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1/11/2013 Chapter 4 - Force System Resultants 1 STATICS: CE201 Chapter 4 Force System Resultants Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson Dr M. Touahmia & Dr M. Boukendakdji Civil Engineering Department, University of Hail (2012/2013) 4. Force System Resultants ________________________________________________________________________________________________________________________________________________ Main Goals: 1. Determine the moment of a force. 2. Define the moment of a couple. 3. Determine the resultants of force systems. Contents: 4.1 Moment of a Force Scalar Formulation 4.2 Cross Product 4.3 Moment of a Force Vector Formulation 4.4 Principle of Moments 4.5 Moment of a Force about a Specified Axis 4.6 Moment of a Couple 4.7 Simplification of a Force and Couple System 4.8 Further Simplification of a Force and Couple System 4.9 Reduction of a Simple Distributed Loading 2 Chapter 4 - Force System Resultant
Transcript
Page 1: CE201 Statics Chap4

1/11/2013

Chapter 4 - Force System Resultants 1

STATICS: CE201

Chapter 4

Force System Resultants

Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson

Dr M. Touahmia & Dr M. Boukendakdji

Civil Engineering Department, University of Hail

(2012/2013)

4. Force System Resultants ________________________________________________________________________________________________________________________________________________

Main Goals: 1. Determine the moment of a force.

2. Define the moment of a couple.

3. Determine the resultants of force systems.

Contents: 4.1 Moment of a Force – Scalar Formulation

4.2 Cross Product

4.3 Moment of a Force – Vector Formulation

4.4 Principle of Moments

4.5 Moment of a Force about a Specified Axis

4.6 Moment of a Couple

4.7 Simplification of a Force and Couple System

4.8 Further Simplification of a Force and Couple System

4.9 Reduction of a Simple Distributed Loading

2 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 2

4.1 Moment of a Force – Scalar Formulation

When a force is applied to a body it will produce a

tendency for the body to rotate about a point or axis that

is not on the line of action of the force.

This tendency to rotate is called “moment of a force” or

simply the moment.

3 Chapter 4 - Force System Resultant

4.1 Moment of a Force – Scalar Formulation

For example, consider a wrench used to unscrew the bolt

in the figure. If a force is applied to the handle of the

wrench it will tend to turn the bolt about point O (or the

z axis).

The magnitude of the moment is directly proportional to

the magnitude of F and the perpendicular distance or

moment arm d.

4 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 3

4.1 Moment of a Force – Scalar Formulation

The larger the force or the longer the moment arm, the

greater the moment or turning effect.

NOTE: If force F is applied at an angle then it

will be more difficult to turn the bolt since the moment

arm will be smaller than d. If F is applied

along the wrench, its moment arm will be zero and as a

result, the moment of F about O will be zero also.

5 Chapter 4 - Force System Resultant

90

sindd

4.1 Moment of a Force – Scalar Formulation

The moment of a force F about a point O is defined

as the vector product of F and d, where d is the

perpendicular distance from O to the line of action of the

force F.

Clearly the moment is a vector and has both: magnitude

and direction.

Magnitude: The magnitude of the moment is

determined from:

Units: N.m; kN.m; N.mm

OM

FdM O

OM

6 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 4

4.1 Moment of a Force – Scalar Formulation

Direction: The right-hand rule is used

to establish the sense of direction of the

moment .

The moment of a force will be positive

if it is directed along the +z axis and

will be negative if it is directed along

the –z axis.

OM

7 Chapter 4 - Force System Resultant

4.1 Moment of a Force – Scalar Formulation

Resultant Moment: The Resultant Moment

about point O (the z axis) can be determined by finding

the algebraic sum of the moments caused by all the

forces in the system:

As a convention, we will generally consider positive

moments (+) as counterclockwise and negative moments

(-) as clockwise.

ORM

FdMOR

332211 dFdFdFFdMOR

8 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 5

Example 1

For each case below, determine the moment of the force

about point O.

N.m 5.37m 75.0N 50 OM

Answer

Answer

9 Chapter 4 - Force System Resultant

mN. 200m 2N 100 OM

FdM O

Example 1

kN.m 229m 30cos2 m4kN 40 o

OM

kN.m 229m 45sin1kN 60 o

OM

Answer

Answer

10 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 6

Example 2

Determine the resultant moment of the four forces acting

on the rod about point O.

11 Chapter 4 - Force System Resultant

Solution 2

Assuming that positive moments (+) are in the

counterclockwise:

FdMOR

12 Chapter 4 - Force System Resultant

N.m 334m 30cos3 m 4 N 40

m 30sin3 N 200 N 60m 2 N 50

o

o

ROM

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Chapter 4 - Force System Resultants 7

4.2 Cross Product

The cross product of two vectors A and B yields the

vector C written:

Magnitude: The magnitude of C is defined as the

product of the magnitudes of A and B and the sine of the

angle θ between their tails:

Direction: Vector C has a direction perpendicular to the

plane containing A and B and is specified

by the right-hand rule:

defines the direction of C

BAC

Cu

13 Chapter 4 - Force System Resultant

sin ABC

CAB u BAC sin

Laws of Operation:

Commutative law: The commutative law is not valid

If the cross product is multiplied by a scalar a, it obeys

the associative law:

The vector cross product also obeys the distributive law

of addition:

ABBA

CABBA

aaaa BABABABA

DABADBA

14 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 8

Cross products of unit vectors (i, j, k):

The direction is determined using the right hand rule.

