ch03 rigid bodies equivalent systems of forces

8/9/2019 Ch03 Rigid Bodies Equivalent Systems of Forces

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Vector Mechanics for Engineers: StaticsE i gh t h

3 - 1

CE 102 Statics

hapter 3

Rigid Bodies:

Equivalent Systems of Forces





3 - 2

Contents

Introduction

External and Internal Forces

Principle of Transmissibility: Equivalent

Forces

Vector Products of Two Vectors

Moment of a Force About a PointVarion!s T"eorem

#ectanular $omponents of t"e Moment

of a Force

%ample Problem &'(

%calar Product of Two Vectors

%calar Product of Two Vectors:

Applications

Mixed Triple Product of T"ree Vectors

Moment of a Force About a )iven Axis

%ample Problem &'*

Moment of a $ouple

Addition of $ouples

$ouples $an +e #epresented +y Vectors

#esolution of a Force Into a Force at O and a$ouple

%ample Problem &'&

%ystem of Forces: #eduction to a Force and a

$ouple

Furt"er #eduction of a %ystem of Forces

%ample Problem &',

%ample Problem &'-





3 - 3

Introduction. Treatment of a body as a sinle particle is not always possible' In

eneral/ t"e si0e of t"e body and t"e specific points of application of t"eforces must be considered'

. Most bodies in elementary mec"anics are assumed to be riid/ i'e'/ t"e

actual deformations are small and do not affect t"e conditions of

equilibrium or motion of t"e body'

. $urrent c"apter describes t"e effect of forces exerted on a riid body and

"ow to replace a iven system of forces wit" a simpler equivalent system'

. moment of a force about a point

. moment of a force about an axis

. moment due to a couple

. Any system of forces actin on a riid body can be replaced by an

equivalent system consistin of one force actin at a iven point and one

couple'




Vector Mechanics for Engineers: StaticsEi gh t h

3 - 4

External and Internal Forces

. Forces actin on riid bodies are

divided into two roups:1 External forces

1 Internal forces

. External forces are s"own in a

free1body diaram'

. If unopposed/ eac" external

force can impart a motion of

translation or rotation/ or bot"'

E





3 - 5

Principle of Transmissibility: Equivalent Forces

. Principle of Transmissibility 1

$onditions of equilibrium or motionare not affected by transmitting a force

alon its line of action'

23TE: F and F! are equivalent forces'

. Movin t"e point of application of

t"e force F to t"e rear bumper

does not affect t"e motion or t"e

ot"er forces actin on t"e truc4'

. Principle of transmissibility may

not always apply in determinin

internal forces and deformations'

E





3 - 6

Vector Product of Two Vectors

. $oncept of t"e moment of a force about a point is

more easily understood t"rou" applications oft"e vector product or cross product '

. Vector product of two vectors P and Q is defined

as t"e vector V w"ic" satisfies t"e followin

conditions:

(' 5ine of action of V is perpendicular to plane

containin P and Q'

*' Manitude of V is

&' 6irection of V is obtained from t"e ri"t1"and

rule'

θ sinQ P V =

. Vector products:

1 are not commutative/

1 are distributive/

1 are not associative/

( )Q P P Q ×−=×

( ) *(*( Q P Q P QQ P ×+×=+×

( ) ( ) S Q P S Q P ××≠××

E


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3 - 7

Vector Products: Rectanular Components

. Vector products of $artesian unit vectors/

7

7

7

=×=×−=×

−=×=×=×

=×−=×=×

k k ik j jk i

i jk j jk ji

jik k i jii

. Vector products in terms of rectanularcoordinates

k Q jQiQk P j P i P V z y x z y x ++×++=

( )

( )k Q P Q P

jQ P Q P iQ P Q P

x y y x

z x x z y z z y

−+

−+−=

z y x

z y x

QQQ

P P P

k ji

=

E




3 - 8

!oment of a Force "bout a Point. A force vector is defined by its manitude and

direction' Its effect on t"e riid body also dependson it point of application'

. T"e moment of F about O is defined as

F r M O ×=

. T"e moment vector M O is perpendicular to t"e plane containin O and t"e force F '

. Any force F’ t"at "as t"e same manitude and

direction as F / is equivalent if it also "as t"e same line

of action and t"erefore/ produces t"e same moment'

. Manitude of M O measures t"e tendency of t"e force

to cause rotation of t"e body about an axis alon

M O'

