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VECTOR MECHANICS FOR ENGINEERS:

STATICS

Ninth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

© 2010 The McGraw-Hill Companies, Inc. All rights reserved.

3Rigid Bodies:

Equivalent Systems

of Forces


Vector Mechanics for Engineers: Statics

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Contents & Objectives

3 - 2

Introduction

Define external and internal forces

Understand the principle of

transmissibility

Be able to calculate vector products of

two vectors

Understand and calculate the moment

of a force about a point (2D and 3D)

Varigon’s theorem

Rectangular components of the moment

of a force

Scalar product of two vectors

Scalar product of two vectors:

Applications

Mixed triple product of three vectors

Define and calculate the moment of a

force about a given axis

Define and calculate the moment of a

couple

Resolution of a force into a force at O and

a couple

Be able to find and equivalent force-

couple system of forces

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Introduction

3 - 3

• We need to consider:

• the size of the body and

• the specific points of application of the forces.

• We assume that bodies are rigid, (actual deformations are

small and do not affect the conditions of equilibrium or

motion of the body).



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system consisting of:

one force acting at a given point and

one couple.

Introduction (main concepts)

• 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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External and Internal Forces

3 - 5

• Forces acting on rigid bodies are

divided into two groups:

- External forces

- 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: Equivalent Forces

3 - 6

• Principle of Transmissibility -

Conditions of equilibrium or motion are not

affected by transmitting a force along its line

of action. Forces are sliding vectors.

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.

• Principle of transmissibility may

not always apply in determining

internal forces and deformations.

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Vector Product of Two Vectors

3 - 7

• The moment of a force about a point is a 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 is

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

rule.

sinQPV

• Vector products:

- are not commutative,

- are distributive,

- are not associative,

QPPQ

2121 QPQPQQP

SQPSQP



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Vector Products: Rectangular Components

3 - 8

• Vector products of Cartesian unit vectors,

0

0

0

kkikjjki

ijkjjkji

jikkijii

• Vector products in terms of rectangular

coordinates

kQjQiQkPjPiPV zyxzyx

zyx

zyx

QQQ

PPP

kji

To be mastered!!!

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Calculate the following vector products:



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Moment of a Force About a Point

Where do you think a moment is

applied?

Beams are often used to bridge

gaps in walls. We have to

know what the effect of the

force on the beam will have on

the beam supports.

What do you think those impacts are at points A and B?

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The moment of a force about a point provides a measure of the

tendency for rotation (sometimes called a torque).



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1. What is the moment of the 12 N force

about point A (MA)?

A) 3 N·m B) 36 N·m C) 12 N·m

D) (12/3) N·m E) 7 N·m • Ad = 3 m

F = 12 N

2. The moment of force F about point O

is defined as MO = ___________ .

A) r x F B) F x r

C) r • F D) r * F

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3 - 13

• A force vector is defined by its magnitude and

direction. Its effect on the rigid body depends on its

point of application.

• The moment of F about O is defined as

O

M Fr

• The moment vector MO is perpendicular to the

plane containing O and the force F.

• 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.

• 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 sin



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3 - 14

• Two-dimensional structures have length and breadth but

negligible depth and are subjected to forces contained in

the plane of the structure.

• 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.

• If the force tends to rotate the structure anticlockwise, the

sense of the moment vector is out of the plane of the

structure and the magnitude of the moment is positive.

• 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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Varignon’s Theorem

3 - 15

• 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.

• 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.

1 2 1 2r rF F F Fr



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Rectangular Components of the Moment of a Force

3 - 16

kyFxFjxFzFizFyF

FFF

zyx

kji

kMjMiMM

xyzxyz

zyx

zyxO

The moment of F about O,

kFjFiFF

kzjyixrFrM

zyx

O

,

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3 - 17

The moment of F about B,

FrM BAB

/

kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

/

zyx

BABABAB

FFF

zzyyxx

kji

M



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3 - 18

For two-dimensional structures,

O y x

O Z

y x

M xF yF k

M M

xF yF

B A B A B

B Z

A B A B

y x

y x

M x x y y k

M M

x x y

F F

F Fy

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Sample Problem 3.1

3 - 19

A 500-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 1200-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.



