engineering mechanics: statics in si units, 12e...1. moment of a force –scalar formation 2. cross...

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Force System Resultants44

Engineering Mechanics:

Statics in SI Units, 12e

Copyright © 2010 Pearson Education South Asia Pte Ltd


Chapter Objectives

• Concept of moment of a force in two and three

dimensions

• Method for finding the moment of a force about a

specified axis.

• Define the moment of a couple.

• Determine the resultants of non-concurrent force

systems

• Reduce a simple distributed loading to a resultant force

having a specified location

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Chapter Outline

1. Moment of a Force – Scalar Formation

2. Cross Product

3. Moment of Force – Vector Formulation

4. Principle of Moments

5. Moment of a Force about a Specified Axis

6. Moment of a Couple

7. Simplification of a Force and Couple System

8. Further Simplification of a Force and Couple System

9. Reduction of a Simple Distributed Loading


4.1 Moment of a Force – Scalar Formation

• Moment of a force about a point or axis – a measure

of the tendency of the force to cause a body to rotate

about the point or axis

• Torque – tendency of rotation caused by Fx or simple

moment (Mo) z

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Magnitude

• For magnitude of MO,

MO = Fd (N.m) or (Ib.ft)

where d = perpendicular distance

from O to its line of action of force

Direction

• Direction using “right hand rule”



Resultant Moment

• Resultant moment, MRo = moments of all the forces

MRo = ∑Fd

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Example 4.1

For each case, determine the moment of the force about

point O.


Solution

Line of action is extended as a dashed line to establish

moment arm d.

Tendency to rotate is indicated and the orbit is shown as

a colored curl.

)(.200)2)(100()( CWmNmNMa o ==

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Solution

)(.5.37)75.0)(50()( CWmNmNMb o ==

)(.229)30cos24)(40()( CWmNmmNMc o =+= o


Solution

)(.4.42)45sin1)(60()( CCWmNmNMd o == o

)(.0.21)14)(7()( CCWmkNmmkNMe o =−=

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4.2 Cross Product

• Cross product of two vectors A and B yields C, which

is written as

C = A X B

and is read C equals A cross B

Magnitude

• Magnitude of C is the product of

the magnitudes of A and B

• For angle θ, 0 ≤ θ ≤ 180°

C = AB sinθ


4.2 Cross Product

Direction

• Vector C has a direction that is perpendicular to the

plane containing A and B such that C is specified by

the right hand rule

• Expressing vector C when

magnitude and direction are known

C = A X B = (AB sinθ)uC

Magnitude of C Direction of C

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4.2 Cross Product

Laws of Operations

1. Commutative law is not valid

A X B ≠ B X A

Rather,

A X B = - B X A

• Cross product A X B yields a

vector opposite in direction to C

B X A = -C


4.2 Cross Product

Laws of Operations

2. Multiplication by a Scalar

a( A X B ) = (aA) X B = A X (aB) = ( A X B )a

3. Distributive Law

A X ( B + D ) = ( A X B ) + ( A X D )

• Proper order of the cross product must be maintained

since they are not commutative

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4.2 Cross Product

Cartesian Vector Formulation

• Use C = AB sinθ on pair of Cartesian unit vectors

• A more compact determinant in the form as

zyx

zyx

BBB

AAA

kji

BXA

rrr

rr=


4.3 Moment of Force - Vector Formulation

• Moment of force F about point O can be expressed

using cross product

MO = r X F

Magnitude

• For magnitude of cross product,

MO = rF sinθ

• Treat r as a sliding vector. Since d = r sinθ,

MO = rF sinθ = F (rsinθ) = Fd

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Direction

• Direction and sense of MO are determined by right-

hand rule

*Note:

- “curl” of the fingers indicates the sense of rotation

- Maintain proper order of r and F since cross product

is not commutative

MO = rF sinθ

• Treat r as a sliding vector. Since d = r sinθ,

MO = rF sinθ = F (rsinθ) = Fd



Principle of Transmissibility

• For force F applied at any point A, moment created

about O is MO = rA x F

• F has the properties of a sliding vector, thus

MO = r1 X F = r2 X F = r3 X F

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Cartesian Vector Formulation

• For force expressed in Cartesian form,

• With the determinant expended,

MO = (ryFz – rzFy)i

– (rxFz - rzFx)j + (rxFy – ryFx)k

zyx

zyxO

FFF

rrr

kji

FXrM

rrr

rrr==



Resultant Moment of a System of Forces

• Resultant moment of forces about point O can be

determined by vector addition

MRo = ∑(r x F)

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Example 4.4

Two forces act on the rod. Determine the resultant

moment they create about the flange at O. Express the

result as a Cartesian vector.


