famous law at all in the 17th centaury and called itLaw of Universal Gravitation.

Statement: any two objects have a masses (M1 and M2 )and far away from each other by a distance (R) will have attraction force (gravitational) related proportionally with the objects masses product (M1 M2 ) and inversely with the square of the distance between them (R2). Mathematically:

Where: G = 66.73 x 10-12 m3/(kg.s2)

Weight: according to Newtown's Law of Gravity, weight is the gravitational force acting between the body has a mass (m) and the earth (of mass Me) and is given as: W=mg. g is the gravitational acceleration and equal to GMe/R2.

constant isG ;2

21

R

MMGF

Units of Measurements

According to Newton's 2nd law, the unit of force is a combination of the

other three quantities: mass, length and time. So, the force units N

(Newtown) and lbf can be written as:22

.s

ftlbmlbfand

s

mkgN

Units of Measurements

Length

(m or ft)

Mass

(kg or lbm)

Time

(sec)

Force

(N or lbf)

SImKgN

Englishft

lbmlbf

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Scalar versus vectors

Physical

Quantity

Scalar Vector

Is any physical quantity can be described fully by a magnitude

only.

Is any physical quantity needs both magnitude and direction

to be fully described.

Examples: mass, length and time.

Examples: force, position and moment.

Scalar versus vectors

Direction (θ)

Sense of Direction

Fixed axis 30o

This vector has a magnitude of5 units and it tilted from thehorizontal axis by +30o

Example

Vector notations:

1. A

2.

3. V = M∟θA

Sense of Direction

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Vector operations

A

2A

0.5A-A

Multiplication by a scalar

30o

2A =10∟30o

0.5A =2.5∟30o

-A =-5∟30o = 5∟180+30o

Example

Vector operations

A

Vector addition (A+B) Vector subtraction (A-B)

B

R=A+B

A

B

R=A+B

A

B

A

-B

R=A-B

A

-B

R=A-B

A-B

OR

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Vector operations

Example:

Assume the following vectors: A = 10∟30o and B = 7∟-20 o

Find: 1. A+B2. A-B

Solution:

A+B

AB

A-B A

-B

Vector operations

Special cases:Collinear vectors:

Multiple addition:

A+B+C+…: in such case, the addition can be in successive order or in multiple steps.

A B

A + B

A + B

B - AA B

B-A

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Vector operations

Example:

Assume the following vectors: A = 10∟30o ,B = 7∟-20 o and C = 6∟135o . Find: A+B+(A+C)

Solution:

A+B

A+CA

BC

A

A+B+(A+C)

A+B

A+C

Vector operations

Example (cont): Find A+B+(A-C)

A+B

AB A+B

A

A+B+A

A+B+A-C

-C

A+B+A

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Examples

Examples

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Examples

Examples

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Coplanar forces

Coplanar forcesCoplanar forces are the forces that share the same plane.

These forces can be represented by their components on x and yaxes which are called the rectangular components.

The components can be represented by scalar and Cartesian notations.

Scalar notation

Cartesian notation

Fx = F cos(θ)

y

x

Fy

Fx

F

θ

Fy= F sin(θ)

Fy

Fx

F

x

i

y

j F = Fx i + Fy j

Scalar notation cases

y

x

Fy

Fx

F

θ

Fx = F cos(θ)

Fy= F sin(θ)

1 2

y

xFy

Fx

c

Fx /F= a/c

Fy /F= b/c

abF

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Scalar notation cases

Cartesian notation cases

Fy

Fx

F

x

i

y

jF2,y

F2

F2,x

F1,y

F1

F1,x

F4,xF4

F4,y

x

y

F3,x

F3,y

F3

F 1= F1,x i + F1,y j

F 2 = -F2,x i + F2,y j

F 3= -F3,x i - F3,y j F 4= F4,x i - F4,y j

Force summation

For both cases of notations, the magnitude of the resultant force is found by:

And the direction is found by:

FR,x =∑Fx

FR,y =∑Fy

2

yR,

2

xR,R FFF FR,y

FR,x

FR

θ

xR

yR

F

F

,

,1tan

+ve

+ve

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Examples

Examples

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Examples

Examples

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Cartesian Vectors

Forces

Three

dimensions

Two

dimensions

One

dimension

Cartesian Vectors

• In three dimensions system, new component appeared (Az).