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Sample Problem 3.1

3 - 20

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.

m 0.3500

3.030060cos600

NM

mmmmmd

FdM

O

O

m . N 150OM

SOLUTION:

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Sample Problem 3.1

3 - 21

b) Horizontal force at A that produces the same

moment,

m 0.5196

m . N 150

5196.0m . N 150

m 0.5196 mm 519.660sinmm 600

F

mF

FdM

d

O

N 288.68 F



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Sample Problem 3.1

3 - 22

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 0.6

m . N 150

m 0.6m . N 150

F

F

FdM O

N 250F

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Sample Problem 3.1

3 - 23

d) To determine the point of application of a 1200 N

force to produce the same moment,

mm 125 cos60

mm 125 N 1200

m . N 150

N 1200m . N 150

OB

d

d

FdMO

mm 250OB



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Sample Problem 3.1

3 - 24

e) Although each of the forces in parts b), c), and d)

produces the same moment as the 500-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

500-N force.

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Quiz

2. If r = { 5 j } m and F = { 10 k } N, the moment

r x F equals { _______ } N·m.

A) 50 i B) 50 j C) –50 i

D) – 50 j E) 0

10 N3 m P 2 m

5 N

1. Using the CCW direction as positive, the net moment of the

two forces about point P is

A) 10 N ·m B) 20 N ·m C) - 20 N ·m

D) 40 N ·m E) - 40 N ·m



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Sample Problem 3.4

3 - 26

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:

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

3 - 27

SOLUTION:

12896120

08.003.0

kji

M A

kjiM A

mN 8.82mN 8.82mN 68.7

kirrr ACAC

m 08.0m 3.0

FrM ACA

kji

kji

r

rFF

CD

CD

N 128N 69N 120

m 5.0

m 32.0m 0.24m 3.0N 200

N 200



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Scalar Product of Two Vectors

3 - 28

• The scalar product or dot product between

two vectors P and Q is defined as

• Scalar products:

- are commutative,

- are distributive,

- are not associative,

PQQP

2121 QPQPQQP

undefinedP Q S

• Scalar products with Cartesian unit components,

000111 ikkjjikkjjii

kQjQiQkPjPiPQP zyxzyx

2222 PPPPPP

QPQPQPQP

zyx

zzyyxx

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Scalar Product of Two Vectors: Applications

3 - 29

• Angle between two vectors:

PQ

QPQPQP

QPQPQPPQQP

zzyyxx

zzyyxx

cos

cos

• Projection of a vector on a given axis:

OL

OL

PPQ

QP

PQQP

OLPPP

cos

cos

along of projection cos

zzyyxx

OL

PPP

PP

coscoscos

• For an axis defined by a unit vector:



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Example:

Find the angle between and vectors:

Find the unit vector in the direction of .

3 2 2

1 2 1

V

P

i j k

i j k

P

V

V

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Mixed Triple Product of Three Vectors

3 - 31

• 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,

SPQQSPPQS

PSQSQPQPS

zyx

zyx

zyx

xyyxz

zxxzyyzzyx

QQQ

PPP

SSS

QPQPS

QPQPSQPQPSQPS

• Evaluating the mixed triple product,



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Moment of a Force About a Given Axis - Applications

With the force P, a person is creating a moment MA. Does all

of MA act to turn the socket? How would you calculate an

answer to this question?

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Sleeve A of this bracket can provide a maximum resisting

moment of 125 N·m about the x-axis. How would you

determine the maximum magnitude of F before turning

about the x axis occurs?



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Moment of a Force About a Given Axis

3 - 34

• Moment MO of a force F applied at the point A

about a point O,

OrM F

• Scalar moment MOL about an axis OL is the

projection of the moment vector MO onto the

axis,

is the unit vector along OL

OL OM M r F

• Moments of F about the coordinate axes,

xyz

zxy

yzx

yFxFM

xFzFM

zFyFM

Component of on OLO

rM F

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Moment of a Force About a Given Axis

3 - 35

• Moment of a force about an arbitrary axis,

BL B

A B

A B A B

M M

r F

r r r

• The result is independent of the point B

along the given axis.



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Quiz

1. When determining the moment of a force about a specified

axis, the axis must be along _____________.

A) the x axis B) the y axis C) the z axis

D) any line in 3-D space E) any line in the x-y plane

2. The triple scalar product u • ( r F ) results in

A) a scalar quantity ( + or - ). B) a vector quantity.