As shown in the diagram, for this case the direction is k

and the Magnitude is:

| i j |=(1)(1)(sin90°) = (1)(1)(1)=1

so: i j = (1) k = k

and:

i j = k i k = -j i i = 0

j k = i j i = -k j j = 0

k i = j k j = -i k k = 0

Alphabetical order +

15 Chapter 4 - Force System Resultant

Cartesian Vector Formulation

The cross product of two vectors A and B expressed in

Cartesian vector form is:

16 Chapter 4 - Force System Resultant

kkjkik

kjjjij

kijiii

kjikjiBA

zzyzxz

zyyyxy

zxyxxx

zyxzyx

BABABA

BABABA

BABABA

BBBAAA

kjiBA xyyxxzzxyzzy BABABABABABA

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Chapter 4 - Force System Resultants 9

Cartesian Vector Formulation

This equation may also be

written in a more compact

determinant form as :

The three minors can be generated with the following

scheme:

Adding the results yields the expanded form of : BA

17 Chapter 4 - Force System Resultant

kjiBA xyyxxzzxyzzy BABABABABABA

4.3 Moment of a Force – Vector Formulation

The moment of a force F about point O, or about an axis

passing through O and perpendicular to the plane

containing O can be expressed using the vector cross

product:

where r is the position vector from O to any point on the

line of action of F.

FrM O

18 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 10

4.3 Moment of a Force – Vector Formulation

Magnitude:

Direction: The direction of Mo follows the right-hand

rule as it applies to the cross product, “r cross F”.

FdrFrFMO sinsin

19 Chapter 4 - Force System Resultant

4.3 Moment of a Force – Vector Formulation

Principle of Transmissibility: We can use any position

vector r measured from point O to any point on the line

of action of the force F:

FrFrFrM 321O

20 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 11

4.3 Moment of a Force – Vector Formulation

Resultant Moment of a System of Forces: If a body is

subjected to a system of forces, the resultant moment of

all the forces about point O is equal to the vector

addition of the moment of each force:

n

i

iiRO

1

FrM

21 Chapter 4 - Force System Resultant

Example 3

Determine the moment MO produced by the force F

about point O. Express the result as a Cartesian vector.

22 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 12

Solution 3

As shown in the figure, either rA or rB can be used to

determine the moment about O:

F expressed as Cartesian vector:

The direction of the unit vector uAB

can be determined from the position

Vector rAB which extends from A to B.

m 12 kr A m 12 4 jir B

23 Chapter 4 - Force System Resultant

ABFuF

m 12 12 4 kjir -AB

Solution 3

The magnitude of rAB which represents the length of AB

is:

Forming the unit vector that defines

The direction of both rAB and F, we

have:

24 Chapter 4 - Force System Resultant

222m 12m 12m 4 ABr

222m 12m 12m 4

12124

kjiru

AB

AB

ABr

kN 376.1 376.1 4588.0

m 12m 12m 4

12124kN 2

222

kji

kjiuF

ABF

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Chapter 4 - Force System Resultants 13

Solution 3

The moment about O: or

, ,

OR

m 12 kr A m 12 4 jir B

FrM AO FrM BO

kN 376.1 376.1 5488.0 kjiF

25 Chapter 4 - Force System Resultant

Example 4

Two forces act on the rod shown below. Determine the

resultant moment they create about the flange at O.

Express the results as a Cartesian vector.

26 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 14

Solution 4

Position vectors rA and rB are directed from point O to

each force:

The resultant moment about O is:

m 5 jr A m 25 4 kjir B

27 Chapter 4 - Force System Resultant

21

1

FrFrFrM

BA

n

i

iiRO

304080

254

204060

050

21

kjikji

FrFrM BARO

Solution 4

The resultant moment about O is:

The coordinate direction

angles were determined

from the unit vector for

The two forces tend to cause the rod to rotate about the

moment axis in the manner shown by the curl indicate

on the moment vector.

ORM

28 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 15

4.3 Principle of Moments

Varignon’s Theorem: The moment of a force about a

point is equal to the sum of the moments of the

components of the force about the point.

For example, consider the moment of the force F which

has two components F1 and F2, therefore:

Then, the moment of F about O is:

2121 FrFrFFrFrM O

21 FFF

29 Chapter 4 - Force System Resultant

4.3 Principle of Moments

For two-dimensional problems (2D), we can use the

principle of moments by resolving the force into its

rectangular components and then determine the moment

using a scalar analysis:

This method is generally easier than finding the same

moment using:

FdM O

xFyFMyxO

30 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 16

Example 3

Determine the moment of the force about point O.