T"e sense of t"e moment may be determined by t"e

ri"t1"and rule'

Fd rF O == θ sin

E




3 - 9

!oment of a Force "bout a Point

. T!o"dimensional structures "ave lent" and breadt" but

neliible dept" and are sub8ected to forces contained int"e plane of t"e structure'

. T"e plane of t"e structure contains t"e point O and t"e

force F ' M O/ t"e moment of t"e force about O is

perpendicular to t"e plane'

. If t"e force tends to rotate t"e structure cloc4wise/ t"e

sense of t"e moment vector is out of t"e plane of t"e

structure and t"e manitude of t"e moment is positive'

. If t"e force tends to rotate t"e structure countercloc4wise/

t"e sense of t"e moment vector is into t"e plane of t"e

structure and t"e manitude of t"e moment is neative'

f SE




3 - 10

Varinon#s T$eorem

. T"e moment about a ive point O of t"eresultant of several concurrent forces is equal

to t"e sum of t"e moments of t"e various

moments about t"e same point O'

. Varion!s T"eorem ma4es it possible to

replace t"e direct determination of t"e

moment of a force F by t"e moments of two

or more component forces of F '

( ) +×+×=++× *(*( F r F r F F r

V t M h i f E i St tiE




3 - 11

Rectanular Components of t$e !oment of a Force

( ) ( ) ( ) k yF xF j xF zF i zF yF

F F F

z y x

k ji

k j i

x y z x y z

z y x

z y xO

−+−+−=

=

++=

T"e moment of F about O/

k F j F i F F

k z j yi xr F r

z y x

O

++=

++=×= /





3 - 12


T"e moment of F about #/

F r # $ # ×= 9

( ) ( ) ( )k F j F i F F

k z z j y yi x x

r r r

z y x

# $ # $ # $

# $ # $

++=−+−+−=

−=9

( ) ( ) ( ) z y x

# $ # $ # $ #

F F F z z y y x x

k ji

−−−=





3 - 13


For two1dimensional structures/

( )

x y

% O

x yO

yF xF

k yF xF

−=

=

−=

( ) ( )[ ]

( ) ( ) x # $ y # $

% #

x # $ y # $ #

F y y F x x

k F y y F x x

−−−=

=

−−−=





3 - 14

%ample Problem &'(

A (771lb vertical force is applied to t"e end of a

lever w"ic" is attac"ed to a s"aft at O'

6etermine:

a moment about O&

b "ori0ontal force at $ w"ic" creates t"e samemoment/

c smallest force at A w"ic" produces t"e same

moment/

d location for a *,71lb vertical force to produce

t"e same moment/

e w"et"er any of t"e forces from b/ c/ and d is

equivalent to t"e oriinal force'





3 - 15

%ample Problem &'(

a Moment about O is equal to t"e product of t"e

force and t"e perpendicular distance between t"eline of action of t"e force and O' %ince t"e force

tends to rotate t"e lever cloc4wise/ t"e moment

vector is into t"e plane of t"e paper'

( )

( ) ( )in'(*lb(77

in'(*;7cosin'*,

=

=°==

O

O

d Fd

inlb(*77 ⋅=O

V t M h i f E i St tiE i




3 - 16

%ample Problem &'(

b <ori0ontal force at $ t"at produces t"e same

moment/

( )

( )

in'='*7in'lb(*77

in'='*7in'lb(*77

in'='*7;7sinin'*,

⋅=

=⋅

=

=°=

F

F

Fd

d

O

lb>'->= F





3 - 17

%ample Problem &'(

c T"e smallest force $ to produce t"e same moment

occurs w"en t"e perpendicular distance is amaximum or w"en F is perpendicular to O$'

( )

in',*

in'lb(*77

in',*in'lb(*77

⋅=

=⋅

=

F

F

Fd O

lb-7= F




Vector Mechanics for Engineers: Staticsi gh t h

3 - 18

%ample Problem &'(

d To determine t"e point of application of a *,7 lb

force to produce t"e same moment/

( )

in'-cos;7

in'-lb,7*

in'lb(*77

lb*,7in'lb(*77

=°

=⋅

=

=⋅

=

O#

d

d

Fd O

in'(7=O#

Vector Mechanics for Engineers StaticsE i



Vector Mechanics for Engineers: Statics gh t h

3 - 19

%ample Problem &'(

e Alt"ou" eac" of t"e forces in parts b/ c/ and d produces t"e same moment as t"e (77 lb force/ none

are of t"e same manitude and sense/ or on t"e same

line of action' 2one of t"e forces is equivalent to t"e

(77 lb force'

