Week 2---Forces+Moments

7/23/2019 Week 2---Forces+Moments

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COUPLESCOUPLES



–

Tutorial Problems: Ch 3 and Ch 4



•• Types of ForcesTypes of Forces

•• na ys s o orcesna ys s o orces

••

•• Computational MechanicsComputational Mechanics



•• Terminology:Terminology:

– –

•• The straight line collinear with the force vectorThe straight line collinear with the force vector



T es of ForcesT es of Forces

– – External & Internal Forces:External & Internal Forces:

•• x erna orce x erna orce – – g ven o ec s su ec e o ag ven o ec s su ec e o a

force exerted by aforce exerted by a different different objectobject

•• Internal forceInternal force – – one part of a given object isone part of a given object is

sub ected to a force b another art of thesub ected to a force b another art of the same same

objectobject

••

considerationconsideration




– –

•• o y orceo y orce – – orce ac ng on e vo ume oorce ac ng on e vo ume oan objectan object

»» e.g. gravitational force on an objecte.g. gravitational force on an object•• Surface ForceSurface Force – – force acting on theforce acting on the

surface of an objectsurface of an object

»» Can be exerted on an object by contact withCan be exerted on an object by contact withanother objectanother object

resulted from electromagnetic effects




•• Gravitational Forces:Gravitational Forces:

•• The force exerted on an object by the earth’sThe force exerted on an object by the earth’s

•• Gravitational force, or weight, ofGravitational force, or weight, of

an object can be represented by a vectoran object can be represented by a vector




•• •• Magnitude of an object’s weight is related toMagnitude of an object’s weight is related to

its mass b :its mass b :

||WW| =| = mg mg

22 ..




Gravitational forces & electromagnetic forcesGravitational forces & electromagnetic forces

act at a distance:act at a distance:

– – necessarily in contact with thenecessarily in contact with theo ec s exer ng e orceso ec s exer ng e orces




•• Contact Forces:Contact Forces:

– – Forces that result from contacts betweenForces that result from contacts between

objectsobjects – – e.g. push on a walle.g. push on a wall exert a contact forceexert a contact force

– –

wallwall

your hand your hand (Newton’s 3(Newton’s 3rdrd Law)Law)




– –

•• Consider 2 plane surfaces in contact:Consider 2 plane surfaces in contact:

•• Force exerted on the right surface by teForce exerted on the right surface by teleft surface Fleft surface F



T es of ForcesT es of Forces – – Surfaces:Surfaces:

•• Force exerted on right surface by left surface FForce exerted on right surface by left surface F

•• Resolve F into:Resolve F into:

– – Normal force NNormal force N (normal to surface)(normal to surface)

– – Friction force f Friction force f (parallel to surface)(parallel to surface)»» Smooth surfacesSmooth surfaces – – friction force assumed to befriction force assumed to be

negligiblenegligible

– –

neglectedneglected




•• perpendicular & parallel to the plane tangentperpendicular & parallel to the plane tangentto the surface at their point of contactto the surface at their point of contact



T es of ForcesT es of Forces – – Ropes & Cables:Ropes & Cables:

•• attaching a rope or cable to the object &attaching a rope or cable to the object &

pulling on itpulling on it



T es of ForcesT es of Forces – – Ropes & Cables:Ropes & Cables:

– – Cable exerts a force T on containerCable exerts a force T on container

– – Magnitude of TMagnitude of T – – tension in cabletension in cable

– – ne o ac on o co near w ca ene o ac on o co near w ca e

– – Cable exerts an equal & opposite forceCable exerts an equal & opposite force TT

on craneon crane




•• Assumption: Assumption:»» Cable is straightCable is straight

»» Tension where cable is connected toTension where cable is connected tocon a ner = ens on near cranecon a ner = ens on near crane

»» Approximately true if weight of cable << Approximately true if weight of cable <<




•• PullePulle – – wheel with rooved rim that can bewheel with rooved rim that can be used to change the direction of a rope orused to change the direction of a rope or

cablecable




»» Tension is the same on both sides of aTension is the same on both sides of a

»» True when pulley can turn freely & theTrue when pulley can turn freely & the

turns at a constant rateturns at a constant rate




– – Springs:Springs:

•• To exert contact forces in mechanical devicesTo exert contact forces in mechanical devices

•• E.g. suspension of carsE.g. suspension of cars



MOMENT OF A FORCEMOMENT OF A FORCE

ObjectivesObjectives :: Moment of a

Students will be able to:Students will be able to:

orce

a)a) understand and defineunderstand and definemomen , an ,momen , an ,

b) determ ne momentsb) determ ne momentsof a force in 2of a force in 2--D andD and

..



