vector mechanics for engineers:...

VECTOR MECHANICS FOR ENGINEERS:

STATICS

Eighth Edition

Ferdinand P. BeerE. Russell Johnston, Jr.

Lecture Notes:J. Walt OlerTexas Tech University

CHAPTER

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Rigid Bodies: Equivalent Systems of Forces


Vector Mechanics for Engineers: Statics

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ContentsIntroduction

External and Internal Forces

Principle of Transmissibility: Equivalent Forces

Vector Products of Two Vectors

Moment of a Force About a Point

Varignon’s Theorem



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ContentsRectangular 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



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ContentsMoment 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 Oand a Couple



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ContentsSample 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



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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.



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Introduction

•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.



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Introduction: Objectives •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



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Introduction

•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.



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External and Internal Forces•Forces acting on rigid bodies are divided into two groups:- External forces

- Internal forces



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External and Internal Forces•External forces are shown in a free-body diagram.

•If unopposed, each external force can impart a motion of translation or rotation, or both.



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•Principle of Transmissibility -Conditions of equilibrium or motion are not affected by transmitting a force along its line of action.



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•NOTE: F and F’ are equivalent forces.



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•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.



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



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Vector classifications:

-Fixed or bound vectors have well defined points of application that cannot be changed

-Free vectors may be freely moved in space

-Sliding vectors may be applied anywhere along their line of action

without affecting an analysis.



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•A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application.•The moment of F about O is defined as FrMO ×=



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•The moment vector MO is perpendicular to the plane containing O and the force F.

FrMO ×=



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•Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO.

The sense of the moment may be determined by the right-hand rule.

FdrFMO == θsin



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A human hand can comfortably apply a torque (torsional moment) of 1.5 Nm. For a bottle lid of diameter 4 cm how much uniformly distributed force (N/m) do we apply do we apply to open the lid?



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Ans : f(2 * 22/7 * 2*2) = 150f=3 N/cm. Total force =f(2*22/7*2) = 40 N or 4 kg.



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How much torque in turning the steering wheel of the car?

How much torque in tightening the nut of the car wheel?



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•The moment about a give 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.( ) +×+×=++× 2121 FrFrFFr



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•Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force Fby the moments of two or more component forces of F.



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Rectangular Components of the Moment of a Force

The moment of Fabout O,

kFjFiFFkzjyixrFrM

zyx

O++=

++=×= ,



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( ) ( ) ( )kyFxFjxFzFizFyF

FFFzyxkji

kMjMiMM

xyzxyz

zyx

zyxO

−+−+−=

=

++=



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The moment of Fabout B,

FrM BAB ×= /

( ) ( ) ( )kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

++=

−+−+−=

−=/



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( ) ( ) ( )zyx

BABABAB

FFFzzyyxx

kjiM −−−=



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•Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment.



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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,



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Sample Problem 3.1

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

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



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Sample Problem 3.1

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

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



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Sample Problem 3.1

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.



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Sample Problem 3.1

( )( )( )cm 12N 100

m 1260cosm24=

=°==

O

O

Md

FdM

m N 1200 ⋅=OM



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Sample Problem 3.1

c) Horizontal force at Athat produces the same moment,

( )

( )

m 8.20m N 1200

m 8.20m N 1200

m 8.2060sinm 24

⋅=

=⋅=

=°=

F

FFdM

d

O

N7.57=F



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Sample Problem 3.1

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.

( )

m42m N 1200

m 42m N 1200⋅

=

=⋅=

F

FFdM O

N50=F



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Sample Problem 3.1

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

( )

cm 5cos60

cm 5N 402

m N 1200N 240m N 1200

=°

=⋅

=

=⋅=

OB

d

dFdM O

m 10=OB



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Sample Problem 3.1e) 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.