Vector Mechanics for Engineers: StaticsE i



Vector Mechanics for Engineers: Staticsgh t h

3 - 20

%ample Problem &')

T"e rectanular plate is supported by

t"e brac4ets at $ and # and by a wire

()' ?nowin t"at t"e tension in t"e

wire is *77 2/ determine t"e moment

about $ of t"e force exerted by t"e

wire at ( '

%35@TI32:

T"e moment $ of t"e force F exerted

by t"e wire is obtained by evaluatin

t"e vector product/

F r $( $ ×=

Vector Mechanics for Engineers: StaticsE i g




3 - 21

%ample Problem &')%35@TI32:

(*=;(*7

7=77&7

= ''

k ji

F F F

z z y y x x

k ji

z y x

$( $( $( $

( ) ( ) ( )k ji $ m 2='=*m 2='=*m 2;='> ⋅+⋅+⋅−=

F r $( $ ×=

( )

( ) ( ) ( ) ( )

( ) ( ) ( )k ji

k jir

r F F

)(

)(

2(*= 2; 2(*7

m-'7

m&*'7m7'*,m&'7 2*77

2*77

−+−=

−+−=

== λ

→

k 'i'r r r $( $( m7=7m&7

B7'&/ 7/ 7',

B7/7/7'&*





3 - 22

%calar Product of Two Vectors

. T"e scalar product or dot product between

two vectors P and Q is defined as( )resultscalar cosθ PQQ P =•

. %calar products:

1 are commutative/

1 are distributive/1 are not associative/

P QQ P •=•

( ) *(*( Q P Q P QQ P •+•=+•( ) undefined=•• * Q P

. %calar products wit" $artesian unit components/

777((( =•=•=•=•=•=• ik k j jik k j jii

( ( k Q jQiQk P j P i P Q P z y x z y x ++•++=•

**** P P P P P P

Q P Q P Q P Q P

z y x

z z y y x x

=++=•

++=•





3 - 23

%calar Product of Two Vectors: "pplications

. Anle between two vectors:

PQ

Q P Q P Q P

Q P Q P Q P PQQ P

z z y y x x

z z y y x x

++=

++==•

θ

θ

cos

cos

. Pro8ection of a vector on a iven axis:

O+

O+

P P Q

Q P

PQQ P

O+ P P P

==•

=•

==

θ

θ

θ

cos

cos

alonof pro8ectioncos

z z y y x x

O+

P P P

P P

θ θ θ

λ

coscoscos ++=

•=

. For an axis defined by a unit vector:





3 - 24

!ixed Triple Product of T$ree Vectors

. Mixed triple product of t"ree vectors/

( ) resultscalar=×• Q P *

. T"e six mixed triple products formed from S / P / and

Q "ave equal manitudes but not t"e same sin/

( ) ( ) ( )* P QQ* P P Q*

P * Q* Q P Q P *

×•−=×•−=×•−=

×•=×•=×•

( ) ( )

( )

z y x

z y x

z y x

x y y x z

z x x z y y z z y x

QQQ

P P P

* * *

Q P Q P *

Q P Q P * Q P Q P * Q P *

=

−+

−+−=×•

. Evaluatin t"e mixed triple product/





3 - 25

!oment of a Force "bout a *iven "xis

. Moment M O of a force F applied at t"e point A

about a point O/ F r O ×=

. %calar moment O+ about an axis OL is t"e

pro8ection of t"e moment vector M O onto t"e

axis/( ) F r OO+ ×•=•= λ λ

. Moments of F about t"e coordinate axes/

x y z

z x y

y z x

yF xF

xF zF zF yF

−=

−=−=






3 - 26

!oment of a Force "bout a *iven "xis

. Moment of a force about an arbitrary axis/

( )

# $ # $

# $

# #+

r r r

F r

−=

×•=

•=

λ

λ

. T"e result is independent of t"e point #

alon t"e iven axis'






3 - 27

%ample Problem &'&

a about $

b about t"e ede $# and

c about t"e diaonal $, of t"e cube'd 6etermine t"e perpendicular distance

between $, and F( '