UIZUIZ1. What is the moment of the 10 N1. What is the moment of the 10 N

= A A

A) 10 N A) 10 N·m;·m; B) 30 NB) 30 N·m·m

C) 13 NC) 13 N·m;·m; D) (10/3) ND) (10/3) N·m·m

··• A

d = 3 m

2. Moment of force2. Moment of force F F aboutabout

point O is defined aspoint O is defined as

O O = ___________ .= ___________ .

A) A) rr x x F F B)B) FF x x r r

C)C) rr • F • F D) r * FD) r * F



What is the netWhat is the net

e ect o t e twoe ect o t e twoorces on eorces on e

w eew ee



30 N force on30 N force on

the lu nut?the lu nut?



MOMENT IN 2MOMENT IN 2--DD

The moment of a force about a pointThe moment of a force about a pointrovides a measure of therovides a measure of the tendenc fortendenc for

rotation (sometimes called a torque).rotation (sometimes called a torque).



MOMENT IN 2MOMENT IN 2--DD

--

thethe magnitudemagnitude ofof

the momentthe moment

is Mis Moo = F d= F d

As shown, d is the As shown, d is the perpendicular perpendicular distance fromdistance frompoint O to thepoint O to the line of actionline of action of the force.of the force.

In 2In 2--D, theD, the directiondirection of Mof MOO is either clockwise oris either clockwise or--

rotation.rotation.



For example, MFor example, MOO = F d and= F d andthe direction is counterthe direction is counter--

ab

clockwise.clockwise.d

O

F

Often it is easier toOften it is easier todetermine Mdetermine MOO by using theby using the abF x

y

..O

, , OO X X ..on the terms! The typical sign convention for a moment in 2on the terms! The typical sign convention for a moment in 2--D isD isthat counterthat counter--clockwise is considered positive. We can determineclockwise is considered positive. We can determine

e rec on o ro a on y mag n ng e o y p nne a ane rec on o ro a on y mag n ng e o y p nne a an

deciding which way the body would rotate because of the force.deciding which way the body would rotate because of the force.



EXAMPLE 1EXAMPLE 1

Given:A 400 N force isGiven:A 400 N force isapplied to theapplied to the

°° ..

Find: The moment of theFind: The moment of theorce a .orce a .

PlanPlan::

1) Resolve the force along x and y axes.1) Resolve the force along x and y axes.

e erm nee erm ne A A us ng sca ar ana ys s.us ng sca ar ana ys s.



EXAMPLE 1EXAMPLE 1

Solution:Solution:



1. If a force of magnitude F can be applied in four different 21. If a force of magnitude F can be applied in four different 2--DDcon gura ons , , , , se ec e cases resu ng n econ gura ons , , , , se ec e cases resu ng n emaximum and minimum torque values on the nut. (Max,maximum and minimum torque values on the nut. (Max,

Min).Min).

A) (Q, P) A) (Q, P)

B R SB R S

S

C) (P, R)C) (P, R) R

.. , ,

A) 0 A) 0 B) 1B) 1

rr one o e a ove.one o e a ove.



GROUP PROBLEM SOLVINGGROUP PROBLEM SOLVING

Given: A 40 N force is appliedGiven: A 40 N force is applied. .

Find: The moment of theFind: The moment of the

force at O.force at O.

Plan: 1) Resolve the forcePlan: 1) Resolve the forcealong x and y axes.along x and y axes.

2) Determine M2) Determine MOO usingusing..

↑

°° ↑ y y --

++ → FF x x == -- 40 sin 2040 sin 20°° NN

+ M+ MOO = {= {--(40 c(40 cos 20os 20°°)(200) + (40 sin 20)(200) + (40 sin 20°°)(30)}N·mm)(30)}N·mm

== --7107 N·mm =7107 N·mm = -- 7.11 N·m7.11 N·m



3 m P 2 m3 m P 2 m 5 N5 N

1. Using the CCW direction as positive, the net1. Using the CCW direction as positive, the net

A) 10 N A) 10 N ··m B) 20 Nm B) 20 N ··m C)m C) -- 20 N20 N ··mm

·· -- ··

r r x x F F equals { _______ } Nequals { _______ } N·m.·m.

A) 50 A) 50 i i B) 50B) 50 j j C)C) ––5050 i i

– –



MOMENT ABOUT AN AXISMOMENT ABOUT AN AXIS

Students will be able toStudents will be able to

determine the moment of adetermine the moment of a

aa scalar anal sis andscalar anal sis and

b) vector analysis.b) vector analysis.



