rigid bodies: equivalent systems of forces...

Statics

Rigid Bodies: Equivalent Systems of Forces

3 - 1

School of Mechanical Engineering

기계공학부 최해진

Statics Contents 3 - 2

Introduction

External and Internal Forces

Principle of Transmissibility: Equivalent Forces

Vector Products of Two Vectors

Moment of a Force About a Point

Varigon’s Theorem

Rectangular Components of the Moment of a Force

Sample Problem 3.1

Scalar Product of Two Vectors

Scalar Product of Two Vectors: Applications

Mixed Triple Product of Three Vectors

Moment of a Force About a Given Axis

Sample Problem 3.5

Moment of a Couple

Addition of Couples

Couples Can Be Represented By Vectors

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

Sample Problem 3.6

System of Forces: Reduction to a Force and a Couple

Further Reduction of a System of Forces

Sample Problem 3.8

Sample Problem 3.10


Introduction

External and Internal Forces

Principle of Transmissibility: Equivalent Forces

Vector Products of Two Vectors

Moment of a Force About a Point

Varigon’s Theorem

Rectangular Components of the Moment of a Force

Sample Problem 3.1

Scalar Product of Two Vectors

Scalar Product of Two Vectors: Applications

Mixed Triple Product of Three Vectors

Moment of a Force About a Given Axis

Sample Problem 3.5

Moment of a Couple

Addition of Couples

Couples Can Be Represented By Vectors

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

Sample Problem 3.6

System of Forces: Reduction to a Force and a Couple

Further Reduction of a System of Forces

Sample Problem 3.8

Sample Problem 3.10

Statics 3.1 Introduction• Treatment of a body as a single particle is not always possible. In

general, the size of the body and the specific points of application of the forces must be considered.

• Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body.

• Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system.

• moment of a force about a point• moment of a force about an axis• moment due to a couple

3 - 3


• Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with 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 acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple.

Statics Principle of Transmissibility: Equivalent Forces

• Principle of Transmissibility -Conditions of equilibrium or motion are not affected by transmitting a force along its line of action.NOTE: F and F’ are equivalent forces.

• Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck.

3 - 4


• Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck.

• Principle of transmissibility may not always apply in determining internal forces and deformations.

내력, 변형을 결정할 때는 주의!

(a) : 인장상태 , (d) : 압축상태

Statics Vector Product of Two Vectors

• Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product.

• 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.

3 - 5


• 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.

qsinQPV =

• Vector products:- are not commutative(비가역성, 교환법칙성립안됨),- are distributive(분배법칙성립),- are not associative(비결합법칙),

( )QPPQ ´-=´( ) 2121 QPQPQQP ´+´=+´

( ) ( )SQPSQP ´´¹´´

Statics Vector Products: Rectangular Components• Vector products of Cartesian unit vectors,

00

0

=´=´-=´-=´=´=´

=´-=´=´

kkikjjkiijkjjkji

jikkijii

rrrrvrrr

rrrrrrrr

rrrrrrrr

• Vector products in terms of rectangular coordinates

( ) ( )

kQPQP

jQPQPiQPQP

kQjQiQkPjPiPV

xyyx

zxxzyzzy

zyxzyx

r

rr

rrrrrrr

)–(

)–()–(

+

+=

++´++=

3 - 6


( ) ( )

kQPQP

jQPQPiQPQP

kQjQiQkPjPiPV

xyyx

zxxzyzzy

zyxzyx

r

rr

rrrrrrr

)–(

)–()–(

+

+=

++´++=

zyx

zyx

QQQPPPkjirrr

=

Statics Moment of a Force About a Point

강체에 가해지는 힘은 그 물체를 병진운동을 하게 할 수도 있으며 회전운동을 하게 할 수도 있다.물체를 회전시키는 힘의 영향 즉 한 점에 대한 힘의 모멘트(moment of a force about a point)를 소개한다

정 의

축에 대한 힘의 모멘트(moment of a force about an axis)

그림과 같이 임의의 힘 F와 F의 작용선 상에 있지 않은 임의의 점O를 생각해 보자

힘의 모멘트(moment) or 토크(torque)