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•Mixed triple product of three vectors,

( ) resultscalar =×• QPS

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



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( ) ( ) ( )( )

zyx

zyx

zyx

xyyxz

zxxzyyzzyx

QQQPPPSSS

QPQPS

QPQPSQPQPSQPS

=

−+

−+−=×••Evaluating the mixed triple product,



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Moment about an axis

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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 ×=•Scalar moment MOL about an axis OLis the projection of the moment vector MO onto the axis,

( )FrMM OOL ×•=•= λλ



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Moment of a Force About a Given Axis

•Moments of F about the coordinate axes,

xyz

zxy

yzx

yFxFM

xFzFM

zFyFM

−=

−=

−=



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Moment of a Force About a Given Axis•Moment of a force about an arbitrary axis,

( )BABA

BA

BBL

rrr

Fr

MM

−=

×•=

•=

λ

λ

•The result is independent of the point B along the given axis.



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Sample Problem 3.4

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.



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Sample Problem 3.4

SOLUTION:

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



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Sample Problem 3.4

SOLUTION:

( ) ( )0.3 m 0.08 mC A C Ar r r i k= − = +

FrM ACA ×=



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Sample Problem 3.4

( )

( ) ( ) ( ) ( )

( ) ( ) ( )

200 N

0.3 m 0.24 m 0.32 m200 N

0.5 m120 N 96 N 128 N

C D

C D

rF F

r

i j k

i j k

λ= =

− + −=

= − + −



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Sample Problem 3.4

1289612008.003.0

−−=

kjiM A

( ) ( ) ( )kjiMA mN 8.82mN 8.82mN 68.7 ⋅+⋅+⋅−=



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Determine the moment about z-axis due to force at C.

( ) ( ) ( )kjiMA mN 8.82mN 8.82mN 68.7 ⋅+⋅+⋅−=

MA . k = 28.8 Nm



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Sample Problem 3.5

a)about Ab)about the edge AB

and

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



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Sample Problem 3.5

c)about the diagonal AG of the cube.

d)Determine the perpendicular distance between AG and FC.



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Sample Problem 3.5

•Moment of Pabout A,

( )( ) ( )( ) ( )jiPjiaM

jiPjiPP

jiajaiar

PrM

A

AF

AFA

+×−=

+=+=

−=−=

×=

2

222

( )( )kjiaPM A ++= 2



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Sample Problem 3.5

•Moment of Pabout AB,

( )( )kjiaPi

MiM AAB

++•=

•=

2

2aPM AB =



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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

λ

λ

6aPM AG −=



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Sample Problem 3.5

•Perpendicular distance between AG and FC,

( ) ( ) ( )

0

11063

12

=

+−=−−•−=•PkjikjPP λ



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Moment of a Couple•Two forces F and -Fhaving the same magnitude, parallel lines of action, and opposite sense are said to form a couple.



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Moment of a Couple•Moment of the couple,

( )( )

FdrFMFr

FrrFrFrM

BA

BA

==×=

×−=

−×+×=

θsin



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Moment of a Couple•The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vectorthat can be applied at any point with the same effect.



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Equivalent couples

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If the wheel was connected to the shaft at a point other than the centre, the wheel would still turn as 12 Nm is a couple and a free vector.



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•

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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.



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Addition of Couples

•Consider two intersecting planes P1 and P2 with each containing a couple

222

111

planein planein PFrMPFrM

×=

×=



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Addition of Couples

•Resultants of the vectors also form a couple

( )21 FFrRrM +×=×=



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Addition of Couples

•By Varigon’s theorem

21

21

MMFrFrM

+=

×+×=

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



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Couples Can Be Represented by Vectors

•Couple vectors may be resolved into component vectors.



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Resultant of couples (free vector)

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The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles.