MOMENT OF A COUPLEMOMENT OF A COUPLE

a) define a couple, and,a) define a couple, and,

b) determine the moment ofb) determine the moment ofa couple.a couple.



1. In statics, a couple is defined as __________1. In statics, a couple is defined as __________. .

A) two forces in the same direction. A) two forces in the same direction.

wo orces o equa magn u e.wo orces o equa magn u e.

C) two forces of equal magnitude acting in the sameC) two forces of equal magnitude acting in the same..

D) two forces of equal magnitude acting in oppositeD) two forces of equal magnitude acting in oppositedirections.directions.

2. The moment of a cou le is called a2. The moment of a cou le is called a _________ _________vector.vector.

C) romanticC) romantic D) slidingD) sliding



A torque or moment of 12 N A torque or moment of 12 N · m is required to rotate· m is required to rotatethe wheel. Which one of the two grips of the wheelthe wheel. Which one of the two grips of the wheela ove w requ re ess orce o ro a e e w eea ove w requ re ess orce o ro a e e w ee



APPLICATIONS APPLICATIONS

The crossbar lug wrench is being used to loosen a lugThe crossbar lug wrench is being used to loosen a lugnet. What is the effect of changing dimensions a, b, or cnet. What is the effect of changing dimensions a, b, or con e orce a mus e app eon e orce a mus e app e





A couple is defined as two A couple is defined as two

parallel forces with theparallel forces with thesame magnitude butsame magnitude butopposite in directionopposite in directionse arated b ase arated b a

perpendicular distance d.perpendicular distance d.

e momen o a coup e s e ne ase momen o a coup e s e ne as

MMOO

= F= F d (using a scalar analysis) or asd (using a scalar analysis) or as

M M O O == rr × F F (using a vector analysis).(using a vector analysis).

– –F F to the line of action of to the line of action of F F ..

GROUP PROBLEM SOLVINGGROUP PROBLEM SOLVING SCALARSCALAR



GROUP PROBLEM SOLVINGGROUP PROBLEM SOLVING -- SCALAR SCALAR

GivenGiven: Two couples act on the: Two couples act on thebeam. The resultantbeam. The resultantcouple is zero.couple is zero.

forces P and F and theforces P and F and thedistance d.distance d.

PLANPLAN::

1) Use definition of a couple to find P and F.1) Use definition of a couple to find P and F.

2) Resolve the 300 N force in x and y directions.2) Resolve the 300 N force in x and y directions.

3) Determine the net moment.3) Determine the net moment.

4) Equate the net moment to zero to find d.4) Equate the net moment to zero to find d.



SolutionSolution::



50 N

351m 2m

4

1. A1. A couplecouple is applied to the beam as shown. Itsis applied to the beam as shown. Itsmoment equals _____ Nmoment equals _____ N·m.·m.

A) 50 A) 50

C) 80C) 80

D) 100D) 100



--



--COUPLE SYSTEMSCOUPLE SYSTEMS

Ob ectivesOb ectives::Students will be able to:Students will be able to:

1)1)Determine the effect ofDetermine the effect ofmoving a force.moving a force.

2) Find an equivalent2) Find an equivalent

--a system of forces anda system of forces andcouples.couples.



1. A general system of forces and couple moments1. A general system of forces and couple momentsacting on a rigid body can be reduced to a ___ .acting on a rigid body can be reduced to a ___ .

A) single force. A) single force.B) single moment.B) single moment.

C sin le force and two moments.C sin le force and two moments.

D) single force and a single moment.D) single force and a single moment.

2. The original force and couple system and an2. The original force and couple system and an

equivalent forceequivalent force--couple system have the samecouple system have the same _____ effect on a body. _____ effect on a body.

A internal A internal B externalB external

C) internal and externalC) internal and external D) microscopicD) microscopic

APPLICATIONSAPPLICATIONS




What is the resultant effect on the person’s handWhat is the resultant effect on the person’s hand

APPLICATIONSAPPLICATIONS




Several forces and a coupleSeveral forces and a couplemoment are acting on thismoment are acting on this

..

Can you replace them with justCan you replace them with just| | ??

moment at point O that willmoment at point O that will

have the same external effect?have the same external effect?If yes, how will you do that?If yes, how will you do that?

