6-Moments Couples and Force Couple Systems

7/24/2019 6-Moments Couples and Force Couple Systems

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Lecture #6

Moments, Couples, and Force

Couple Systems


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Equivalent Forces

We defined equivalent forces as being forces

with the same magnitude acting in the same

direction and acting along the same line of

action (this is through the Principle of

ransmissibility!, but why do the forces need

to act along the same line"


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4.1 Introduction to Moments

he tendency of a force to rotate a rigid body

about any defined a#is is called the Moment

of the force about the a#is


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MOMENT OF A FORE ! "ALAR FORMLATION $"ection 4.1%

he moment, M, of a force about a point provides a measure of the tendency for

rotation (sometimes called a torque!$

M

M % F & d


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Moment caused &' a Force

he Moment of Force (F! about an a#is

through Point ('! or for short, the Moment of

Fabout ', is the product of the magnitude of

the force and the perpendicular distance

between Point ('! and the line of action of

Force (F!

M'% Fd


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nits o( a Moment

he units of a Moment are Nmin the S) system

ftlbsor inlbsin the *S Customary system


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A))LIATION"

+eams are often used to bridge gaps in walls$ We have tonow what the effect of the force on the beam will have on

the beam supports$

What do you thin those impacts are at points ' and +"


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A))LIATION"

Carpenters often use a hammer in this way to pull a stubborn nail$hrough what sort of action does the force F- at the handle pull the nail"

-ow can you mathematically model the effect of force F- at point ."


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)ro*erties o( a Moment

Moments not only have

a magnitude, they also

have a sense to them$

he sense of a moment

is clocwise or counter/

clocwise depending

on which way it willtend to mae the ob0ect

rotate


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)ro*erties o( a Moment

he sense of a Moment is defined by the

direction it is acting on the '#is and can be

found using 1ight -and 1ule$


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+ari,non-s Teorem

The moment of a Force about any axis is

equal to the sum of the moments of its

components about that axis

his means that resolving or replacing forces

with their resultant force will not affect the

moment on the ob0ect being analy2ed


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MOMENT OF A FORE ! "ALAR FORMLATION (continued!

's shown, d is theperpendiculardistance from point . to the line of actionof

the force$

)n 3/4, the directionof M.is either clocwise or

counter/clocwise, depending on the tendency for rotation$

)n the 3/4 case, the magnitudeof the moment is Mo% F d


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REA/IN0 I2

5$ What is the moment of the 56 7 force about point '

(M'!"

'! 8 79m +! 8: 79m C! 53 79m

4! (53;8! 79m '

d % 8 m

F % 53 7


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E3am*le #1 ' 566/lb vertical force is applied to

the end of a lever which is attachedto a shaft at O$

4eterminea! Moment about O,

b! -ori2ontal force atA whichcreates the same moment,

c! Smallest force at ' whichproduces the same moment,

d! ?ocation for a 3@6/lb vertical

force to produce the samemoment,

e! Whether any of the forces fromb, c, and d is equivalent to theoriginal force$


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E3am*le #1

( )

( ) ( )in.12lb100

in.1260cosin.24

=

==

=

O

O

M

d

FdM

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.

inlb1200 =OM


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E3am*le #1

( )

( )

in.!.20

in.lb1200

in.!.20in.lb1200

in.!.2060sinin.24

=

==

==

F

F

FdM

d

O

b) "ori#ontal force atAthat produces the same

moment

lb$.%$=F


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E3am*le #1

( )

in.42

in.lb1200

in.42in.lb1200

=

=

=

F

F

FdMO

c) &he smallest force atAto produce the same

moment occurs when the perpendicular distance is

a ma'imum or whenFis perpendicular to OA.

lb%0=F


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E3am*le #1

( )

in.%cos60

in.%lb402

in.lb1200lb240in.lb1200

=

=

=

=

=

OB

d

d

FdMO

d) &o determine the point of application of a 240 lb

force to produce the same moment

in.10=OB


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E3am*le #1

e) (lthouh each of the forces in parts b) c) and d)

produces the same moment as the 100 lb force none

are of the same manitude and sense or on the same

line of action. *one of the forces is equivalent to the

100 lb force.


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4.4 )rinci*le o( Moments

AarignonBs heorem he moment of a force

about a point is equal to the sum of moments

of the components of the force about the

point


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MOMENT OF A FORE ! "ALAR FORMLATION (continued!

.ften it is easier to determine M.by using the components of Fas shown

(AarignonBs heorem!$

hen M.% (Fa! D (FEb!$ 7ote the different signs on the terms he typical

sign convention for a moment in 3/4 is that counter/clocwise is considered

positive$ We can determine the direction of rotation by imagining the body

pinned at . and deciding which way the body would rotate because of the

force$

For e#ample, M.

% F d and the

direction is counter/clocwise$

Fa

b

d

.

ab

.