3 - 7


그림과 같이 임의의 힘 F와 F의 작용선 상에 있지 않은 임의의 점O를 생각해 보자

O를 중심으로 하는 힘 F의 모멘트는

기하학적 해석

모멘트의 크기에 대한 스칼라 값 계산

수직거리 d는 모멘트 팔(moment arm)이라고 부르며, 모멘트 크기는

sinr dq =

O = ´M r F

qsinrFM o =´== FrMo FdM =0:크기

Statics Moment of a Force About a Point

MO의 크기는 힘의 크기와 수직거리 d 에 의존하므로 이 힘(미끄럼벡터)은 모멘트의 변화 없이

작용선을 따라 어디든지 옮겨 놓을 수 있다.

3 - 8


Statics Moment of a Force About a Pointsliding vector

line of actionF

A

O

daQ

Mr

3 - 9


) ( :)mN(

팔모멘트armmomentddFM ×=

dFαrFM ===´=

sin|| MFrM

Statics Moment of a Force About a Point 3 - 10


Statics Varignon’s Theorem

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

( ) Lrrrr

Lrrr

+´+´=++´ 2121 FrFrFFr

3 - 11


• Varigon’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.

Æ 임의의 점에 대한 힘의 모멘트는

그 점에 대한 힘의 분력의 모멘트의 합과 같다. : Varignon 이론

Statics Varignon’s Theorem

Æ 임의의 점에 대한 힘의 모멘트는

그 점에 대한 힘의 분력의 모멘트의 합과 같다. : Varignon 이론

QrPr)QP(rM

QPR

RrM

o

o

rrrrrrrr

rrr

rrr

´+´=+´=

+=

´=

그림(a)와 같이 힘 R에 의한 모멘트를 생각해 보자.

3 - 12


QqPpdRM o +-==

QrPr)QP(rM

QPR

RrM

o

o

rrrrrrrr

rrr

rrr

´+´=+´=

+=

´=

그림(b)와 같이 d 대신에 p 나 q를 쉽게 구할 경우에 힘R 대신에 힘 P, Q을 사용하는것이 편리함.

Statics Rectangular Components of the Moment of a Force

( ) ( ) ( )kyFxFjxFzFizFyF

FFFzyxkji

kMjMiMM

xyzxyz

zyx

zyxO

rrr

rrr

rrrr

-+-+-=

=

++=

The moment of F about O,

kFjFiFFkzjyixrFrM

zyx

O rrrr

rrrrrrr

++=++=´= ,

3 - 13


( ) ( ) ( )kyFxFjxFzFizFyF

FFFzyxkji

kMjMiMM

xyzxyz

zyx

zyxO

rrr

rrr

rrrr

-+-+-=

=

++=


The moment of F about B,

FrM BAB

rrr´= /

( ) ( ) ( )kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

rrrr

rrr

rrr

++=

-+-+-=

-= 기준 점을B:/

3 - 14


( ) ( ) ( )kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

rrrr

rrr

rrr

++=

-+-+-=

-= 기준 점을B:/

( ) ( ) ( )zyx

BABABAB

FFFzzyyxx

kjiM ---=

rrr

r


For two-dimensional structures,

( )

zy

ZO

zyO

yFxFMM

kyFxFM

-==

-=rr

3 - 15


( ) ( )

( ) ( ) zBAyBA

ZO

zBAyBAO

FyyFxxMM

kFyyFxxM

---==

---=rr

][

Statics Sample Problem 3.1

A 100-N vertical force is applied to the end of a lever which is attached to a shaft at O.

Determine:

a) moment about O,

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

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

d) location for a 240-N vertical force to produce the same moment,

e) whether any of the forces from b, c, and d is equivalent to the original force.

24 m

3 - 16


A 100-N vertical force is applied to the end of a lever which is attached to a shaft at O.

Determine:

a) moment about O,

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

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

d) location for a 240-N vertical force to produce the same moment,

e) whether any of the forces from b, c, and d is equivalent to the original force.


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.