C) zero. D) a unit vector.

E) an imaginary number.

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Sample Problem 3.5

3 - 37

A cube is acted on by a force P as

shown. Determine the moment of P

a) about A

b) about the edge AB and

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

3 - 38

• Moment of P about AB,Note that:

1.

2. Vector is i

r is AF

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Sample Problem 3.5

• Moment of P about the diagonal AG,

3 - 39



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Sample Problem 3.5

0

11063

1

2

P

kjikjP

P

PdaP

M AG 6

• Perpendicular distance between AG and FC,

Therefore, P is perpendicular to AG.

6

ad

3 - 40

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Example

1) We need to use

2) Find rB/A

3) Find F in Cartesian vector form.

4) Complete the triple scalar product & solve for F !

Given: Sleeve A can provide a

maximum resisting moment of

125 N·m about the x-axis.

Find: The maximum magnitude of F

before slipping occurs at A (the

sleeve rotating around the x-axis).

Plan:

.xM AB F



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F = F {(– cos 60° i + cos 60° j + cos 45° k)} N

= { – 0.5 F i + 0.5 F j + 0.707 F k} N

rB/A = {(– 0.15 – 0) i + (0.30 – 0) j + (0.10 – 0) k} m

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Mx = 1 {0.3 (0.707F) – 0.1 (0.5F)} + 0 + 0 = 0.162 F N.m

Mx = 125 N.m = maximum moment along x-axis

125 = 0.162 F

Fmax = 771 N

Now find the triple product,

MX =

1 0 0

-0.15 0.3 0.1

-0.5F 0.5F 0.707F

N.m

.xM AB F



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Couples: Quiz

1. In statics, a couple is defined as __________ separated by a

perpendicular distance.

A) two forces in the same direction

B) two forces of equal magnitude

C) two forces of equal magnitude acting in the same direction

D) two forces of equal magnitude acting in opposite directions

2. The moment of a couple is called a _________ vector.

A) Free B) Spin

C) Romantic D) Sliding

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Couples - Applications

A torque or moment of 12 N.m is required to rotate the wheel.

Why does one of the two grips of the wheel above require less

force to rotate the wheel?



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Couples - Applications

Would older vehicles without power steering have

larger or smaller steering wheels than new X5?

When you grip a vehicle’s steering wheel with both

hands, a couple moment is applied to the wheel.

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Moment of a Couple

3 - 47

• 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,

sin

A B

A B

M r F r F

r r F

r F

M F Fdr



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Moment of a Couple

3 - 48

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.

• The moment vector of the couple is

independent of the choice of the origin of the

coordinate axes (depends only on d), i.e., it is

a free vector that can be applied at any point

with the same effect.

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Addition of Couples

3 - 49

• Consider two intersecting planes P1 and

P2 with each containing a couple

222

111

planein

planein

PFrM

PFrM

• Resultants of the vectors also form a

couple

21 FFrRrM

• By Varigon’s theorem

21

21

MM

FrFrM

• 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

3 - 50

• 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.

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Quiz

2. If three couples act on a body, the overall result is that

A) The net force is not equal to 0.

B) The net force and net moment are equal to 0.

C) The net moment equals 0 but the net force is not

necessarily equal to 0.

D) The net force equals 0 but the net moment is not

necessarily equal to 0 .

1. F1 and F2 form a couple. The moment

of the couple is given by ____ .

A) r1 F1 B) r2 F1

C) F2 r1 D) r2 F2

F1

r1

F2

r2



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Example

Given: A 35 N force couple

acting on the rod.

Find: The couple moment

acting on the rod in

Cartesian vector notation.

Plan:

1) Use M = r F to find the couple moment.

2) Set r = rB/A and F = {35 k} N .

3) Calculate the cross product to find M.

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= {(– 8.75 – 0) i – (0 – 0) j – (0 – 0) k} N·m

= {– 8.75 i + 0 j + 0 k} N·m

M = rB/A F

0 0 35

rB/A = { 0 i – (0.25) j + (0.25 tan 30) k} m

rB/A = {– 0.25 j + 0.1443 k} m

F = {0 i + 0 j + 35 k} N

3500

1443.025.00

kji

M A



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

What are the resultant effects on the person’s hand

when the force is applied in these four different ways?