F

F #

F y


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0RO) )ROLEM "OL+IN0

Since this is a 3/4 problem

5! 1esolve the 36 lb force along the

handleBs # and y a#es$

3! 4etermine M'using a scalar

analysis$

0iven5' 36 lb force is applied to the

hammer$Find5 he moment of the force at '$

)lan5

#

y


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0RO) )ROLEM "OL+IN0 $cont.%

"olution

G Fy % 36 sin 86H lb

G F# % 36 cos 86H lb

#

y

G M'% I(36 cos 86H!lb (5J in!(36 sin 36H!lb (K in!L

% 8K5$== lb9in % 8K3 lb9in (clocwise!


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Moments in /

4.7 Moment o( a Force a&out a

"*eci(ic A3is

)n 34 bodies the moment is due to a force

contained in the plane of action perpendicular

to the a#is it is acting around$ his maes the

analysis very easy$

)n 84 situations, this is very seldom found to

be the case$


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Moments in /

he moment about an a#is is still calculatedthe same way (by a force in the planeperpendicular to the a#is! but most forces are

acting in abstract angles$ +y resolving the abstract force into its

rectangular components (or into itscomponents perpendicular to the a#is of

concern! the moment about the a#is can thenbe found the same way it was found in 34 DM % Fd (where d is the distance between theforce and the a#is of concern!


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Notation (or Moments

)n simpler terms the Moment of a Force about

the y/a#is (My! can be found by using the

pro0ection of the Force on thex-zPlane

he 7otation used to denote Moments about

the Cartesian '#es are (Mx, My, and Mz!


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>iven the tension in cable +C is =66 7, find M#, My, and M2 about point '$

84 Moments


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Force ou*les

' Couple is defined as two Forces having the

same magnitude, parallel lines of action, and

opposite sense

)n this situation, the sum of the forces in each

direction is 2ero, so a couple does not affect

the sum of forces equations

' force couple will however tend to rotate thebody it is acting on


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Moment /ue to a Force ou*le

+y multiplying the magnitude of one Force by the

distance between the Forces in the Couple, the

moment due to the couple can be calculated$ M % Fdc

he couple will create a moment around an a#is

perpendicular to the plane that the couple falls in$

Pay attention to the sense of the Moment (1ight

-and 1ule!


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Moment o( a ou*le

&wo couples will have equal moments if

+ 2211 dFdF =

+ the two couples lie in parallel planes and

+ the two couples have the same sense orthe tendenc, to cause rotation in the same

direction.


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8' do 9e use Force ou*les:

he reason we use Force Couples to analy2e

Moments is that the location of the a#is the

Moment is calculated about does not matter

he Moment of a Couple is constant over the

entire body it is acting on


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ou*les are Free +ectors

he point of action of a Couple does notmatter

he plane that the Couple is acting in does

not matter'll that matters is the orientation of the plane

the Couple is acting in

herefore, a Force Couple is said to be aFree Aector and can be applied at any pointon the body it is acting


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Resolution o( +ectors

he Moment due to the Force Couple is

normally placed at the Cartesian Coordinate

.rigin and resolved into its #, y, and 2

components (Mx, My, and Mz!$


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+ector Addition o( ou*les

+y applying AarignonBs heorem to the Forces in the

Couple, it can be proven that couples can be added

and resolved as Aectors$


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Force ou*le "'stem

wo opposing force can be added to a rigid bodywithout affecting the equilibrium of it$

)f there is a force acting at a distance from an a#is,

two forces of equal magnitude and oppositedirection can be added at the a#is with out affectingthe equilibrium of the rigid body$

he original force and its opposing force at the a#ismae a couple that equates to a moment on therigid body$

he other force at the a#is results in the same forceacting on the body


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Force ou*le "'stems

's a result of this it can be stated that any

force (F) actin on a rii! bo!y may be

mo"e! to any i"en point on the rii! bo!y as

lon as a moment equal to moment of (F)about the axis is a!!e! to the rii! bo!y#


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Force ou*le "'stems

he reverse of this is also true, any force an!

moment actin at a point on a rii! bo!y can

be represente! as one force at a !istance

from the oriinal point $here the moment ofthe force about the oriinal point is equal to

the oriinal moment

R lt t ( " t (


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Resultant o( a "'stem o(

o*lanar Forces

iven a system of forces acting on a rigid body,

each force can be resolved into a force/couple

system acting at one point$

he Forces can then be added as Aectors and theMoments can be added as scalars as long as the

sense of each moment is taen into account$

his will reduce the system of forces to a single

force/moment system acting on the rigid body$ his can be reduced further to one force at a

distance Mr;Fr from the point of interest$

R l ti ( F i t


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Resolution o( a Force into a

Force ou*le "'stem in /

he same principles used in 34 apply in 84

also

he method used to resolve a force in 84 into

a Force/Couple system acting at one point is

to draw the rectangular coordinate system at

the point of interest$

1esolve the force in the plane it is in into theforce acting on one a#is and moment about

the a#is for each a#is one at a time$

R l ti ( " t (


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Resolution o( a "'stem o(

Forces in /


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6-Moments Couples and Force Couple Systems