( )( )( )24m cos60 12 m

100 N 12 m

O

O

M Fdd

M

=

= ° =

=

3 - 17


( )( )( )24m cos60 12 m

100 N 12 m

O

O

M Fdd

M

=

= ° =

=

m N 1200 ×=OM


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

( )

( )

m 8.20m N 1200

m 8.20m N 1200

m 8.2060sinm 24

×=

=×=

=°=

F

FFdM

d

O

N 7.57=F

24 m

3 - 18


( )

( )

m 8.20m N 1200

m 8.20m N 1200

m 8.2060sinm 24

×=

=×=

=°=

F

FFdM

d

O

N 7.57=F


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

( )

m 42m N 1200

m 42m N 1200×

=

=×=

F

FFdM O

N 50=F

24 m

3 - 19


( )

m 42m N 1200

m 42m N 1200×

=

=×=

F

FFdM O

N 50=F


d) To determine the point of application of a 240 N force to produce the same moment,

( )1200 N m 240 N1200 N m 5 m

240 Ncos60 5 m

OM Fdd

d

OB

=

× =

×= =

° =

3 - 20


( )1200 N m 240 N1200 N m 5 m

240 Ncos60 5 m

OM Fdd

d

OB

=

× =

×= =

° = m 10=OB


e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 N force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 N force.

3 - 21



SOLUTION:

The moment MA of the force F exerted by the wire is obtained by evaluating the vector product,

FrM ACA

rrr´=

3 - 22


The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C.


SOLUTION:

( ) ( )kirrr ACAC

rrrrr m 08.0m 3.0 +=-=

FrM ACA

rrr´=

( )

( ) ( ) ( ) ( )

( ) ( ) ( )kji

kji

rr

FFDC

DC

rrr

rrr

rrr

N 128N 69N 120m 5.0

m 32.0m 0.24m 3.0N 200

N 200

-+-=

-+-=

== l

5.0)32.024.03.0(

)4.0,0,3.0(,)08.0,24.0,0(

/

/

=

-+-=

==

DC

DC

rkjir

CD

r

rrrr

3 - 23


1289612008.003.0

--=

kjiM A

rrr

r

( ) ( ) ( )kjiM A

rrrvmN 8.82mN 8.82mN 68.7 ×+×+×-=

( )

( ) ( ) ( ) ( )

( ) ( ) ( )kji

kji

rr

FFDC

DC

rrr

rrr

rrr

N 128N 69N 120m 5.0

m 32.0m 0.24m 3.0N 200

N 200

-+-=

-+-=

== l

Statics Scalar Product of Two Vectors

• The scalar product or dot product between two vectors P and Q is defined as

( )resultscalarcosqPQQP =·rr

• Scalar products:- are commutative, (교환법칙성립),- are distributive, (분배법칙성립),- are not associative, (결합법칙 무의미),

PQQPrrrr

·=·( ) 2121 QPQPQQP

rrrrrrr·+·=+·

( ) undefined =·· SQPrrr

3 - 24


( ) undefined =·· SQPrrr

• Scalar products with Cartesian unit components,

000111 =·=·=·=·=·=· ikkjjikkjjiirrrvrrrrrrrr

( ) ( )kQjQiQkPjPiPQP zyxzyx

rrrrrrrr++·++=·

2222 PPPPPP

QPQPQPQP

zyx

zzyyxx

=++=·

++=·rr

rr

Statics Scalar Product of Two Vectors: Applications• Angle between two vectors:

PQQPQPQP

QPQPQPPQQP

zzyyxx

zzyyxx

++=

++==·

q

q

cos

cosrr

• Projection of a vector on a given axis: 주어진 축에 벡터의 투영

OL

OL

PPQ

QPPQQP

OLPPP

==·

=·

==

q

q

q

cos

cos

along of projection cos

rr

rr

3 - 25


OL

OL

PPQ

QPPQQP

OLPPP

==·

=·

==

q

q

q

cos

cos

along of projection cos

rr

rr

zzyyxx

OL

PPPPP

qqql

coscoscos ++=·=rr

• For an axis defined by a unit vector: 축의 단위벡터에 대한 벡터의 투영

( ) ( )kjikPjPiP zyxzyx

rrrrrrqqq coscoscos ++·++=

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

( ) resultscalar =´· QPSrrr

• The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign,

( ) ( ) ( )( ) ( ) ( )SPQQSPPQS

PSQSQPQPSrrrrrrrr

rrrrrrrrr

´·-=´·-=´·-=

´·=´·=´·

스칼라 삼중곱 (scalar triple product)은

두 벡터의 외적을 다른 세 번째 벡터와 내적을 한 것이다.