Solution

Position vectors are directed from point O to each force

as shown.

These vectors are

The resultant moment about O is

{ }{ }m 254

m 5

kjir

jr

B

A

−+=

=

( )

{ } mkN 604030

304080

254

204060

050

⋅+−=

−

−+

−

=

×+×=×=∑

kji

kjikji

FrFrFrM BAO

r

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4.4 Principles of Moments

• Also known as Varignon’s Theorem

“Moment of a force about a point is equal to the sum of

the moments of the forces’ components about the point”

• Since F = F1 + F2,

MO = r X F

= r X (F1 + F2)

= r X F1 + r X F2


Example 4.5

Determine the moment of the force about point O.

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Solution

The moment arm d can be found from trigonometry,

Thus,

Since the force tends to rotate or orbit clockwise about

point O, the moment is directed into the page.

( ) m 898.275sin3 =°=d

( )( ) mkN 5.14898.25 ⋅=== FdMO


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Examples and Problems

• Resolve the following problems for Homework:

4-29, 4-37 and 4-49



4.5 Moment of a Force about a Specified Axis

• For moment of a force about a point, the moment and

its axis is always perpendicular to the plane

• A scalar or vector analysis is used to find the

component of the moment along a specified axis that

passes through the point

yy FdM =

θcosddy =

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Scalar Analysis

• According to the right-hand rule, My is directed along

the positive y axis

• For any axis, the moment is

• Force will not contribute a moment

if force line of action is parallel or

passes through the axis

aa FdM =



Vector Analysis

• For magnitude of MA,

MA= MO·ua

where ua = unit vector

• In determinant form,

zyx

zyx

azayax

axa

FFF

rrr

uuu

FXruM =⋅= )(rrrr

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Example


Example 4.8

Determine the moment produced by the force F which

tends to rotate the rod about the AB axis.

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Solution

Unit vector defines the direction of the AB axis of the rod,

where

For simplicity, choose rD

The force is

{ }m6.0 irD =

{ }ji

ji

r

ru

B

BB 4472.08944.0

2.04.0

2.04.0

22+=

+

+==

r

{ }N 300kF −=


Solution

{ }ji

ji

r

ru

B

BB 4472.08944.0

2.04.0

2.04.0

22+=

+

+==

r

{ }m6.0 irD = { }N 300kF −=

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Examples and Problems

• Resolve the following problems for Homework:

4-51, 4-57 and 4-63


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4.6 Moment of a Couple

• Couple

– two parallel forces

– same magnitude but opposite direction

– separated by perpendicular distance d

• Resultant force = 0

• Tendency to rotate in specified direction

• Couple moment = sum of moments of both couple

forces about any arbitrary point

Page148

Slide 85



Scalar Formulation

• Magnitude of couple moment

M = Fd

• Direction and sense are determined by right hand rule

• M acts perpendicular to plane containing the forces

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Vector Formulation

• For couple moment,

M = r X F

• If moments are taken about point A, moment of –F is

zero about this point

• r is crossed with the force to which it is directed



Equivalent Couples

• 2 couples are equivalent if they produce the same

moment

• Forces of equal couples lie on the same plane or plane

parallel to one another

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Resultant Couple Moment

• Couple moments are free vectors and may be applied

to any point P and added vectorially

• For resultant moment of two couples at point P,

MR = M1 + M2

• For more than 2 moments,

MR = ∑(r X F)


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Example 4.12

Determine the couple moment acting on the pipe.

Segment AB is directed 30° below the x–y plane.


SOLUTION I (VECTOR ANALYSIS)

Take moment about point O,

M = rA X (-250k) + rB X (250k)

= (0.8j) X (-250k) + (0.6cos30ºi

+ 0.8j – 0.6sin30ºk) X (250k)

= {-130j}N.cm

Take moment about point A

M = rAB X (250k)

= (0.6cos30°i – 0.6sin30°k)

X (250k)

= {-130j}N.cm

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SOLUTION II (SCALAR ANALYSIS)

Take moment about point A or B,

M = Fd = 250N(0.5196m)

