MEM202 Engineering Mechanics - Statics 2.5 Resolution of A ...cac542/L2.pdf3 MEM202 Engineering...

1

MEM202 Engineering Mechanics - Statics MEM

2.5 Resolution of A Force into Components(Parallelograms and Laws of Sines and Cosines)

Ar

Br

Ar

Br

DC

BARrr

rrr

+=

+=

Cr

Dr

Rr

αβ γ φ

α

Rr

βαφ += φπβαπγ −=−−=

γβα sinsinsin :Sines ofLaw RBA

==

RAγα

sinsin

= RBγβ

sinsin

=

2


2.5 Resolution of A Force into Components(Parallelograms and Laws of Sines and Cosines - Examples)

Resolve 900 N into u- and v-components as shownooo 110sin

90025sin45sin

== vu FF

N 405110sin

25sin900

N 677110sin

45sin900

==

==

o

o

o

o

v

u

F

F

Determine F2 and α so that the resultant of F1 and F2 is a 1,000 lb force in the x-direction

lb 81838cos2 122

12 =−+= oRFRFF

oo

o85.1038sin

818250sin

38sin818

sin250 1 =⎟

⎠⎞

⎜⎝⎛=⇒= −α

α

Homework: Problems 2-31, 2-33, 2-37, 2-40

3


2.6 Rectangular Components of A ForceTwo Dimensions

x

yFr

iFF xx

rr=

jFF yy

rr=

θ

ir

jr

direction- inr unit vecto A

direction- inr unit vecto A

yj

xi

=

=r

r

FyθFF

FxθFF

y

xr

r

ofcomponent scalar :sin

ofcomponent scalar :cos

=

=

( ) ( ) jFiFjFiFF yx

rrrrrθθ sincos +=+=

x

y

FF1tan−=θ

4


2.6 Rectangular Components of A ForceTwo Dimensions - Example

lb 4.16669.33sin300

sinlb 25069.33cos300

cos

111

111

==

===

=

o

o

θ

θ

FF

FF

y

x

o69.3332tan 1

1 == −θ

( )

( ) lb 4.13725sin325

sinlb 29525cos325

cos

222

222

−=−=

==−=

=

o

o

θ

θ

FF

FF

y

x

o252 −=θ

jiFrrr

4.1662501 += jiFrrr

4.1372952 −=

Determine the xand y scalar components for the two forces shown.

5


2.6 Rectangular Components of A ForceTwo Dimensions

x

yFr

iFF xx

rr=

jFF yy

rr=

θ xθ

yθ Fe

F FFFF

F

yxrr

rr

of direction theinr unit vecto A :

of Magnitude:22 +==

θθπθ

θθ

sin2

coscos

coscos

FFFF

FFF

yy

xx

=⎟⎠⎞

⎜⎝⎛ −==

==Fyx eFjFiFF rrrr

=+=

ir

jr

Fer

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛=

+=+=

−−

FF

θFFθ

jθiθjFF

iFFe

yy

xx

yxyx

F

11 coscos

coscosrrrrr

jFiF

jFiF

eFjFiFF

yx

Fyx

rr

rr

vrrr

θθ

θθ

coscos

sincos

+=

+=

=+=

6


2.6 Rectangular Components of A ForceTwo Dimensions - Example

lb 4.166coslb 250cos

111

111

====

yy

xx

FFFF

θθ

oo 31.5669.33 11 == yx θθ

lb 4.137coslb 295cos

222

222

−====

yy

xx

FFFF

θθ

( ) oooo 1152590 25 22 =−−=−= yx θθ

jieFF F

rrrr4.166250

111 +== jieFF F

rrrr4.137295

222 −==

ji

jie yxFrr

Determine the xand y scalar components for the two forces shown. rrr

555.0832.0

coscos 111

+=

+= θθ

ji

jie yxFrr

rrr

423.0906.0

coscos 222

−=

+= θθ

7


x

y

z

Fr

xF

yF

zF

xθyθ

zθ

ir j

rkr

Fer

2.6 Rectangular Components of A ForceThree Dimensions

Fzyx eFkFjFiFF rrrrr=++=

⎟⎠⎞

⎜⎝⎛==

⎟⎟⎠

⎞⎜⎜⎝

⎛==

⎟⎠⎞

⎜⎝⎛==

−

−

−

FFFF

FF

FF

FFFF

zzzz

yyyy

xxxx

1

1

1

coscos

coscos

coscos

θθ

θθ

θθ

222zyx FFFF ++=

axes coordinate positive and between Angles:,,

alongr unit vecto A:

componentsScalar :,, of Magnitude:

F

Fe

FFFFF

zyx

F

zyx

r

rr

r

θθθkji

kFFj

FF

iFFe

zyx

zyxF

rrr

rrrr

θθθ coscoscos ++=

++=

1coscoscos :NOTE 222 =++ zyx θθθ

8


2.6 Rectangular Components of A ForceThree Dimensions - Example

lb 6340.65cos500,1cos

lb 278,16.31cos500,1cos

lb 4640.72cos500,1cos

===

===

===

o

o

o

zz

yy

xx

FF

FF

FF

θ

θ

θ

lb 500,1222 =++= zyx FFFF

kji

kjie zyxFrrr

rrrr

42268517.03090.0

coscoscos

++=

++= θθθ

lb 634278,1464 kjieFkFjFiFF Fzyx

rrrrrrrr++==++=

9


2.6 Rectangular Components of A ForceThree Dimensions - Example

( ) ( ) ( ) m 123.4223 222 =+−+−=d

o

o

o

0.61123.42coscos

0.119123.42coscos

7.136123.43coscos

11

11

11

===

=−

==

=−

==

−−

−−

−−

dzdydx

z

y

x

θ

θ

θ

kN 3.240.61cos50cos

kN 3.240.119cos50cos

kN 4.367.136cos50cos

===

−===

−===

o

o

o

zz

yy

xx

FF

FF

FF

θ

θ

θ

lb 3.243.244.36 kjikFjFiFF zyx

rrrrrrr+−−=++=

10


2.6 Rectangular Components of A ForceAlternative Method: Azimuth and Elevation Angles

x

y

z

Fr

xF

yF

zF

θ

φ

Angle Elevation:Angle Azimuth :

φθ

θφθ

θφθφ

φ

sincossincoscoscos

sincos

FFFFFF

FFFF

xyy

xyx

z

xy

==

===

=

xyF

11


2.6 Rectangular Components of A ForceAlternative Method: Azimuth and Elevation Angles – Example

oo 47 56 == φθ

lb 42456sin512sin lb 28656cos512cos

lb 54947sin750sin lb 51247cos750cos

======

======oo

oo

θFFθFF

FFFF

xyyxyx

zxy φφ

lb 549424286 kjikFjFiFF zyx

rrrrrrr++=++=

12


2.6 Rectangular Components of A ForceA Force Defined by Two Points, A(xA, yA, zA) and B(xB, yB, zB)

x

y

z

Fr

AB

AxAy

Az

Bx

By

Bz

( ) ( ) ( )222ABABAB zzyyxxF −+−+−=

kjie zyxF

rrrr θθθ coscoscos ++=

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )222

222

222

cos

cos

cos

ABABAB

ABz

ABABAB

ABy

ABABAB

ABx

zzyyxx

zzzzyyxx

yyzzyyxx

xx

−+−+−

−=

−+−+−

−=

−+−+−

−=

θ

θ

θ

ABzz

AByy

ABxx

zzFFyyFFxxFF

−==

−==−==

θ

θθ

coscoscos

FeFF rr=

13


2.6 Rectangular Components of A ForceA Force Defined by Two Points, A(xA, yA, zA) and B(xB, yB, zB) - Example

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )2233.0cos

8930.0cos

3907.0cos

222

222

222

=−+−+−

−=

=−+−+−

−=

=−+−+−

−=

ABABAB

ABz

ABABAB

ABy

ABABAB

ABx

zzyyxx

zzzzyyxx

yyzzyyxx

xx

θ

θ

θ

N 223cosN 893cosN 391cos

==

====

zz

yy

xx

FFFFFF

θ

θθ

mm 500mm 750mm 475mm 400mm 350mm 250

======

BBB

AAA

zyxzyx

lb 223893391 kjikFjFiFF zyx

rrrrrrr++=++=

14


2.6 Rectangular Components of A ForceA Force Defined by Two Points, A(xA, yA, zA) and B(xB, yB, zB)

If A is at origin, xA = yA = zA = 0

Bzz

Byy

Bxx

zFFyFFxFF

==

====

θ

θθ

coscoscos

x

y

z

Fr

A

B

Bx

By

Bz

222BBB zyxF ++=

kjie zyxF


222

222

222

cos

cos

cos

BBB

Bz

BBB

By

BBB

Bx

zyxz

zyxy

zyxx

++=

++=

++=

θ

θ

θ

FeFF rr=

15


2.6 Rectangular Components of A ForceAlong an arbitrary direction n

x

y

z

Fr

Ferα

FeFF rr= nen of Direction:r

( )nFnFnn eeFeeFeFF rrrrrr⋅=⋅=⋅=

ner( ) ( ) nnFnnnnn eeeFeeFeFF rrrrrrrr

⋅=⋅==

( )( )