A cube is acted on by a force P as

s"own' 6etermine t"e moment of P






3 - 28

%ample Problem &'&

. Moment of P about $/

( )

( ) ( )( ) ( )k j P jia

k j P k j P P

jia jaiar

P r

$

$ F

$ F $

−×−=

−=−=

−=−=

×=

*9

*9*9(*9(

( )( )k jiaP $ ++= *9

. Moment of P about $#&

( )( )k jiaP i

i $ $#

++•=

•=

*9

*9aP $# =






3 - 29

%ample Problem &'&

. Moment of P about t"e diaonal $,/

( )

( )

( ) ( )

( )(((;

*&

(*

&

(

&

−−=

++•−−=

++=

−−=−−

==

•=

aP

k jiaP

k ji

k jiaP

k jia

k a jaia

r

r

$,

$

, $

, $

$ $,

λ

λ

;

aP $, −=





Vector Mechanics for Engineers: Staticsh t h

3 - 30

%ample Problem &'&

. Perpendicular distance between $, and F(&

( ) ( ) ( )

7

((7;&

(

*

=

+−=−−•−=• P

k jik j P

P

λ

T"erefore/ P is perpendicular to $,'

Pd aP

$, ==;

;

ad =






3 - 31

!oment of a Couple

. Two forces F and 1F "avin t"e same manitude/

parallel lines of action/ and opposite sense aresaid to form a couple'

. Moment of t"e couple/

( )

Fd rF

F r

F r r

F r F r

# $

# $

==

×=

×−=

−×+×=

θ sin

. T"e moment vector of t"e couple isindependent of t"e c"oice of t"e oriin of t"e

coordinate axes/ i'e'/ it is a free vector t"at

can be applied at any point wit" t"e same

effect'

Vector Mechanics for Engineers: StaticsE i gh





3 - 32

!oment of a Couple

Two couples will "ave equal moments if

. **(( d F d F =

. t"e two couples lie in parallel planes/ and

. t"e two couples "ave t"e same sense ort"e tendency to cause rotation in t"e same

direction'






3 - 33

"ddition of Couples

. $onsider two intersectin planes P ( and

P * wit" eac" containin a couple

***

(((

planein

planein

P F r

P F r

×=

×=

. #esultants of t"e vectors also form a

couple

( )*( F F r -r +×=×=

. +y Varion!s t"eorem

*(

*(

F r F r

+=×+×=

. %um of two couples is also a couple t"at is

equal to t"e vector sum of t"e two couples






3 - 34

Couples Can +e Represented by Vectors

. A couple can be represented by a vector wit" manitudeand direction equal to t"e moment of t"e couple'

. (ouple vectors obey t"e law of addition of vectors'

. $ouple vectors are free vectors/ i'e'/ t"e point of applicationis not sinificant'

. $ouple vectors may be resolved into component vectors'






3 - 35

Resolution of a Force Into a Force at O and a Couple

. Force vector F can not be simply moved to O wit"out modifyin its

action on t"e body'

. Attac"in equal and opposite force vectors at O produces no net

effect on t"e body'

. T"e t"ree forces may be replaced by an equivalent force vector and

couple vector/ i'e/ a force"couple system'






3 - 36

Resolution of a Force Into a Force at O and a Couple

. Movin F from $ to a different point O. requires t"e

addition of a different couple vector M O’

F r O ×′=C

. T"e moments of F about 3 and O. are related/

( )

F s

F s F r F sr F r

O

O

×+=

×+×=×+=×= CC

. Movin t"e force1couple system from O to O. requires t"e

addition of t"e moment of t"e force at O about O. '






3 - 37

%ample Problem &',

6etermine t"e components of t"esinle couple equivalent to t"e

couples s"own'

%35@TI32:

. Attac" equal and opposite *7 lb forces int"e D x direction at $/ t"ereby producin &

couples for w"ic" t"e moment components

are easily computed'

. Alternatively/ compute t"e sum of t"e

moments of t"e four forces about an

arbitrary sinle point' T"e point ) is a

ood c"oice as only two of t"e forces will

produce non10ero moment contributions''






3 - 38

%ample Problem &',

. Attac" equal and opposite *7 lb forces

in t"e D x direction at $

. T"e t"ree couples may be represented by

t"ree couple vectors/

( ) ( )