• The new dimension direction is represented by unit vector (k)

• The position of the vector has to be located by three angles one from each axis

• The projection of vector on x-y plane represent a new

vector called A’

A

Az

z

y

x

Ax

Ay

A’

Note that

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Cartesian Vectors

Right hand coordinate system

• This method is used to describe the rectangular coordinate system.

• The system is said to be right handed if the thumb points to the positive z-axis and the fingers are curled about this axis and points from the positive x-axis to the positive y-axis.

Cartesian Vectors

Notation • To find the components of a

vector oriented in three dimensions, two successive applications of the parallel-ogram must be done.

• One of the parallelogram applications is to resolve A to A’and Az and the other is used to resolve A’ into Ax and Ay.

• A = A’ + Az = Ax + Ay + Az

A

Az

z

y

x

Ax

A y

A’

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Cartesian Vectors

Angles

• α is the angle between the vector and x-axis

• β is the angle between the vector and y-axis

• γ is the angle between the vector and z-axis

αβ

γ

Cartesian Vectors

Cartesian unit vectors

k

j

i

A = Ax i + Ay j + Az k

Ax = Acos(α)

Ay = Acos(β)

Az = Acos(γ)

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Cartesian Vectors

Magnitude and Direction

Magnitude

Direction

222

zyx AAAA

A

A

A

A

A

A

z

y

x

cos

cos

cos

1coscoscos 222

Examples

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Examples

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Cartesian Vectors

Other direction definition

A = Ax i + Ay j + Az k

kjiuA

A

A

A

A

A

A

zyxA

A

uA = cos(α) i + cos(β) j + cos(γ) k

A = |A| uA

= |A| cos(α) i + |A| cos(β) j + |A| cos(γ) k

= Ax i + Ay j + Az k

Cartesian Vectors

Other direction definition cont

Az = A cos(ϕ)

A’ = A sin(ϕ)

Ax = A’ cos(θ) = A sin(ϕ) cos(θ)

Ay = A’ sin(θ) = A sin(ϕ) sin(θ) θ

ϕ

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Cartesian Vectors

Cartesian Vectors Addition

• A = Ax i + Ay j + Az k

• B = Bx i + By j + Bz k

• R = A+ B

= (Ax + Bx ) i + (Ay + By) j +(Az + Bz ) k

General rule:

FR = ∑F = ∑Fx i + ∑Fy j + ∑Fz k

Cartesian Vectors

Example [1]:Question: assume the following forces:

F1 = 5 i + 6j -4k

F2 = -3i +3j +3k

F3=7i-12j+2k

Find the resultant force F = F1 + F2 +F3 and represent it in both Cartesian and scalar notations

Solution:

Given: F1, F2 and F3

Required: find resultant force F and represent it in both Cartesian and scalar notation

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Cartesian Vectors

Example [1]:

Solution:

F = (5-3+7)i + (6+3-2)j +(-4+3+2)k = 9i +7j +10k (Cartesian notation)

Scalar notation:

Magnitude:

Unit vector u:

NF 17.151079 222

kjikjiF

u 66.046.06.015.17

10 7 9

F

7.48 66.0cos

61.62 46.0cos

13.53 6.0cos

o

o

o

Cartesian Vectors

Position vector

Assume: r = ai + bj + ck

• At this point, r is called position vector.

• Position vector is a vectorlocates a point in space withrespect to other point.

• in this case, vector r is aposition vector relates point P

to point O.

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Cartesian Vectors

Position vector

• Assume vectors rA and rB are used to locate points A and B from the origin (ie. Point 0,0,0).

• Define a position vector r to relate point A to point B.

• rB = r + rA → r = rB – rA

• r = (xB – xA)i + (yB – yA)j + (zB – zA)k

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