이는 P, Q, S를 변으로 하는 평행육면체의 부피와 같다.

3 - 26


( ) ( ) ( )( ) ( ) ( )SPQQSPPQS

PSQSQPQPSrrrrrrrr

rrrrrrrrr

´·-=´·-=´·-=

´·=´·=´·

( ) ( ) ( )( )

zyx

zyx

zyx

xyzxyz

zxxzyyzzyx

QQQPPPSSS

QPQPSQPQPSQPQPSQPS

=

-+

-+-=´·rrr

• Evaluating the mixed triple product,

( ) 부피평행육면체의 : QPSrrr

´·

Statics Moment of a Force About a Given Axis• Moment MO of a force F applied at the point A

about a point O,FrM O

rrr´=

• Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis, 주어진 축 OL에 대한 힘 F의 모멘트 크기 MOL

: 축 OL 에 관한 모멘트 MO 의 투영 OC 이다.

( )FrMM OOL

rrrrr´·=·= ll

• Moments of F about the coordinate axes,

3 - 27


• Moments of F about the coordinate axes,

xyz

zxy

yzx

yFxFMxFzFMzFyFM

-=

-=

-=

zzyyxx

zyx

zyx

zyx

oOL

MMMFFFzyx

FFFzyxkji

)Fr(λMM

llllll

l

l

++==·=

´·=·=rrr

r

rrrrr

• Fx , Fy , Fz : 힘 F가 강체를 x, y 및 z 축 방향으로 작용하는 힘의 크기

• Mx , My , Mz : 힘 F가 강체를 x, y 및 z 축 대하여 회전시키려는 모멘트 크기

Statics Moment of a Force About a Given Axis

• Moment of a force about an arbitrary axis,: 임의의 주어진 축에 힘의 모멘트

( )BABA

BA

BBL

rrr

Fr

MM

rrr

rrr

rr

-=

´·=

·=

l

l

3 - 28


• The result is independent of the point Balong the given axis.

zyx

BABABA

zyx

BL

FFFzyxM ///

lll=

C점을 기준으로 구하여도 동일한 결과로 유도됨.


a) about Ab) about the edge AB andc) about the diagonal AG of the cube.d) Determine the perpendicular distance

between AG and FC.

A cube is acted on by a force P as shown. Determine the moment of P

3 - 29


a) about Ab) about the edge AB andc) about the diagonal AG of the cube.d) Determine the perpendicular distance

between AG and FC.


• Moment of P about A ?

)(2

)22

()–(

22

–

kjiaPkPjPjaiaM

kPjPP

jaiar

PrM

A

AF

AFA

rrrrrrrr

rrr

rrr

rrr

++=-´=

-=

=

´=

3 - 30


)(2

)22

()–(

22

–

kjiaPkPjPjaiaM

kPjPP

jaiar

PrM

A

AF

AFA

rrrrrrrr

rrr

rrr

rrr

++=-´=

-=

=

´=

( )( )kjiaPM A

rrrr++= 2/

• Moment of P about AB ?

( ) ( )kjiiaP

MiM AABrrrr

rr

++·=

·=

2/

2aPM AB =

Statics Sample Problem 3.5• Moment of P about the diagonal AG ?

( )

( )

( ) ( )

( )1116

2312

31

3

--=

++·--=

++=

--=--

==

·=

aP

kjiaPkjiM

kjiaPM

kjia

kajaiarr

MM

AG

A

GA

GA

AAG

rrrrrr

rrrr

rrrrrrr

r

rr

l

l

3 - 31


( )

( )

( ) ( )

( )1116

2312

31

3

--=

++·--=

++=

--=--

==

·=

aP

kjiaPkjiM

kjiaPM

kjia

kajaiarr

MM

AG

A

GA

GA

AAG

rrrrrr

rrrr

rrrrrrr

r

rr

l

l

6aPM AG -=

6/

2/2/00

3/13/13/1

)(

///

aP

PPaa

FFFzyxM

zyx

AFAFAF

zyx

AG

-=

--

--==

lll

별해

),,0(,)0,0,( aaAaG ==

)( / PrMM AFAAG

rrrrr´·=·= ll


• Perpendicular distance d between AG and FC ?