( ) ( )

( ) ( ) in'lb(=7in'lb*7

in'lb*,7in'(*lb*7

in'lb-,7in'(=lb&7

⋅+=+=

⋅+=+=

⋅−=−=

z

y

x

( ) ( )

( )k

ji

in'lb(=7

in'lb*,7in'lb-,7

⋅+

⋅+⋅−=






3 - 39

%ample Problem &',

. Alternatively/ compute t"e sum of t"e

moments of t"e four forces about )'

. 3nly t"e forces at ( and / contribute to

t"e moment about )'

( ) ( )

( ) ( )[ ] ( ) ik j

k j )

lb*7in'(*in'

lb&7in'(=

−×−+

−×==

( ) ( )

( )k

ji

in'lb(=7

in'lb*,7in'lb-,7

⋅+

⋅+⋅−=





Vector Mechanics for Engineers: Statics t h

3 - 40

%ystem of Forces: Reduction to a Force and Couple

. A system of forces may be replaced by a collection of

force1couple systems actin a iven point O

. T"e force and couple vectors may be combined into a

resultant force vector and a resultant couple vector/

( )∑∑ ×== F r F - -

O

. T"e force1couple system at O may be moved to O. wit" t"e addition of t"e moment of R about O. /

- s -

O

-

O ×+=C

. Two systems of forces are equivalent if t"ey can be

reduced to t"e same force1couple system'





Vector Mechanics for Engineers: Staticsth

3 - 41

Furt$er Reduction of a %ystem of Forces

. If t"e resultant force and couple at O are mutually

perpendicular/ t"ey can be replaced by a sinle force actinalon a new line of action'

. T"e resultant force1couple system for a system of forces

will be mutually perpendicular if:

( t"e forces are concurrent/

* t"e forces are coplanar/ or& t"e forces are parallel'

Vector Mechanics for Engineers: StaticsE i gh t





3 - 42

Furt$er Reduction of a %ystem of Forces

. %ystem of coplanar forces is reduced to a

force1couple system t"at ismutually perpendicular'

-O - and

. %ystem can be reduced to a sinle force

by movin t"e line of action of until

its moment about O becomes -

O

-

. In terms of rectanular coordinates/ -

O x y y- x- =−






3 - 43

%ample Problem &'-

For t"e beam/ reduce t"e system of

forces s"own to Ba an equivalent

force1couple system at $/ Bb an

equivalent force couple system at #/

and Bc a sinle force or resultant'

2ote: %ince t"e support reactions are

not included/ t"e iven system will

not maintain t"e beam in equilibrium'

%35@TI32:

a $ompute t"e resultant force for t"e

forces s"own and t"e resultant

couple for t"e moments of t"e

forces about $'

b Find an equivalent force1couple

system at # based on t"e force1

couple system at $'

c 6etermine t"e point of application

for t"e resultant force suc" t"at itsmoment about $ is equal to t"e

resultant couple at $'






3 - 44

%ample Problem &'-

%35@TI32:

a $ompute t"e resultant force and t"eresultant couple at $'

( ) ( ) ( ) ( ) j j j j

F -

2*-7 2(77 2;77 2(-7 −+−=

=∑

( ) j - 2;77−=

( )( ) ( ) ( ) ( )

( ) ( ) ji

ji ji

F r - $

*-7=',

(77='*;77;'(

−×+

×+−×=

×=∑

( )k - $ m 2(==7 ⋅−=






3 - 45

%ample Problem &'-

b Find an equivalent force1couple system at #

based on t"e force1couple system at $'T"e force is unc"aned by t"e movement of t"e

force1couple system from $ to #'

( ) j - 2;77−=

T"e couple at # is equal to t"e moment about #

of t"e force1couple system found at $'

( ) ( ) ( )( ) ( )k k

jik

-r $ #

-

$

-

#

m 2*==7m 2(==7

2;77m=',m 2(==7

⋅+⋅−=

−×−+⋅−=

×+=

( )k - # m 2(777 ⋅+=






3 - 46

%ample Problem &'.

T"ree cables are attac"ed to t"e brac4et as s"own' #eplace t"e

forces wit" an equivalent force1

couple system at $'

%35@TI32:

. 6etermine t"e relative position vectors

for t"e points of application of t"e

cable forces wit" respect to $'

. #esolve t"e forces into rectanular

components'

. $ompute t"e equivalent force/

∑= F -

. $ompute t"e equivalent couple/

( )∑ ×= F r - $





Vector Mechanics for Engineers: Staticsh

3 - 47

%ample Problem &'.