= 129.9N.cm

Apply right hand rule, M acts in the –j direction

M = {-130j}N.cm


4.7 Simplification of a Force and Couple System

• An equivalent system is when the external effects are

the same as those caused by the original force and

couple moment system

• External effects of a system is the translating and

rotating motion of the body

• Or refers to the reactive forces at the supports if the

body is held fixed

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• Equivalent resultant force acting at point O and a

resultant couple moment is expressed as

• If force system lies in the x–y plane

and couple moments are

perpendicular to this plane,

( ) ∑∑∑

+=

=

MMM

FF

OOR

R

( )( )( ) ∑∑

∑∑

+=

=

=

MMM

FF

FF

OOR

yyR

xxR



Procedure for Analysis

1. Establish the coordinate axes with the origin located at

point O and the axes having a selected orientation

2. Force Summation

3. Moment Summation

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Example 4.16

A structural member is subjected to a couple moment M

and forces F1 and F2. Replace this system with an

equivalent resultant force and couple moment acting at its

base, point O.

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Solution

Express the forces and couple moments as Cartesian

vectors.

mNkjkjM

Njiji

r

rNuNF

NkF

CB

CBCB

.}300400{5

3500

5

4500

}4.1666.249{)1.0()15.0(

1.015.0300

)300()300(

}800{

22

2

1

rrrr

rrrr

r

rrr

rr

+−=

+

−=

+−=

+

+−=

==

−=


Solution

Force Summation.

mNkji

kji

kXkkj

FXrFXrMMMM

Nkji

jikFFF

FF

BCOCRo

R

R

.}300650166{

04.1666.249

11.015.0)800()1()300400(

}8004.1666.249{

4.1666.249800

;

21

21

rrr

rrr

rrrr

rrrrrrrr

rrr

rrrrrr

rr

+−−=

−

−+−++−=

++=Σ+Σ=

−+−=

+−−=+=

Σ=

mNkjkjM .}300400{5

3500

5

4500

rrrr+−=

+

−=

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4.8 Further Simplification of a Force and Couple System

Concurrent Force System

• A concurrent force system is where lines of action of

all the forces intersect at a common point O

∑= FFR



Coplanar Force System

• Lines of action of all the forces lie in the same plane

• Resultant force of this system also lies in this plane

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Parallel Force System

• Consists of forces that are all parallel to the z axis

• Resultant force at point O must also be parallel to this

axis



Reduction to a Wrench

• 3-D force and couple moment system have an

equivalent resultant force acting at point O

• Resultant couple moment not perpendicular to one

another

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Example 4.18

The jib crane is subjected to three coplanar forces.

Replace this loading by an equivalent resultant force and

specify where the resultant’s line of action intersects the

column AB and beam BC.

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Solution

Force Summation

↓=−=

−

−=

Σ=→+

←=−=

−

−=

Σ=→+

NkN

kNNF

FF

kNkN

kNkNF

FF

Ry

yRy

Rx

xRx

60.260.2

6.05

45.2

;

25.325.3

75.15

35.2

;


Solution

For magnitude of resultant force,

For direction of resultant force,

16.4

)60.2()25.3()()( 2222

=

+=+=RyRxR

kN

FFF

o7.38

25.3

60.2tantan 11

=

÷

=

÷

÷

= −−

Rx

Ry

F

Fθ

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Solution

Moment Summation

� Summation of moments about point A,

my

mkNmkN

mkNmkn

kNykN

MM ARA

458.0

)6.1(5

450.2)2.2(

5

350.2

)6.0(6.0)1(75.1

)0(60.2)(25.3

;

=

−

+

−=

+

Σ=


Solution

Moment Summation

� Principle of Transmissibility

mx

mkNmkN

mkNmkn

xkNmkN

MM ARA

177.2

)6.1(5

450.2)2.2(

5

350.2

)6.0(6.0)1(75.1

)(60.2)2.2(25.3

;

=

−

+

−=

−

Σ=

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4.9 Reduction of a Simple Distributed Loading

• Large surface area of a body may be subjected to

distributed loadings

• Loadings on the surface is defined as pressure

• Pressure is measured in Pascal (Pa): 1 Pa = 1N/m2

Uniform Loading Along a Single Axis

• Most common type of distributed

loading is uniform along a

single axis



Magnitude of Resultant Force

• Magnitude of dF is determined from differential area dA

under the loading curve.

• For length L,

• Magnitude of the resultant force is equal to the total

area A under the loading diagram.