( ) ( ) ( )

0

11063

12

=

+-=--·-=·PkjikjPP

rrrrrrrl

Therefore, P is perpendicular to AG.

힘 P 방향방향l

3 - 32


PdaPM AG ==6

6ad =

Statics Moment of a Couple• 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,

( )( )

FdrFMFr

Frr

FrFrM

o

BA

BAo

==´=

´-=

-´+´=

qsin

rr

rrr

rrrrr

: 크기가 같고 작용선이 평행하며 방향이 반대인 두 힘은 우력(couple)을 구성함.합력은 제로이지만 회전시키려는 모멘트가 존재함.

3 - 33


( )( )

FdrFMFr

Frr

FrFrM

o

BA

BAo

==´=

´-=

-´+´=

qsin

rr

rrr

rrrrr

• 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.

: 우력의 모멘트

Statics 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.

3 - 34


• the two couples have the same sense or the tendency to cause rotation in the same direction.

Statics

단일 힘으로 합할 수 없으며, 그 효과는 전적으로 회전만을 일으킴

- the moment produced by two forces

equal oppositenon_colinear

FdFadaFMo =-+= )(

우 력 ( couple)

우력에 의한 모멘트는 선택한 모멘트 중심 a 에 무관하다.

Moment of a Couple 3 - 35


F)r(rF)(rFrM

BA

BAo

´-=-´+´=

rrr BA =-

FrM o ´=

Statics

우력에 의한 모멘트는 선택한 모멘트 중심 a 에 무관하므로우력 모멘트 M 를 자유벡터라고 할 수 있으며방향은 우력이 작용하는 평면에 수직이고, 오른손 법칙에준한다.

Equivalent Couples (등가우력)

Moment of a Couple 3 - 36


Fd 가 일정하면 우력에 의한 모멘트는 변함이 없다.

임의의 어느 평면에 힘들이 작용하여도 우력 모멘트는 동일한 자유벡터를 갖는다.

Equivalent Couples (등가우력)

Statics Addition of Couples (우력의 합성)• Consider two intersecting planes P1 and

P2 with each containing a couple

222

111

planein planein PFrMPFrM

rrr

rrr

´=

´=

• Resultants of the vectors also form a couple

( )21 FFrRrMrrrrrr

+´=´=

3 - 37


• By Varigon’s theorem

21

21

MMFrFrM

rr

rrrrr

+=

´+´=

• Sum of two couples is also a couple that is equal to the vector sum of the two couples

Statics Couples Can Be Represented by Vectors

• A couple can be represented by a vector with magnitude and direction equal to the moment of the couple.

3 - 38


• 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.

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

• Force vector F can not be simply moved to O without modifying its action on the body.

3 - 39


• Force vector F can not be simply moved to O without modifying its action on the body.

• Attaching equal and opposite force vectors at O produces no net effect on the body.

• The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system.

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

• Moving F from A to a different point O' requires the addition of a different couple vector MO’

3 - 40


• Moving F from A to a different point O' requires the addition of a different couple vector MO’

FrMrrr

´¢=¢0

• The moments of F about O and O' are related,

( )FsM

FsFrFsrFrM

O

rrr

rrrrrrrrrr

O

´+=

´+´=´+=´¢=¢


SOLUTION:

• Attach equal and opposite 90 N forces in the +x direction at A, thereby producing 3 couples for which the moment components are easily computed.

• Alternatively, compute the sum of the moments of the four forces about an arbitrary single point. The point D is a good choice as only two of the forces will produce non-zero moment contributions.

3 - 41


Determine the components of the single couple equivalent to the couples shown.

• Alternatively, compute the sum of the moments of the four forces about an arbitrary single point. The point D is a good choice as only two of the forces will produce non-zero moment contributions.


