Rectangular Components in Space Notes/ME101-Lecture03-KD.pdf · Rectangular Components in Space Example: The tension in the guy wire is 2500 N. Determine: a) components F x, F y,

Rectangular Components in Space

• Vector F is

contained in the

plane OBAC

• Resolve F into

horizontal and

vertical

components.

yh FF sin

yy FF cos

• Resolve Fh into

rectangular

components

sinsin

sin

cossin

cos

y

hz

y

hx

F

FF

F

FF

1ME101 - Division III Kaustubh Dasgupta

Spatial Components (Direction Cosines)


Spatial Components (Direction Cosines)

• With the angles between F and the axes,

kji

F

kjiF

kFjFiFF

FFFFFF

zyx

zyx

zyx

zzyyxx

coscoscos

coscoscos

coscoscos


• is a unit vector along the line of

action of F;

are the direction cosines

zyx cos and,cos,cos


• Direction of force F

– Defined by location of two points

• M (x1, y1 and z1) and N (x2, y2 and z2)

d

FdF

d

FdF

d

FdF

kdjdidd

FF

zzdyydxxd

kdjdid

NMd

zz

yy

xx

zyx

zyx

zyx

1

and joining vector

121212


Rectangular Components in SpaceExample: 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

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.


Rectangular Components in SpaceSolution

• Determine the unit vector pointing from Atowards B.

m 3.94

m30m80m40

m30m80m40

222

AB

kjiAB

• Determine the components of the force.

kji

kji

FF

N 795N 2120N1060

318.0848.0424.0N 2500

kji

kji

318.0848.0424.0

3.94

30

3.94

80

3.94

40


Solution

• Noting that the components of the unit

vector are the direction cosines for the

vector, calculate the corresponding angles.

kji

kji zyx

318.0848.0424.0

coscoscos

5.71

0.32

1.115

z

y

x



Product of 2 Vectors: Dot Product

• Dot Product (Scalar product)

– A.B = AB cos θ

– Applications

• Determination of the angle between two vectors

A

B

A.B = (Axi+Ayj+Azk).(Bxi+Byj+Bzk) = AxBx+AyBy+AzBz

A.B = ABcos

Obtain


Product of 2 Vectors: Dot Product

• Applications

– Determination of the projection of a vector on a

given axis

PAB

A.B = AxBx+AyBy+AzBz

PAB = Acos = (A.B)/B

AB


Product of 2 Vectors: Cross Product

• Cross Product (Vector Product)

– C = A B

• C = AB sin θ



• Cross Product

Cartesian Vector



• Cross Product

– Distributive property

• C (A+B) = C A + C B

i j k

Ax AY AZ

BX BY BZ

=


AB = (Axi+Ayj+Azk) (Bxi+Byj+Bzk)

= (AyBz- AzBy)i + (….)j + (….)k

Moment of a Force (Torque)

• Moment of a Force (F) @ point A

– Mo = r F


r = position

vector

directed from

O to any

point on the

line of

action of F

Moment of a Force

• Magnitude of Moment

– tendency of F to cause rotation of the body

about an axis along MO

FdrFrFMO )sin(sin


Moment of a Force

• Characteristic

– Moment arm (d = r sinα) does not depend on the

particular point on the line of action of F to

which the vector r is directed

• Sliding vector

– Line of action same as the moment axis


Moment of a Force

• Principle of Transmissibility

– Two forces are equivalent

• Same magnitude, direction and lines of action

• Same magnitude, direction and equal moments

about a given point


Moment of a Force

• Varignon’s Theorem

– Moment of a force about a point is equal to the

sum of the moments of the components of

the force about the same point


Moment of a System of Concurrent Forces

• Varignon’s Theorem

– Moment of the resultant of a system of

concurrent forces about a point is equal to the

sum of the moments of the of the individual

forces about the same point


kyFxFjxFzFizFyF

FFF

zyx

kji

kMjMiMM

xyzxyz

zyx

zyxO

The moment of F about O,

kFjFiFF

kzjyixrFrM

zyx

O

,

Rectangular Components of Moment


The moment of F about B,

FrM BAB

/

zyx

BABABAB

FFF

zzyyxx

kji

M

kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

/

Rectangular Components of Moment


• Moment MO of a force F applied at

the point A about a point O,

FrMO

• Scalar moment MOL about an axis

OL is the projection of the

moment vector MO onto the axis,

FrMM OOL

Moment of a Force About a Given Axis


Application of Scalar Product


• Significance of MOL

– Resolve into components



• Significance of MOL

– MOL is a measure of the tendency of the F to

impart a rigid body rotation about the axis OL

• Moments of F about the coordinate axes

xyz

zxy

yzx

yFxFM

xFzFM

zFyFM


Moment: Example


Moment: Example


Moment produced by two equal, opposite and

noncollinear forces is called a couple.

Magnitude of the combined moment of the two

forces about O:

M = F(a+d) – Fa = Fd

Vector Algebra Method:

Moment of the couple

about point O:

FdrFM

Fr

Frr

FrFrM

BA

BA

sin

Moment of a Couple

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.


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.

Examples:

Moment of a Couple


• 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

• By Varignon’s theorem

21

21

MM

FrFrM

• Sum of two couples is also a couple that is

equal to the vector sum of the two couples

Addition of Couples


• 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.

Representation of Couples by Vectors


Couple: Example

Moment reqd to turn the shaft connected at center of

the wheel = 12 Nm

• First case: Couple Moment

produced by 40 N forces = 12 Nm

• Second case: Couple Moment

produced by 30 N forces = 12 Nm

If only One hand is used F = 60N

Same couple moment will be produced

even if the shaft is not connected at the

center of the wheel

Couple Moment is a Free VectorL04

Equivalent Systems (Force-Couple Systems)

At support O:

WR = W1 + W2

(MR)O = W1d1 + W2d2


Equivalent Systems: Resultants

FR = F1 + F2 + F3

How to find d?

Moment of the Resultant force about the grip must be equal

to the moment of the forces about the grip

FRd = F1d1 + F2d2 + F3d3

Equilibrium Conditions


Rectangular Components in Space Notes/ME101-Lecture03-KD.pdf · Rectangular Components in Space Example: The tension in the guy wire is 2500 N. Determine: a) components F x, F y,

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