%35@TI32:

. 6etermine t"e relative position

vectors wit" respect to $'( )

( )

( )m(77'7(77'7

m7-7'77>-'7

m7-7'77>-'7

jir

k ir

k ir

$ )

$(

$ #

−=

−=

+=

. #esolve t"e forces into rectanular

components'( )

( ) 2 *77;77&77

*='7=->'7,*'7

(>-

-7(-7>-

2>77

k ji F

k ji

k ji

r

r

F

#

# /

# /

#

+−=+−=

+−==

=

λ

λ

( ) ( )

( ) 2 (7&;77

&7cos;7cos 2(*77

ji

ji F )

+=

+=

( ) ( )

( ) 2 >7>>7>

,-cos,-cos 2(777

ji

ji F (

−=

−=





ecto ec a cs o g ee s Stat csh

3 - 48

%ample Problem &'.. $ompute t"e equivalent force/

( )

( )

( )k

j

i

F -

>7>*77

(7&;77

;77>7>&77

−+

+−+

++=

=∑

( ) 2 -7>,&(;7> k ji - −+=

. $ompute t"e equivalent couple/

( )

k

k ji

F r

j

k ji

F r

k i

k ji

F r

F r

) $ )

c $(

# $ #

-

$

'(;&

7(7&;77

7(77'7(77'7

;='(>

>7>7>7>

7-7'777>-'7

,-&7

*77;77&77

7-7'777>-'7

=−=×

=

−

−=×

−=

−

=×

×=∑

k ji - $ '((=;='(>&7 ++=



,/

Problem &'0

T$e (-1ft boom $# $as a fixed

end $' " steel cable is stretc$ed

from t$e free end # of t$e boom

to a point ( located on t$e

vertical wall' If t$e tension in t$e

cable is -02 lb3 determine t$e

moment about $ of t$e force

exerted by t$e cable at #'

(

#

$

x

y

z

(- ft

; ft(7 ft

( y Problem &'0



-2

(' Determine the rectangular components of a force defined by

its magnitude and direction. If t$e direction of t$e force isdefined by two points located on its line of action3 t$e force can

be expressed by:

F F λ Bd x i D d y j D d z k F

d

(

#

$

x

y

z

(- ft

; ft(7 ft %olvin Problems on 4our 5wn


end $' " steel cable is stretc$edfrom t$e free end # of t$e boom




moment about $ of t$e forceexerted by t$e cable at #'

( y Problem &'0



-(

)' Compute the moment of a force in three dimensions. If r is a

position vector and F is t$e force t$e moment M is iven by:

M r x F

(

#

$

x

z

(- ft

; ft(7 ft %olvin Problems on 4our 5wn


end $' " steel cable is stretc$edfrom t$e free end # of t$e boom




moment about $ of t$e forceexerted by t$e cable at #'

Problem &'0 %olution(

y



-)

Determine the rectangular

components of a force defined

by its magnitude and direction.

First note:

d #( B (-* D B; * D B (7 *

d #( ( ft

T$en:

T #( B (- i D ; j (7 k B,-7 lb i D B(=7 lb j B&77 lbk ->7 lb

(

->7 2

(

#

$

x

z

(- ft

; ft(7 ft

Problem &'0 %olution(

y



-&

Compute the moment of a

force in three dimensions.

6ave:

M $ r #0$ x T #(

7$ere: r #0$ B(- ft i

T$en:

M $ (- i x B ,-7 i D (=7 j &77 k

M $

B,-77 lb'ft j D B*>77 lb'ft k

->7 2

(

#

$

x

z

(- ft

; ft(7 ft



-,

Problem &'8

9nowin t$at t$e tension in

cable $( is ().2 3 ;etermine

<a= t$e anle between cable $( and t$e boom $# 3 <b= t$e

pro>ection on $# of t$e force

exerted by cable $( at point $'

y

& m

*', m

('* m

*'; m

*', m z

x

$ #

(

)

('= mP

Problem &'8 y



--

(' Calculate the angle formed by two vectors. Express t$e vectors

in terms of t$eir components' T$e cosine of t$e desired anle is

obtained by dividin t$e scalar product of t$e two vectors by t$e

product of t$eir manitudes'

%olvin Problems on 4our 5wn


cable $( is ().2 3 ;etermine<a= t$e anle between cable $(

and t$e boom $# 3 <b= t$e



& m

*', m

('* m

*'; m

*', m z

x

$ #

(

)

('= mP

Problem &'8 y

* ,



-.