• Attach equal and opposite 90 N forces in the +x direction at A

• The three couples may be represented by three couple vectors,

( )( )( )( )( )( ) mN 25.20mm 225N 90

mN 0.27mm 300N 90mN 75.60–mm 450N 135

×+=+=

×+=+=×=-=

z

y

x

MMM

3 - 42


( )( )( )( )( )( ) mN 25.20mm 225N 90

mN 0.27mm 300N 90mN 75.60–mm 450N 135

×+=+=

×+=+=×=-=

z

y

x

MMM

k m)N 25.20( ×+=zM

j m)N 0.27( ×+=yM

im)N 75.60( ×-=xM( ) ( ) ( )kjiM

rrrrmN 25.20mN0.27mN 75.60 ×+×+×-=


• Alternatively, compute the sum of the moments of the four forces about D.

• Only the forces at C and E contribute to the moment about D.

( ) ( )( ) ( )[ ] ( )ikj

kjMM Drrr

rrrr

N 90–mm 003mm 225

N 135–mm 450

´-+

´==

별해 임의의 점 D에서의 모멘트 계산하여 결정

3 - 43


( ) ( )( ) ( )[ ] ( )ikj

kjMM Drrr

rrrr

N 90–mm 003mm 225

N 135–mm 450

´-+

´==

( ) ( ) ( )kjiMrrrr

mN 25.20mN0.27mN75.60 ×+×+×-=

두 힘 90N, 135N 는 D점에서모멘트 제로

Statics System of Forces : Reduction to a Force and Couple

• A system of forces may be replaced by a collection of force-couple systems acting a given point O

• The force and couple vectors may be combined into a resultant force vector and a resultant couple vector,

3 - 44


• The force and couple vectors may be combined into a resultant force vector and a resultant couple vector,

( )ååå ´=== FrMMFR oRO

rrrrrr

• The force-couple system at O may be moved to O' with the addition of the moment of R about O' ,

RsMM RO

RO

rrrr´+=¢

• Two systems of forces are equivalent if they can be reduced to the same force-couple system.

Statics Further Reduction of a System of Forces

• If the resultant force and couple at O are mutually perpendicular, they can be replaced by a single force acting along a new line of action.

• The resultant force-couple system for a system of forces will be mutually perpendicular if:1) the forces are concurrent, (동일점에 작용하는 힘들)2) the forces are coplanar, (동일평면상의 힘들)3) the forces are parallel. (평행한 힘들)

주어진 힘계가 단일 힘으로 변환될 수 있는 조건 :

3 - 45


• The resultant force-couple system for a system of forces will be mutually perpendicular if:1) the forces are concurrent, (동일점에 작용하는 힘들)2) the forces are coplanar, (동일평면상의 힘들)3) the forces are parallel. (평행한 힘들)

합우력이 제로이고, 힘-우력계는 합력벡터만 남아서합력만으로 변환되어진다. (2장 참조)

( ) 0RO oR F M M r F= = = ´ =å å å

r r r r rr

1) the forces are concurrent, (동일점에 작용하는 경우)

Statics

• System of coplanar forces is reduced to a force-couple system that is mutually perpendicular.

ROMR

rr and

• System can be reduced to a single force by moving the line of action of until its moment about O becomes R

OMrRr

2) the forces are coplanar, (동일평면상의 힘들인 경우)

Further Reduction of a System of Forces3 - 46


• System can be reduced to a single force by moving the line of action of until its moment about O becomes R

OMr

• In terms of rectangular coordinates,ROxy MyRxR =-


For the beam, reduce the system of forces shown to (a) an equivalent force-couple system at A, (b) an equivalent force couple system at B, and (c) a single force or resultant.

Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium.

SOLUTION:

a) Compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A.

b) Find an equivalent force-couple system at B based on the force-couple system at A.

3 - 47


For the beam, reduce the system of forces shown to (a) an equivalent force-couple system at A, (b) an equivalent force couple system at B, and (c) a single force or resultant.

Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium.

b) Find an equivalent force-couple system at B based on the force-couple system at A.

c) Determine the point of application for the resultant force such that its moment about A is equal to the resultant couple at A.


SOLUTION:

a) Compute the resultant force and the resultant couple at A.