Presenter

Presentation Notes

This is Problem 4-94 of Statics book, page no. 159. Figure is taken from 04PPT-08 (27 October 2002)



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{ }

{ }

{ }{ }

1

2

1 2

50 N.m20cos 20sin 30 20cos 20cos30

20sin 20 N.m= 9.397 16.276 6.840 N.m

9.397 16.276 6.840 50 N.m= 9.397 16.276 56.840 N.m

R

M kM i j

ki j k

M M Mi j k ki j k

=

= − −+

− − +

= +

= − − + +

− − +



EighthEdition

( ) ( ) ( )

( )( )( )

2 2

1 9.37959.867

1 16.27659.867

1 56.84059.867

9.379 16.276 + 56.840 N.m59.867 N.m=59.9 N.m

cos 99.0

cos 106.0

cos 18.3

RM

α

β

ν

− −

− −

−

= − + −

=

= =

= =

= =

Presenter

Presentation Notes

This is Problem 4-94 of Statics book, page no. 159. Figure is taken from 04PPT-08 (27 October 2002)



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Sample Problem 3.6

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



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Sample Problem 3.6

•Compute the sum of the moments of the four forces about an arbitrary single point. D is a good choice as only the forces at C and E contribute to the moment about D.



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Sample Problem 3.6



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Sample Problem 3.6

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

kjMM D

N 20m 12m 9

N 30m 18

−×−+

−×==

( ) ( )( )k

jiM

mN 180

mN240mN 540

⋅+

⋅+⋅−=



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Introduction

•Any force acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple.



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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.



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



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



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•Moving F from A to a different point O’



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

FrM O ×′='



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

( )FsM

FsFrFsrFrM

O

O

×+=

×+×=×+=×= ''



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•Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’.



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Sample Problem 3.4

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 equivalent force at A.



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( ) ( ) ( )120 N 96 N 128 NRF i j k= − + −

( ) ( ) ( )kjiMA mN 8.82mN 8.82mN 68.7 ⋅+⋅+⋅−=



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Equivalent force system at B ?

MB= MA +BA x FR = MA +80 k x FR

( ) ( ) ( )120 N 96 N 128 NRF i j k= − + −



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Introduction

•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.



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System of Forces: Reduction to a Force and Couple

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



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

( )∑∑ ×== FrMFR RO



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

RsMM RO

RO ×+='



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



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Special cases

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•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.



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Further Reduction of a System of Forces•The resultant force-couple system for a system of forces will be mutually perpendicular

•1) the forces are concurrent, 2) the forces are coplanar, 3) the forces are parallel.



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Concurrent forces

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•System of coplanar forces is reduced to a force-couple system

that is mutually perpendicular.

ROMR and



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

OM

R



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•In terms of rectangular coordinates,R

y x OxR yR M− =

x ,y coordinates of point of application



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Rx OyR M− =

Ry OxR M=



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Sample Problem 3.8For 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.



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Sample Problem 3.8

SOLUTION:

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



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Sample Problem 3.8

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.



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Sample Problem 3.8

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

( ) ( ) ( ) ( ) jjjjFR

N 250N 100N 600N 150 −+−=

=∑

( ) jR N600−=



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( )( ) ( ) ( ) ( )

( ) ( )jijiji

FrM RA

2508.41008.26006.1

−×+

×+−×=

×= ∑

( )kM RA mN 1880 ⋅−=



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Sample Problem 3.8

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



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Sample Problem 3.8

( ) ( ) ( )( ) ( )kk

jik

RrMM ABRA

RB

mN 2880mN 1880

N 600m 8.4mN 1880

⋅+⋅−=

−×−+⋅−=

×+=

( )kM RB mN 1000 ⋅+=



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Sample Problem 3.8

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



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Sample Problem 3.8



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RAM

Resultant of given system of forces is equal to R, and its point of application must besuch that the moment of R about A is equal to .



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( )jR N600−=

( ) ( )( ) ( )kk

kjiMRr

m.N1800N600m.N1800N600

−=−−−=−×

=×

xx

RA

m133.x=



EighthEdition

-(600 N )j

A B

Single Force or Resultant.