FF

FeF

FFeF

nn

nn

11 coscos

cos1

−− =⋅

=⇒

==⋅rr

rr

α

α

16


2.6 Rectangular Components of A ForceAlong an arbitrary direction n - Example

Determine the rectangular scalar component Fn of F along line OA

( ) ( )( )( ) ( )( ) ( )( ) lb 5313245.08738112.04364867.0218

3245.08112.04867.0873436218=++−=

++−⋅++=

⋅=

kjikji

eFF nnrrrrrr

rr

( ) ( ) ( )kji

kjien

rrr

rrrr

3245.08112.04867.0

253

253222

++−=

++−

++−=

( )( ) orr9.57

000,1531coscoscos1 11 ===⇒==⋅ −−

FFFFeF n

nn αα

lb 873436218 kjieFF F

rrrrr++==

( ) ( ) ( ) lb 000,1873436218 222 =++=F

17


2.7 Resultants by Rectangular ComponentsExample

( ) ( ) ( )kji

kjie

rrr

rrrr

6674.04767.05721.0

5.35.23

5.35.232221

+−=

+−+

+−=

( ) ( ) ( )kji

kjie

rrr

rrrr

5433.06985.04657.0

5.35.43

5.35.432222

+−−=

+−+−

+−−=

( ) ( ) ( )kji

kjie

rrr

rrrr

5657.07071.04243.0

453

4532223

++=

++

++=

kip 628.22284.28972.16

kip 299.16955.20971.13

kip 348.13534.9442.11

333333

222222

111111

kFjFiFkjieFF

kFjFiFkjieFF

kFjFiFkjieFF

zyx

zyx

zyx

rrrrrrrr

rrrrrrrr

rrrrrrrr

++=++==

++=+−−==

++=+−==

18


2.7 Resultants by Rectangular ComponentsExample

kip 628.22284.28972.16

kip 299.16955.20971.13

kip 348.13534.9442.11

333

222

111

kjieFF

kjieFF

kjieFF

rrrrr

rrrrr

rrrrr

++==

+−−==

+−==

( ) ( ) ( )kip 275.52205.2443.14 kji

kFjFiF

kRjRiRR

iziyix

zyx

rrr

rrr

rrrr

+−=

++=

++=

∑∑∑

kip 3.54222 =++= zyx RRRR

R

ooo 6.15cos 3.92cos 6.74cos 111 ====== −−−

RR

RR

RR z

zy

yx

x θθθ

19


Concurrent Force Systems – A Summary (2-D)

xF4

r

yF1

r

yF4

r

043214123

321312123

2112

=+++=+=

++=+=

+=

FFFFFRR

FFFFRR

FFR

rrrrrrr

rrrrrr

rrr

1Fr

2Fr

3Fr

4Fr

04321 =+++== ∑ FFFFFRrrrrrr

Parallelogram and Laws of Sines and Cosines

Rectangular Components

1F Vector Additionr

2Fr

3Fr

4Fr

12Rr

123Rr

00

0

4321

4321

=+++==+++=

=+=

yyyyy

xxxxx

yx

FFFFRFFFFR

jRiRRrrr

1Fr

2Fr

4Fr

xF1

rxF2

r

yF2

r

xFF 33

rr=

04321 =+++== ∑ FFFFFRrrrrrr

1Fr

2Fr

3Fr

4Fr

xF4

r

yF1

r

yF4

r xF1

r xF2

r

yF2

r

20


kFjFiFF

kFjFiFFkFjFiFF

nznynxn

zyx

zyx

vrrrM

vrrr

vrrr

++=

++=++=

2222

1111

∑∑∑

=+++=

=+++=

=+++=

++=

znzzzz

ynyyyy

xnxxxx

xxx

FFFFR

FFFFR

FFFFRiRiRiRR

L

L

L

rrrr

21

21

21

222zyx RRRR ++=

RF

RR

RF

RR

RF

RR

zzz

yyy

xxx

∑

∑

∑

−−

−−

−−

==

==

==

11

11

11

coscos

coscos

coscos

θ

θ

θ

ReRR rr =

kjie zyxR


Concurrent Force Systems – A Summary (3-D)Rectangular Components

Forces in components

Resultant in components

Magnitude and direction of the resultant

MEM202 Engineering Mechanics - Statics 2.5 Resolution of A ...cac542/L2.pdf3 MEM202 Engineering...

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