( ) ( ) ( ) ( ) jjjjFR

rrrr

rr

N 250N 100N 600N 150 -+-=

=å

( ) jRrr

N600-=

3 - 48


( ) jRrr

N600-=

( )( ) ( ) ( ) ( )

( ) ( )jijiji

FrM RA

rr

rrrr

rrr

2508.41008.26006.1

-´+

´+-´=

´=å

( )kM RA

rrmN 1880 ×-=


b) Find an equivalent force-couple system at Bbased on the force-couple system at A.

The force is unchanged by the movement of the force-couple system from A to B.

( ) jRrr

N 600-=

The couple at B is equal to the moment about Bof the force-couple system found at A.

( ) ( ) ( )( ) ( )

/

1880 N m 4.8 m 600 N

1880 N m 2880 N m

R RB A A BM M r R

k i j

k k

= + ´

= - × + - ´ -

= - × + ×

r r rr

r r r

r r

3 - 49


( ) ( ) ( )( ) ( )

/

1880 N m 4.8 m 600 N

1880 N m 2880 N m

R RB A A BM M r R

k i j

k k

= + ´

= - × + - ´ -

= - × + ×

r r rr

r r r

r r

( )kM RB

rrmN 1000 ×+=

c) 단일 힘 또는 합력 : 합력은 R과 같고, 그 작용점은R의 A점에 대한 모멘트가 다음같다.

( ) ( ) ( ) m13.31880600 =Þ-=-´Þ=´ xkjixMRr RA

rrrrrr

( ) m13.3,N 600 =-= xjRrr


SOLUTION:

• Determine the relative position vectors for the points of application of the cable forces with respect to A.

• Resolve the forces into rectangular components.

3 - 50


Three cables are attached to the bracket as shown. Replace the forces with an equivalent force-couple system at A.

• Compute the equivalent force,

å= FRrr

• Compute the equivalent couple,

( )å ´= FrM RA

rrv


SOLUTION:

• Determine the relative position vectors with respect to A.

• Resolve the forces into rectangular components.

( )

( )N 200600300

289.0857.0429.0

1755015075

N 700

kjiFkji

kjirr

F

B

BE

BE

B

rrrr

rrr

rrrrr

rr

+-=

+-=

+-==

=

l

l

3 - 51


SOLUTION:

• Determine the relative position vectors with respect to A.

( )( )( )m 100.0100.0

m 050.0075.0

m 050.0075.0

jir

kir

kir

AD

AC

AB

rrr

rrr

rrr

-=

-=

+=

( )

( )N 200600300

289.0857.0429.0

1755015075

N 700

kjiFkji

kjirr

F

B

BE

BE

B

rrrr

rrr

rrrrr

rr

+-=

+-=

+-==

=

l

l

( )( )( )N 1039600

30cos60cosN 1200ji

jiFDrr

rrr

+=

+=

( )( )( )N 707707

45cos45cosN 1000

jijiFC

rr

rrr

-=

-=


• Compute the equivalent force,

( )( )( )k

ji

FR

r

r

r

rr

707200

1039600600707300

-+

+-+

++=

=å

( )N 5074391607 kjiRrrrr

-+=

• Compute the equivalent couple,

( )

kkji

Fr

jkji

Fr

kikji

Fr

FrM

DAD

cAC

BAB

RA

r

rrr

rr

r

rrr

rr

rr

rrr

rr

rrv

9.163010396000100.0100.0

68.177070707050.00075.0

4530200600300050.00075.0

=-=´

=--=´

-=-

=´

´=å

3 - 52


( )

kkji

Fr

jkji

Fr

kikji

Fr

FrM

DAD

cAC

BAB

RA

r

rrr

rr

r

rrr

rr

rr

rrr

rr

rrv

9.163010396000100.0100.0

68.177070707050.00075.0

4530200600300050.00075.0

=-=´

=--=´

-=-

=´

´=å

kjiM RA

rrrr9.11868.1730 ++=

StaticsStatics Practice Problems

q 3.72, 3.80, 3.96, 3.106, 3.119, 3.126, 3.147

3 - 53


rigid bodies: equivalent systems of forces...

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