)' Compute the projection of a vector P on a given line O+.Express t$e vector P and t$e unit vector

λ

in t$e direction of t$e

line in component form' T$e required pro>ection is equal to t$e

scalar product Pλ' If t$e anle θ formed by P and λ is ?nown3 t$e

pro>ection is also iven by P cos θ.



cable $( is ().2 3 ;etermine<a= t$e anle between cable $(

and t$e boom $# 3 <b= t$e



& m

*', m

('* m

*'; m

*', m z

x

$ #

(

)

('= mP

Problem &'8 %olution y* ,



-0

Calculate the angle formed by

two vectors.

<a= First note:

$( B *',* D B7'=* D B('**

*'= m

$# B *',* D B ('=* D B 7 *

&'7 m

$( B*', m i D B7'= m j D B('* m k

$# B*', m i B('= m j

and

& m

*', m

('* m

*'; m

*', m z

x

$ #

(

)

('= mP




-8

+y definition:

$( $# B $( B $# cos θ

B *', i D 7'= j D ('* k B *', i ('= j B*'=B&'7 cos θ

B *',B *', D B7'=B ('= D B('*B7 =', cos θ

cos θ 7'-(,* θ -'(o

& m

*', m

('* m

*'; m

*', m z

x

$ #

(

)

('= mP




-/

& m

*', m

('* m

*', m z

$ #

(

)Compute the projection of a

vector on a given line.

<b=

BT $( $# T $( λ

$#

T $( cos θ

B(*;7 2 B7'-(,*

BT $( $# ;,= 2

T $(

λ

$#

*'; m

x

('= mP



.2

Problem &'/

T$e frame $() is $ined at $

and ) and is supported by a

cable w$ic$ passes t$rou$ a

rin at # and is attac$ed to

$oo?s at , and 1 ' 9nowin

t$at t$e tension in t$e cable is

,-2 3 determine t$e moment

about t$e diaonal $) of t$e

force exerted on t$e frame byportion #1 of t$e cable'

y

z

x

7'=>- m

7'>- m

7 '- m

7 '- m

7'&- m

7'*- m

$

#

O

( )

P

1

,

7'>- m

y7'&- m

Problem &'/



.(

(' Determine the moment MO+ of a force about a given axis O+.

MO+ is defined as

MO+ λ

MO λ

B r x F

w$ereλ

is t$e unit vector alon O+ and r is a position vector from

z x

7'=>- m

7'>- m

7 '- m

7 '- m

7'*- m

$

#

O

( )

P

1

,

7'>- m


T$e frame $() is $ined at $

and ) and is supported by acable w$ic$ passes t$rou$ a

rin at # and is attac$ed to

$oo?s at , and 1 ' 9nowin

t$at t$e tension in t$e cable is

,-2 3 determine t$e momentabout t$e diaonal $) of t$e

force exerted on t$e frame by

portion #1 of t$e cable'



Problem &'/ %olution y7'&- m



.&

$) λ $) B r #0$ x T #1

$) 2

(

-

, 7 &

7'- 7 7(-7 &77

&77

GB &B7'-B&77H

(

-

$) 2 7 2m

Finally:

$)

z

7'=>- m

7'>- m

7 '- m 7 '- m

7'*- m

$

#

O

( )

P

1

,

T #1 $)

r #0$ x

7'>- m



.,

Problem &'(2

T$e force P $as a manitude

of )-2 and is applied at t$e

end ( of a -22 mm rod $(

attac$ed to a brac?et at $ and

#' "ssumin α @ &2o and

β @ .2o3 replace P wit$ <a= an

equivalent force1couple

system at # 3 <b= an equivalent

system formed by two parallelforces applied at $ and #'

α

β

P

(

#

$*77 mm

&77 mm

Problem &'(2



.-


T$e force P $as a manitude

of )-2 and is applied at t$eend ( of a -22 mm rod $(

attac$ed to a brac?et at $ and

#' "ssumin α @ &2o and

β @ .2o3 replace P wit$ <a= an

equivalent force1couplesystem at # 3 <b= an equivalent

system formed by two parallel

forces applied at $ and #'

(' Replace a force with an equivalent force-couple system at aspecified point. T$e force of t$e force1couple system is equal to

t$e oriinal force3 w$ile t$e required couple vector is equal to t$e

moment of t$e oriinal force about t$e iven point'

α

β

P

(

#

$

*77 mm

&77 mm

Problem &'(2 %olution(



..