MOVING A FORCE ON ITSMOVING A FORCE ON ITS



MOVING A FORCE ON ITSMOVING A FORCE ON ITS

LINE OF ACTIONLINE OF ACTION

Moving a force from A to O, when both points are on the vectors’Moving a force from A to O, when both points are on the vectors’ , , . ,. ,

vector is called avector is called a sliding vectorsliding vector. (But the internal effect of the. (But the internal effect of theforce on the body does depend on where the force is applied).force on the body does depend on where the force is applied).

MOVING A FORCE NOT ONMOVING A FORCE NOT ON



MOVING A FORCE NOT ONMOVING A FORCE NOT ON

ITS LINE OF ACTIONITS LINE OF ACTION

Moving a force from point A to O (as shown above) requirescreating an additional couple moment.

Since this new couple moment is a “free” vector, it can be appliedat any point P on the body.

RESULTANT OF A FORCE ANDRESULTANT OF A FORCE AND



RESULTANT OF A FORCE ANDRESULTANT OF A FORCE AND

COUPLE SYSTEMCOUPLE SYSTEM

If the force system lies in the xIf the force system lies in the x--y plane (the 2 y plane (the 2--D case),D case),

following three scalar equations.following three scalar equations.

iizRAd FMM Σ+Σ=

EXAMPLE 1EXAMPLE 1



EXAMPLE 1EXAMPLE 1

-- system as shown.system as shown.

Find: The equivalent resultantFind: The equivalent resultant2mforce and couple momentforce and couple momentacting at A and then theacting at A and then thee uivalent sin le forcee uivalent sin le forcelocation along the beam AB.location along the beam AB.

PlanPlan::

2m

1) Sum all the x and y components of the forces to find F1) Sum all the x and y components of the forces to find FRARA..

2) Find and sum all the moments resulting from moving each2) Find and sum all the moments resulting from moving eachforce to A.force to A.

== y y

EXAMPLE 1EXAMPLE 1



EXAMPLE 1EXAMPLE 14m

FR

2m 3m3m

2m

EXAMPLE 2EXAMPLE 2



EXAMPLE 2EXAMPLE 2

columns. Fcolumns. F11 and Fand F22 = 0.= 0.o

force and couple momentforce and couple momentat the origin O. Also findat the origin O. Also find

, ,

single equivalentsingle equivalentresultant force.resultant force.

PlanPlan::

== ∑

== ∑

zozo

2) Find2) Find M M RO RO == ∑ ((r r i i F F i i ) = M) = MRxORxO i i + M+ MRyORyO j j

given as x =given as x = --MM

RyORyO/F/F

RzORzOand y = Mand y = M

RxORxO/F/F

RzORzO

EXAMPLE 2EXAMPLE 2



EXAMPLE 2EXAMPLE 2

UIZUIZ



UIZUIZ1. The forces on the pole can be reduced1. The forces on the pole can be reduced

to a single force and a single momentto a single force and a single moment Z•

•

at point ____ .at point ____ .

A) P A) PR

S•

B) QB) Q

C) R C) R

Q

P

•

D) SD) SE An of these oints.E An of these oints.

X

2. Consider2. Consider two couplestwo couples acting on a body. The simplest possibleacting on a body. The simplest possibleequivalent system at any arbitrary point on the body will haveequivalent system at any arbitrary point on the body will have

A) one force and one couple moment. A) one force and one couple moment.

B) one force.B) one force.

C) one couple moment.C) one couple moment.

D) two couple moments.D) two couple moments.



system at P is ___________ .system at P is ___________ .

RPRP = a ong x= a ong x-- r. anr. an RPRP = .m= .m

B) FB) FRPRP = 0 N and M= 0 N and MRPRP = +30 N.m= +30 N.m

C) FC) FRPRP = 30 N (along +y= 30 N (along +y--dir.) and Mdir.) and MRPRP == --30 N.m30 N.m

D) FD) FRPRP = 40 N (along +x= 40 N (along +x--dir.) and Mdir.) and MRPRP = +30 N.m= +30 N.m

1m 1m

30 N y

P 30 N40 N•



2. Consider three couples acting on a body.2. Consider three couples acting on a body.

Equivalent systems will be _______ at differentEquivalent systems will be _______ at differentpoints on the body.points on the body.

A different when located A different when located

B) the same even when locatedB) the same even when located

zero w en oca ezero w en oca e

D) None of the above.D) None of the above.



REDUCTION OF DISTRIBUTEDREDUCTION OF DISTRIBUTED



LOADINGLOADING

iviv ::

You will be You will be

able toable todetermine andetermine anequivalentequivalentforce for aforce for a

load.load.



