Vector_mechanics [Compatibility Mode]

7/29/2019 Vector_mechanics [Compatibility Mode]

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Vectors

Vector classifications:

- Fixedorboundvectors have well defined points of

application that cannot be changed without affecting

an analysis.

- Free vectors may be freely moved in space without

changing their effect on an analysis.

Equal vectors have the same magnitude and direction.

Negative vector of a given vector has the same magnitude

and the opposite direction.

Engineers Mechanics- Review of Vector Algebra/Applications

Addition of Vectors

Parallelogram rule for vector addition

Triangle rule for vector addition

B

B

C

C

QPR

BPQQPR

cos2222 Law of cosines,

Law of sines,

P

C

R

B

Q

A sinsinsin

Vector addition is commutative,

PQQP

Vector subtraction



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Addition 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

Multiplication of a vector by a scalar



Addition of Vectors

A quantity which has magnitude and

direction, but doesnt follow

parallelogram law, cannot be a vector.

Can you name such a quantity?

Think in terms of associative property!!

Answer: Finite Rotation


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Application: Resultant of Several Concurrent Forces Concurrent forces: set of forces which all

pass through the same point.

A set of concurrent forces applied to a

particle may be replaced by a single

resultant force which is the vector sum of the

applied forces.

Vector force components: two or more force

vectors which, together, have the same effect

as a single force vector.


Rectangular Components of a Vector

cosFFx

Vector components may be expressed as products of

the unit vectors with the scalar magnitudes of the

vector components.

Fx andFy are referred to as the scalar components of

jFiFF yx

F

May resolve a vector into perpendicular components so

that the resulting parallelogram is a rectangle.

are referred to as rectangular vector components and

yx FFF

yx FF

and

Define perpendicularunit vectors which are

parallel to thex andy axes.ji

and

sinFFy



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Application: Addition of Concurrent Forces

SQPR

Wish to find the resultant of 3 or more

concurrent forces,

jSQPiSQPjSiSjQiQjPiPjRiR

yyyxxx

yxyxyxyx

Resolve each force into rectangular components

x

xxxx

F

SQPR

The scalar components of the resultant are

equal to the sum of the corresponding scalar

components of the given forces.

y

yyyy

F

SQPR

x

yyx

R

RRRR

122 tan

To find the resultant magnitude and direction,


Example-1

Two structural members A and B are bolted to

a bracket as shown. Knowing that both

members are in compression and that the force

is 20 kN in member A and 30 kN in member B,

determine, using trigonometry, the magnitude

and direction of the resultant of the forces

applied to the bracket by members A and B.

SOLUTION KEY

o Construct the force triangle

and apply the sine and

cosine rules.


Example of a timber truss joint


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SOLUTION


Example-1

A collar that can slide on a vertical rod is subjected to

the three forces shown. Determine (a) the value of the

angle for which the resultant of the three forces is

horizontal, (b) the corresponding magnitude of the

resultant.

SOLUTION KEY

o Since the resultant (R) is to be

horizontal, sum of the vertical

comp. of the forces, i.e., Ry = 0.Example of an Umbrella


Example-2


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1 - 11

0y yR F

90 lb 70 lb sin 130 lb cos 0

13 cos 7 sin 9

213 1 sin 7 sin 9

2 2169 1 sin 49 sin 126 sin 81 2218 sin 126 sin 88 0

sin 0.40899

1.24

(a) Since R is to be horizontal,Ry = 0

Then,

Squaringboth sides:

Solving by quadratic formula:

or,

SOLUTION


Example-2

1 -12

117.0 lbR or,


Example-2


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Rectangular Components of a Vector in Space

1coscoscos222 zyx

Components of the vectorF

kji

FkjiF

kFjFiFF

FFFFFF

zyx

zyx

zyx

zzyyxx

coscoscos

coscoscos

coscoscos

is a unit vector along the line of action of

and are the direction

cosines for

F

F

zyx cosand,cos,cos222

zyx FFFF


Application: Rectangular Components of a Force Vector in Space

222

zyx dddd

The magnitude of the force vector is

F and the direction of the force is

defined by the location of two

points,

222111 ,,and,, zyxNzyxM

d

FdF

d

FdF

d

FdF

kdjdidd

FF

zzdyydxxdkdjdid

NMd

zz

yy

xx

zyx

zyx

zyx

1

andjoiningvector

121212

d

d

d

d

d

d zz

y

yx

x cos;cos;cos



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Scalar Product of Two Vectors The scalar productordot productbetween

two vectors P andQ is defined as

resultscalarcosPQQP

Scalar products:

- are commutative,

- are distributive,

- are not associative,

PQQP

2121 QPQPQQP

undefined SQP

Scalar products with Cartesian unit components,

000111 ikkjjikkjjii

kQjQiQkPjPiPQP zyxzyx

2222 PPPPPP

QPQPQPQP

zyx

zzyyxx


Applications: Scalar Product of Two Force Vectors

Angle between two force vectors:

PQ

QPQPQP

QPQPQPPQQP

zzyyxx

zzyyxx

cos

cos

Projection of a force vector on a given axis:

OL

OL

PPQ

QP

PQQP

OLPPP

cos

cos

alongofprojectioncos

zzyyxx

OL

PPP

PP

coscoscos

- For an axis defined by a unit vector ():

Q

Q

Note:



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Vector Product of Two Vectors

Vector product of two vectorsP andQ is defined

as the vectorVwhich satisfies the following

conditions:

1. Line of action ofVis perpendicular to plane

containingP andQ.