( ) AdAdxxwFAL

R === ∫∫

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Location of Resultant Force

• MR = ∑MO

• dF produces a moment of xdF = x w(x) dx about O

• For the entire plate,

• Solving for

∫=Σ=L

RORo dxxxwFxMM )(

x

∫

∫

∫

∫==

A

A

L

L

dA

xdA

dxxw

dxxxw

x)(

)(


Example 4.21

Determine the magnitude and location of the equivalent

resultant force acting on the shaft.

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Solution

For the colored differential area element,

For resultant force

dxxwdxdA 260==

N

x

dxxdAF

FF

A

R

R

160

3

0

3

260

360

60

;

332

0

3

2

0

2

=

−=

=

==

Σ=

∫∫


Solution

For location of line of action,

Checking,

mmax

mNmabA

5.1)2(4

3

4

3

1603

)/240(2

3

===

===

m

xdxxx

dA

xdA

x

A

A

5.1

160

4

0

4

260

160

460

160

)60(44

2

0

42

0

2

=

−

=

===∫

∫

∫

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QUIZ

1. What is the moment of the 10 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

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

• Ad = 3 m

F = 12 N

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QUIZ

3. If a force of magnitude F can be applied in 4 different

2-D configurations (P,Q,R, & S), select the cases

resulting in the maximum and minimum torque values

on the nut. (Max, Min).

A) (Q, P) B) (R, S)

C) (P, R) D) (Q, S)

4. If M = r ×××× F, then what will be the value of M • r ?

A) 0 B) 1

C) r2F D) None of the above.

R

P Q

S


QUIZ

5. 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

6. 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

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QUIZ

7. 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

8. 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.


QUIZ

9. The vector operation (P ×××× Q) • R equals

A) P ×××× (Q • R).

B) R • (P ×××× Q).

C) (P • R) ×××× (Q • R).

D) (P ×××× R) • (Q ×××× R ).

10. The force F is acting along DC. Using the triple

product to determine the moment of F about the bar

BA, you could use any of the following position vectors

except

A) rBC B) rAD C) rAC

D) rDB E) rBD

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QUIZ

11. For finding the moment of the force F about the x-axis,

the position vector in the triple scalar product should

be ___ .

A) rAC B) rBA

C) rAB D) rBC

12. If r = {1 i + 2 j} m and F = {10 i + 20 j + 30 k} N, then

the moment of F about the y-axis is ____ N·m.

A) 10 B) -30

C) -40 D) None of the above.


QUIZ

13. 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

14. The moment of a couple is called a _____ vector.

A) Free B) Spin

C) Romantic D) Sliding

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QUIZ

15. 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

16. 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 .

F1

r1

F2

r2


QUIZ

17. A general system of forces and couple moments

acting on a rigid body can be reduced to a ___ .

A) single force

B) single moment

C) single force and two moments

D) single force and a single moment

18. The original force and couple system and an

equivalent force-couple system have the same

_____ effect on a body.

A) internal B) external

C) internal and external D) microscopic

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QUIZ

18. The forces on the pole can be reduced to a single force and a single moment at point ____ .

A) P B) Q C) R

D) S E) Any of these points.

19. Consider two couples acting on a body. The simplest possible equivalent system at any arbitrary point on the body will have

A) One force and one couple moment.

B) One force.

C) One couple moment.

D) Two couple moments.

R

Z

S

Q

P

X

Y


QUIZ

20. Consider three couples acting on a body. Equivalent

systems will be _______ at different points on the

body.

A) Different when located

B) The same even when located

C) Zero when located

D) None of the above.

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QUIZ

21. The resultant force (FR) due to a distributed load is equivalent to

the _____ under the distributed loading curve, w = w(x).

A) Centroid B) Arc length

C) Area D) Volume

22. The line of action of the distributed load’s equivalent force passes

through the ______ of the distributed load.

A) Centroid B) Mid-point

C) Left edge D) Right edge

x

w

FR

Distributed load curvey


QUIZ

23. What is the location of FR, i.e., the distance d?

A) 2 m B) 3 m C) 4 m

D) 5 m E) 6 m

24. If F1 = 1 N, x1 = 1 m, F2 = 2 N and x2 = 2 m, what

is the location of FR, i.e., the distance x.

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

D) 1.67 m E) 2 m

FR

BAd

BA

3 m 3 m

FRxF2

F1

x1

x2

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QUIZ

25. FR = ____________

A) 12 N B) 100 N

C) 600 N D) 1200 N

26. x = __________.

A) 3 m B) 4 m

C) 6 m D) 8 m

100 N/m

12 m

x

FR

engineering mechanics: statics in si units, 12e...1. moment of a force –scalar formation 2. cross...

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