Distributed load curveDistributed load curve

.. R R distributed load is equivalent to thedistributed load is equivalent to the

________ under the distributed ________ under the distributed

x oa ng curve, w = w x .oa ng curve, w = w x .

A) centroid B) arc length A) centroid B) arc lengthFR

C) areaC) area D) volumeD) volume

2. The line of action of the distributed load’s equivalent force2. The line of action of the distributed load’s equivalent forcepasses through the ___________ of the distributed load.passes through the ___________ of the distributed load.

A centroi A centroi

B) midB) mid--pointpoint

C)C) left edgeleft edge

D) right edgeD) right edge



the beam exists duethe beam exists due

to the wei ht of theto the wei ht of thelumber.lumber.

reduce this forcereduce this forces stem to a sin les stem to a sin leforce that will haveforce that will havethe same externalthe same external

effect?effect?

, ,



The sandbags on the beam create a distributed load.The sandbags on the beam create a distributed load.

How can we determine a single equivalent resultantHow can we determine a single equivalent resultantforce and its location?force and its location?

DISTRIBUTED LOADINGDISTRIBUTED LOADING



In many situations a surface areaIn many situations a surface area

distributed load.distributed load.

Such forces are caused b windsSuch forces are caused b windsfluids, or the weight of items onfluids, or the weight of items onthe body’s surface.the body’s surface.

We will analyse the most commonWe will analyse the most commoncase of a distributed pressurecase of a distributed pressure

..

This is a uniform load along oneThis is a uniform load along one

axis of a flat rectan ular bod .axis of a flat rectan ular bod .

In such cases, w is a function ofIn such cases, w is a function of

length.length.

MAGNITUDE OF RESULTANTMAGNITUDE OF RESULTANT



Consider an element ofConsider an element of..

The force ma nitude dFThe force ma nitude dFacting on it is given asacting on it is given as

is given byis given by

↓ ∫ ∫

R R ∫

LL ∫

LL

HereHere A is the area under A is the area underthe loading curve w(x).the loading curve w(x).

LOCATION OF THE RESULTANTLOCATION OF THE RESULTANT



The force dF will produce aThe force dF will produce a

moment of x dF about oint O.moment of x dF about oint O.

The total moment about point O isThe total moment about point O isgiven asgiven as

+ M+ M == ∫ x dF = x dF = ∫ x w(x) dx x w(x) dx

produce the moment about point Oproduce the moment about point Oasas

+ M+ MRORO

= ( ) (F= ( ) (FR R ) =) = ∫

LL x w(x) dx x w(x) dx

x

LOCATION OF THE RESULTANTLOCATION OF THE RESULTANT



+ F+ FR R ==∫

LL dF =dF =∫

LL w(x) dxw(x) dx

∫

Comparing the above twoComparing the above two

RORO == R R == ∫LL

equat ons, we getequat ons, we get

You will learn later that F You will learn later that F actsacts through a point “C,” which isthrough a point “C,” which iscalled the geometric center orcalled the geometric center orcentroid of the area under thecentroid of the area under the

loading curve w(x).loading curve w(x).



F .. R R , ,i.e., the distance d?i.e., the distance d?

AB

B A

m; m; m; m;m; m; m; m;E) 6 mE) 6 m

d 3 m 3 m

2. If F2. If F11

= 1 N, x= 1 N, x11

= 1 m,= 1 m,

== ==

what is the location of Fwhat is the location of FR R , ,

R xx2

i.e., the distance x.i.e., the distance x.

A) 1 m; B) 1.33 m; C) 1.5 m; A) 1 m; B) 1.33 m; C) 1.5 m;x1

D) 1.67 m; E) 2 mD) 1.67 m; E) 2 m

GROUP PROBLEM SOLVINGGROUP PROBLEM SOLVING



GivenGiven::

e oa ng on e eam ase oa ng on e eam asshown.shown.

The equivalent force and itsThe equivalent force and its. .

PlanPlan::

1)1) Consider the trapezoidal loading as two separate loadsConsider the trapezoidal loading as two separate loads

(one rectangular and one triangular).(one rectangular and one triangular).

2) Find F2) Find FR R and for each of these two distributed loads.and for each of these two distributed loads. x x

3) Determine the overall F3) Determine the overall FR R and for the three point loadings.and for the three point loadings. x x



UIZUIZ



FR

100 N m

12 m x

1. F1. FR R = ________ = ________ 2. x = _______.2. x = _______.

B) 100 NB) 100 N

m m

B) 4 mB) 4 m

C) 600 NC) 600 N C) 6 mC) 6 m

Week 2---Forces+Moments

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