2. Magnitude ofVis

3. Direction ofVis obtained from the right-hand

rule.

sinQPV

Vector products:

- are not commutative,

- are distributive,

- are not associative,

QPPQ

2121 QPQPQQP

SQPSQP


Vector Products: Rectangular Components

Vector products of Cartesian unit vectors,

0

0

0

kkikjjki

ijkjjkji

jikkijii

Vector products in terms of rectangular

coordinates

kQjQiQkPjPiPV zyxzyx

kQPQP

jQPQPiQPQP

xyyx

zxxzyzzy

zyx

zyx

QQQ

PPP

kji



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Application: Moment of a Force About a Point Moment of a force produces a turning action on a

rigid body

The momentof a force F about O is defined as

FrMO


r is the position vector of A from O

The moment vectorMO is perpendicular to the

plane containing O and the forceF.

Magnitude ofMO measures the tendency of the forceto cause rotation of the body about an axis alongMO.

dis the perpendicular distance of the line of action

of Force F from O. The sense of the moment may be

determined by the right-hand rule.

FdrFMO sin


Examples: Application of moments


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Application: Rectangular Components of the Moment of a Force about origin

kyFxFjxFzFizFyF

FFF

zyx

kji

kMjMiMM

xyzxyz

zyx

zyxO

The moment of force F applied at A about O,

kFjFiFF

kzjyixrFrM

zyx

O

,


Appl icati on: Rectangular Components of the Moment of a Force about an

Arbi trary Poi nt

The moment of force F applied at A aboutB,

FrM BAB

/

kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

/

zyx

BABABAB

FFF

zzyyxx

kji

M



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A 36-N force is applied to a wrench to

tighten a showerhead. Knowing that the

centerline of the wrench is parallel to the x

axis. Determine the moment of the force

aboutA.

Example-3

SOLUTION KEY

o Find out the position vector of the

point C .

o Get the components of the appliedforce (F) along X, Y and Z directions.


Example - 3

/A C A M r F

/ 215 mm 50 mm 140 mmC A r i j k

36 N cos45 sin12xF 36 N sin 45yF

36 N cos 45 cos12zF

5.2926 N 25.456 N 24.900 N F i j k

where

, ,

0.215 0.050 0.140 N m

5.2926 25.456 24.900

A

i j k

M

4.8088 N m 4.6125 N m 5.7377 N m i j k

and

4.81 N m 4.61 N m 5.74 N mA M i j k

SOLUTION



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Mixed Triple Product of Three Vectors Mixed triple product of three vectors,

resultscalar QPS

The six mixed triple products formed from S, P, and

Q have equal magnitudes but not the same sign,

SPQQSPPQS

PSQSQPQPS

zyx

zyx

zyx

xyyxz

zxxzyyzzyx

QQQ

PPP

SSS

QPQPS

QPQPSQPQPSQPS

Evaluating the mixed triple product,

P

QS

= volume of parallelepiped


Application: Moment of a Force About an Axis Passing Through the Origin

kzjyixr

MomentMO of a force F applied at the point A

about a point O,

FrMO

Scalar momentMOL about an axis OL is the

projection of the moment vectorMO onto the

axis,

FrMM OOL

Moments ofF about the coordinate axes

(components of )

xyz

zxy

yzx

yFxFM

xFzFM

zFyFM


OM


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Application: Moment of a Force About an Arbitrary Axis

Moment of a force about an arbitrary axis,

BABA

BA

BBL

rrr

Fr

MM

The result is independent of the location

of pointB along the given axis!!


Example - 4


The frame ACD is hinged at A andD and is supported by a cable that

passes through a ring at B and is attached to hooks at G andH. Knowing

that the tension in the cable is 1125 N, determine the moment about the

diagonal AD of the force exerted on the frame by portion BHof the cable.

SOLUTION KEY

o Find unit vector along AD.

o Find the force vector BH.

o Choose either point A or D , get the

position vector AB or DB

/AD AD B A BHM r T


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Example - 4


/AD AD B A BHM r T

2 2

0.8 m 0.6 m0.8 0.6

0.8 m 0.6 mAD

i k i k

/ 0.4 mB A r i

2 2 2

0.3 m 0.6 m 0.6 m1125 N

0.3 0.6 0.6 mBH BH

BHT

BH

i j kT

0.8 0 0.6

0.4 0 0 180 N m

375 750 750

ADM

180.0 N mADM

Moment of a Couple Couple Moment

Two forces F and -F having the same magnitude,

parallel lines of action, and opposite sense are said

to form a couple.

Moment of the couple,

CFdrFM

CFr

Frr

FrFrM

BA

BA

sin

Special Notation

The moment vector of the couple is

independent of the choice of the origin of the

coordinate axes, i.e., it is afree vectorthat can

be applied at any point with the same effect.



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Moment of a Couple

Two couples will have equal moments if

2211 dFdF

the two couples lie in parallel planes, and

the two couples have the same sense or

the tendency to cause rotation in the same

direction.


Addition of Couples

Consider two intersecting planes P1 and

P2 with each containing a couple

222

111

planein

planein

PFrM

PFrM

Resultants of the vectors also form a

couple

21 FFrRrM

Sum of two couples is also a couple that is equal

to the vector sum of the two couples

21

21

MM

FrFrM



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Summary Couple

A couple can be represented by a vector with magnitude

and direction equal to the moment of the couple.

Couple vectors obey the law of addition of vectors.

Couple vectors are free vectors, i.e., the point of application

is not significant.

Couple vectors may be resolved into component vectors.


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Vector_mechanics [Compatibility Mode]