The resultant of the given system of forces is equal to R, and its point of application must be such that the moment of R about Ais equal to . We write

and conclude that X =3.13 m. Thus, the single

force equivalent to the given system is defined as

R = 600 N x = 3.13 m



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Sample Problem 3.10

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



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Sample Problem 3.10

SOLUTION:

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



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Sample Problem 3.10•Resolve the forces into rectangular components.

•Compute the equivalent force,

∑= FR



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Sample Problem 3.10

•Compute the equivalent couple,

( )∑ ×= FrM RA



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Sample Problem 3.10SOLUTION:

•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

−=

−=

+=



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Sample Problem 3.10•Resolve the forces into rectangular components.

( )

( )N 200600300

289.0857.0429.0

1755015075

N 700

kjiF

kji

kjirr

F

B

BE

BE

B

+−=

+−=

+−==

=

λ

λ



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Sample Problem 3.10

( )( )( )N 1039600

30cos60cosN 1200ji

jiFD

+=

+=

( )( )( )N 707707

45cos45cosN 1000ki

jiFC

−=−=



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Sample Problem 3.10•Compute the equivalent force,

( )( )( )k

ji

FR

707200

1039600600707300

−+

+−+

++=

= ∑

( )N 5074391607 kjiR −+=



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Sample Problem 3.10

•Compute the equivalent couple,

( )

0.075 0 0.050 30 45300 600 200

RA

B A B

M r F

i j kr F i k

= ×

× = = −−

∑



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( )

kkji

Fr

jkji

Fr

kikji

Fr

FrM

DAD

cAC

BAB

RA

9.163010396000100.0100.0

68.177070707050.00075.0

4530200600300050.00075.0

=−=×

=−−=×

−=−

=×

×=∑

kjiMRA 9.11868.1730 ++=



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Pauli Effect in Action

3 - 134

It was a standing joke among Wolfgang Pauli's colleagues that the famed theoretical physicist should be kept as far away from experimental equipment as humanly possible. His mere presence in a laboratory, it was said, would cause something to go wrong: the power would fail,



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Pauli Effect in Action

3 - 135

vacuum tubes would suddenly leak, instruments would break or malfunction. Indeed, such was the frequency of Pauli-related incidents that the strange phenomenon came to be known as the 'Pauli Effect'.



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One day, a group of Pauli's mischievous colleagues set up a practical joke to show the strange effect in action: an elaborate contraption was designed and constructed to bring a chandelier crashing down the moment Pauli arrived at a large reception. Pauli duly appeared on schedule and the Pauli effect was demonstrated, though not as the mischievous crew had planned:



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One day, some important experimental equipment in Professor James Frank's laboratory at the Physics Institute at the University of Gottingen unexpectedly blew up for no apparent reason. Moreover, Pauli, who was on his way to Denmark, had not even entered the building.

no sooner was the mechanism activated than a pulley jammed - and the chandelier did not come down.



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Only later was it discovered that the catastrophe had taken place at the precise moment that the train carrying Pauli from Zurich to Copenhagen had stopped to collect passengers at the Gottingen railroad station



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Equivalent force systems: The wrench

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http://www.engin.brown.edu/courses/en3/Notes/Statics/Staticequiv/Staticequiv.htm

This is a good example on how CG shifts due to use of fuel in aircraft.



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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.



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•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.



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2. Magnitude of V is

3. Direction of V is obtained from the right-hand rule.

θsinQPV =



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Vector Products: Rectangular Components

x y z

x y z

i j kV P P P

Q Q Q=



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•Two-dimensional structureshave length and breadth but negligible depth and are subjected to forces contained in the plane of the structure.



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•The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane.



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•If the force tends to rotate the structure counter clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive.



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•If the force tends to rotate the structure clockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.



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( ) ( )[ ]

( ) ( ) zBAyBA

ZO

zBAyBAO

FyyFxxMM

kFyyFxxM

−−−==

−−−=



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For two-dimensional structures,

( )

zy

ZO

zyO

yFxFMM

kyFxFM

−==

−=



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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.

•Couple vectors obey the law of addition of vectors.

vector mechanics for engineers:...

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