α

β

P

#

$*77 mm

&77 mm

Replace a force with an equivalent

force-couple system at a specified

point.

<a= Equivalence requires:

Σ F : F P or F *-7 2 ;7o

Σ M # : M B7'& mB*-7 2 >- 2

T$e equivalent force couple system at # is:

F *-7 2 ;7o/ M >- 2 m

Problem &'(2 %olution(



.0

<b= Require: T$e two force systems are equivalent'

α

β

P

#

$*77 mm

&77 mm

#

$

(

;7o

*-7 2&7o

Φ

Φ

(

#

$ F+

FA

x y

Problem &'(2 %olution



.8

#

$

(

;7o

*-7 2&7o

Φ

Φ

(

#

$ F+

FA

x y

Equivalence t$en requires:

Σ F x : 7 F $ cos Φ D F # cos Φ

F $ F # or cos Φ 7

Σ F y : *-7 F $ sin Φ F # sin Φ

If F $ F # t$en *-7 7 re>ect

o

Problem &'(2 %olution



./

#

$

(

;7o

*-7 2&7o

Φ

Φ

(

#

$ F+

FA

x y

"lso:Σ # : B7'& mB *-7 2 B7'* m F $

or F $ &>- 2

and F # D ;>- 2

F $ &>- 2 ;7o/ F # ;*- 2 ;7o

D



02

Problem &'((

M $ #

(

(* in

= in

;7o

,- lb

&7 lb(7 lb " couple of manitude

@ -, lbin' and t$e t$ree

forces s$own are applied to

an anle brac?et' <a= Find t$e

resultant of t$is system of

forces' <b= Aocate t$e points

w$ere t$e line of action of t$e

resultant intersects line $# and line #( '


Problem &'((



0(

M $ #

(

(* in

= in

;7o

,- lb

&7 lb(7 lb

(' Determine the resultant of two or more forces. ;etermine t$erectanular components of eac$ force' "ddin t$ese components

will yield t$e components of t$e resultant'


" couple of manitude

@ -, lb

in' and t$e t$reeforces s$own are applied to





resultant intersects line $#

and line #( '


Problem &'((



0)

M $ #

(

(* in

= in

;7o

,- lb

&7 lb(7 lb

)' Reduce a force system to a force and a couple at a given point.

T$e force is t$e resultant R of t$e system obtained by addin t$e

various forces' T$e couple is t$e moment resultant of t$e system

M3 obtained by addin t$e moments about t$e point of t$e various

forces'

R Σ F M Σ B r x F



@ -, lb







and line #( '


Problem &'((



0&

M $ #

(

(* in

= in

;7o

,- lb

&7 lb(7 lb

&' Reduce a force and a couple at a given point to a single force.

T$e sinle force is obtained by movin t$e force until its momentabout t$e point < $= is equal to t$e couple vector M $' " position

vector r from t$e point3 to any point on t$e line of action of t$e

sinle force R must satisfy t$e equation

r x R M $

-

-



@ -, lb







and line #( '



Problem &'(( %olution

(* in&7 lb

(7 lb



0-

M $ #

(

(* in

= in

;7o

,- lb

Reduce a force system to

a force and a couple at agiven point.

<b= First reduce t$e iven forces and couple to an equivalent

force1couple system <R / M #= at #'

Σ # : # B -, lbin D B (* in B(7 lb B = in B ,- lb

(=; lbin

D

Problem &'(( %olutiona



Reduce a force and a couple

at a given point to a single force.

$ #

(

c

R

)

/

7it$ R at ):

Σ # : (=; lbin a B (-'=7= lb or a ((';, in

and wit$ R at /

Σ # : (=; lbin c B &7 lb or c ;'*7 inT$e line of action of R intersects line $# (('., in' to t$e left of #

d i t t li #( . )2 i b l #

D

D

ch03 rigid bodies equivalent systems of forces

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