Vector mechanics [compatibility mode]

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Vectors

• Vector classifications:- Fixed or bound vectors 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

QPRBPQQPR

cos2222• Law of cosines,

• Law of sines,

PC

RB

QA 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 doesn’t follow parallelogram law, cannot be a vector.

Can you name such a quantity?

Think in terms of commutative 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 and Fy 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 perpendicular unit vectors which are parallel to the x and y 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

xxxxx

FSQPR

• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

y

yyyyF

SQPR

x

yyx R

RRRR 122 tan

• To find the resultant magnitude and direction,


Example Two structural members A and B are bolted toa bracket as shown. Knowing that bothmembers are in compression and that the forceis 20 kN in member A and 30 kN in member B,determine, using trigonometry, the magnitudeand direction of the resultant of the forcesapplied 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

A collar that can slide on a vertical rod is subjected tothe three forces shown. Determine (a) the value of theangle for which the resultant of the three forces ishorizontal, (b) the corresponding magnitude of theresultant.

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

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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,

Squaring both sides:

Solving by quadratic formula:

or,

SOLUTION


Example

1 - 12

117.0 lbR or,


Example-2

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Rectangular Components of a Vector in Space

1coscoscos 222 zyx

• Components of the vector F

kjiF

kjiF

kFjFiFF

FFFFFF

zyx

zyx

zyx

zzyyxx

coscoscos

coscoscos

coscoscos

• is a unit vector along the line of action ofand are the direction cosines for

F

F

zyx cos and,cos,cos222

zyx FFFF


Application: Rectangular Components of a Force Vector in Space

222zyx 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

dFdF

dFd

Fd

FdF

kdjdidd

FF

zzdyydxxd

kdjdidNMd

zz

yy

xx

zyx

zyx

zyx

1

and joining vector

121212

dd

dd

dd z

zy

yx

x cos;cos;cos


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Example

2 - 15

The tension in the guy wire is 2500 N. Determine:

a) components Fx, Fy, Fz of the force acting on the bolt at A,

b) the angles qx, qy, qz defining the direction of the force (the direction cosines)

SOLUTION:

• Based on the relative locations of the points A and B, determine the unit vector pointing from A towards B.

• Apply the unit vector to determine the components of the force acting on A.

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.


Example

2 - 16

SOLUTION:• Determine the unit vector pointing from A

towards B.

m394m30m80m40

m30m80m40222

.AB

AB

kji

• Determine the components of the force.

kji

kjiλF

N 795N 2120N1060318084804240N 2500

...

F

kji

kjiλ

318084804240394

30394

80394

40

......


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Example

2 - 17

• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

kjikjiλ

318084804240coscoscos

...zyx

5.71

0.32

1.115

z

y

x


2 - 18

What are the components of the force in the wire at point B? Can you find it without doing any calculations?

SOLUTION:

• Since the force in the guy wire must be the same throughout its length, the force at B (and acting toward A) must be the same magnitude but opposite in direction to the force at A.

kjiFF

N 795N 2120N1060 ABBA

FBA

FAB


Example

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Scalar Product of Two Vectors• The scalar product or dot product between

two vectors P and Q 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:

PQQPQPQP

QPQPQPPQQP

zzyyxx

zzyyxx

cos

cos

• Projection of a force vector on a given axis:

OL

OL

PPQ

QPPQQP

OLPPP

cos

cos

along of projection cos

zzyyxx

OL

PPPPP

coscoscos

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

Note:


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Vector Product of Two Vectors

• Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions:1. Line of action of V is perpendicular to plane

containing P and Q.2. Magnitude of V is3. Direction of V is 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,

00

0

kkikjjkiijkjjkji

jikkijii

• Vector products in terms of rectangular coordinates

kQjQiQkPjPiPV zyxzyx

kQPQP

jQPQPiQPQP

xyyx

zxxzyzzy

zyx

zyxQQQPPPkji


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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 moment of a force F about O is defined asFrMO


r is the position vector of A from O

• The moment vector MO is perpendicular to the plane containing O and the force F.

• Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO.

d is 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

Moment of a Force About a Point

3 - 24

• Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained only in the plane of the structure.

• The plane of the structure contains the point O and the force F. MO, the moment of the force about O, is perpendicular to the plane.

• If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive.

• If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.


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Examples: Application of moments

Varignon’s Theorem

• The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various forces about the same point O.

• Varignon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F.

2121 FrFrFFr


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

kyFxFjxFzFizFyF

FFFzyxkji

kMjMiMM

xyzxyz

zyx

zyxO

The moment of force F applied at A about O,

kFjFiFFkzjyixrFrM

zyx

O

,


Application: Rectangular Components of the Moment of a Force about an Arbitrary Point

The moment of force F applied at A about B,

FrM BAB

/

kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

/

zyx

BABABAB

FFFzzyyxx

kjiM


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Rectangular Components of the Moment of a ForceFor two-dimensional structures,

xy

ZO

xyO

yFxFMM

yFxF

kM

xBAyBA

ZB

xBAyBAB

FyyFxxMM

FyyFxx

kM


Example

A 500-N vertical force is applied to the end of a lever which is attached to a shaft (not shown) at O.

Determine:

a) the moment about O,

b) the horizontal force at A which creates the same moment,

c) the smallest force at A which produces the same moment,

d) the location for a 1200-N vertical force to produce the same moment

0.6 m500 N


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Example

a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper which, by our sign convention, would be negative or counterclockwise.

500 N0.6 m

N.m150 m0.3N 500

m3060cosm60

O

O

M..d

FdM

mN 150 OM


Example

Why must the direction of this F be to the right?

b) Horizontal force at A that produces the same moment,

0.6 m

m520 mN 150m520 mN 150

m52060sin m60

.F

.FFdM

..dO

N 5288.F


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Example

What is the smallest force at A which produces the same moment?

F(min) ?

0.6 m

c) The smallest force at A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.

m60 mN 150

m60 mN 150

.F

.FFdMO

N 250F 30o


Example

d) To determine the point of application of a 1200 N force to produce the same moment,

60cos

m1250N 0120 mN 150

N 1200 mN 150

OBd

.d

dFdMO

m250.OB

1200 N


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A 36-N force is applied to a wrench totighten a showerhead. Knowing that thecenterline of the wrench is parallel to the xaxis. Determine the moment of the forceabout A.

Example

SOLUTION KEY

o Find out the position vector of the point ‘C’ .

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


Example /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 m5.2926 25.456 24.900

A

i j kM

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

QQQPPPSSS

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

• Moment MO of a force F applied at the point Aabout a point O,

FrM O

• Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis,

FrMM OOL

• Moments of F about the coordinate axes (components of )

xyz

zxy

yzx

yFxFMxFzFMzFyFM


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

rrrFr

MM

• The result is independent of the location of point B along the given axis!!


Example


The frame ACD is hinged at A and D and is supported by a cable thatpasses through a ring at B and is attached to hooks at G and H. Knowingthat the tension in the cable is 1125 N, determine the moment about thediagonal AD of the force exerted on the frame by portion BH of 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 λ r T

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Example


/AD AD B A BHM λ r T

2 2

0.8 m 0.6 m0.8 0.6

0.8 m 0.6 mAD

i kλ 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

BHTBH

i j kT

0.8 0 0.60.4 0 0 180 N m375 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,

CFdrFMCFr

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 a free vector that 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

PFrMPFrM

• 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

MMFrFrM


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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.


Engineering

Vector mechanics [compatibility mode]