Why is understanding these difference important when

designing various load-bearing structures?

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Several forces and a couple moment are

acting on this vertical section of an I-

beam.

For the process of designing the I-

beam, it would be very helpful if

you could replace the various forces

and moment just one force and one

couple moment at point O with the

same external effect? How will you

do that?

| | ??



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The effect on the vertical pole can be confined to point O.

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3 - 57

• 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. The couple is a free

vector.



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3 - 58

• Moving F from A to a different point O’ requires the

addition of a different couple vector MO’

FrM O

'

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

FsM

FsFrFsrFrM

O

O

''

• 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 3.6

3 - 59

Determine the components of the

single couple equivalent to the

couples shown.

SOLUTION:

• Attach equal and opposite 100-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.



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Sample Problem 3.6

3 - 60

• The three couples may be represented by

three couple vectors,

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Sample Problem 3.6

3 - 61

• Alternatively, compute the sum of the

moments of the four forces about D.



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System of Forces: Reduction to a Force and Couple

3 - 62

• 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,

FrMFR R

O

• The force-couple system at O may be moved to O’

with the addition of the moment of R about O’ ,

RsMM R

O

R

O

'

• Two systems of forces are equivalent if they can be

reduced to the same force-couple system.

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Further Reduction of a System of Forces

3 - 63

• 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,

2) the forces are coplanar, or

3) the forces are parallel.

Conditions under which a given system of forces can

be reduced to a single force.



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Further Reduction of a System of Forces

3 - 64

• System of coplanar forces is reduced to a

force-couple system that is

mutually perpendicular.

R

OMR

and

• 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,

R

Oxy MyRxR

x and y are the coordinates of the point of

application of the new position of R.

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Sample Problem 3.8

3 - 65

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.

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

3 - 66

SOLUTION:

a) Compute the resultant force and the

resultant couple at A.

jjjj

FR

N 250N 100N 600N 150

jR

N600

ji

jiji

FrM R

A

2508.4

1008.26006.1

kM R

A

mN 1880

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Sample Problem 3.8

3 - 67

b) Find an equivalent force-couple system at B

based on the force-couple system at A.

The force is unchanged by the movement of the

force-couple system from A to B.

jR

N 600

The couple at B is equal to the moment about B

of the force-couple system found at A.

kk

jik

RrMM AB

R

A

R

B

mN 2880mN 1880

N 600m 8.4mN 1880

kM R

B

mN 1000



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Sample Problem 3.10

3 - 68

Three cables are attached to the

bracket as shown. Replace the

forces with an equivalent force-

couple system at A.

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.

• Compute the equivalent force,

FR

• Compute the equivalent couple,

FrM R

A

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Sample Problem 3.10

3 - 69

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

• Resolve the forces into rectangular

components.

N 200600300

289.0857.0429.0

175

5015075

N 700

kjiF

kji

kji

r

r

F

B

BE

BE

B

N 1039600

30cos60cosN 1200

ji

jiFD

N 707707

45cos45cosN 1000

ki

kiFC



Nin

thE

ditio

n

Sample Problem 3.10

3 - 70

• Compute the equivalent force,

k

j

i

FR

707200

1039600

600707300

N 5074391607 kjiR

• Compute the equivalent couple,

k

kji

Fr

j

kji

Fr

ki

kji

Fr

FrM

DAD

cAC

BAB

R

A

9.163

01039600

0100.0100.0

68.17

7070707

050.00075.0

4530

200600300

050.00075.0

kjiM R

A

9.11868.1730

36



Nin

thE

ditio

n

Replace each loading with an equivalent force-couple system at end A of the

beam. Which ones are equivalent? Reduce (a) to a single force.



Nin

thE

ditio

n

37



Nin

thE

ditio

n

The resultant is simply:

180 54 36 90 360( ) jR jF

At O we can reduce the system to a single force and couple.



Nin

thE

ditio

n324 378R

OM F i kr

The resultant force and couple are

mutually perpendicular. We can reduce it

further to a single force.

360 324 378

360 360 324

360 378 360 324

1 05 9

378

0. .

R

Oi k

i

M R x z jr i k

kx k

x z

x m m

iz

z

vector mechanics for engineers: statics - …ybu.edu.tr/fozturk/contents/files/154chapter-3.pdf ·...

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