31 Chapter 4 - Force System Resultant

Example 3-(I)

The moment arm d can be found from trigonometry:

The force tends to rotate clockwise about O, the moment

is directed into the page .

m 898.275sinm 3 d

kN.m 5.14m 898.2kN 5 FdM O

32 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 17

Solution 3-(II)

We can apply the principle of moments by resolving the

force into its rectangular components and then determine

the moment using a scalar analysis:

kN.m 5.14

kN.m 5.14

m 30cos3kN 45sin5m 30sin3kN 45cos5

OM

xyyxO dFdFM

33 Chapter 4 - Force System Resultant

Solution 3-(III)

The x and y axes can be set parallel and perpendicular to

the rod’s axis. Fx produces no moment about point O

since its line of action passes through this point:

xyO dFM

kN.m 5.14 OM

kN.m 5.14

m 3kN 75sin5

OM

34 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 18

Example 4

The force F acts at the end of the angle bracket.

Determine the moment of F about point O.

35 Chapter 4 - Force System Resultant

Solution 4-(I) Scalar Analysis

F can be resolved into its x and y components:

N.m 6.98N.m 6.98

m 4.0N 30cos400m 2.0N 30sin400

OM

xFyFM yxO

30sin400 xF

36 Chapter 4 - Force System Resultant

30cos400 yF

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Chapter 4 - Force System Resultants 19

Solution 4-(II) Vector Analysis

Using a Cartesian vector analysis, the force F and the

position vector r can be written as:

m 2.0 4.0 jir

N 4.346 0.200

N 30cos40030sin400

ji

jiF

04.3460.200

02.04.0

kji

FrMO

N.m 6.98

0.2002.04.3464.0 0 0

k

kjiM

O

37 Chapter 4 - Force System Resultant

4.6 Moment of a Couple

A couple is defined as two parallel forces that have the

same magnitude, but opposite directions, and are

separated by a perpendicular distance d.

Since the resultant force is zero, the only effect of a

couple is to produce a rotation or tendency of rotation in

a specified direction.

38 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 20

4.6 Moment of a Couple

The moment produced by a couple is called a couple

moment. We can determine its value by finding the sum

of the moments of both couple force about any arbitrary

point.

The position vectors rA and rB are directed from point O

to points A and B lying on the line of action –F and F.

The couple moment M about O is therefore:

However, or

Therefore:

FrrFrFrM ABAB

)()(

rrr AB AB rrr

FrM

39 Chapter 4 - Force System Resultant

4.6 Moment of a Couple

Scalar Formulation: The moment of a couple, M, is

defined as having a magnitude of:

Vector Formulation: The moment of a couple can also be

expressed by the vector cross product using:

where r is directed from any point on the line of action of

one of the forces to any point on the line of action of the

other force F.

FdM

FrM

40 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 21

4.6 Moment of a Couple

Equivalent Couples: If two couples produce a moment

with the same magnitude and direction, then these two

couples are equivalent.

The two couple moments shown in the figure are

equivalent because each couple moment has a magnitude

of 12 N.m and each is directed into the plane of the page.

N.m 12m 4.0N 30 M

N.m 12m 3.0N 40 M

41 Chapter 4 - Force System Resultant

4.6 Moment of a Couple

Resultant Couple Moment: It is simply the vector sum of

all the couple moments of the system.

For example, consider the couple moments M1 and M2

acting on the pipe. We can join their tails at any arbitrary

point and find the resultant couple moment.

FrMR

21 MMM R

42 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 22

Example 5

Determine the resultant couple moment of the three

couples acting on the plate.

43 Chapter 4 - Force System Resultant

Solution 5

Considering counterclockwise couple moments as

positive (+), we have:

N.m 95N.m 95

m 5.0N 300m 3.0N 450m 4.0N 200

RM

332211 dFdFdFMM R

44 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 23

Example 6

Determine the magnitude and direction of the couple

moment acting on the gear.

45 Chapter 4 - Force System Resultant

Solution 6

Resolving each force into its components. The couple

moment can be determined by summing the moments of

these force components about any point (For example

the center O or point A):

OR

N.m 9.43

2.030sin6002.030cos600

OMM

N.m 9.43

2.030sin6002.030cos600

AMM

46 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 24

Solution 6

The same result can also be obtained using:

where d is the perpendicular distance between the lines

of action of the couple forces. However, the computation

for d is more involved.

FdM

47 Chapter 4 - Force System Resultant

4.7 Simplification of a Force and Couple System

Sometimes it is convenient to reduce a system of forces and

couple moments acting on a body to a simpler form by

replacing it with an equivalent system.

Example:

48 Chapter 4 - Force System Resultant

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Chapter 4 - Force System Resultants 25

System of Forces and Couple Moments:

A system of several forces and couple moments acting

on a body can be reduced to an equivalent single

resultant force acting at a point O and a resultant couple

moment.

Chapter 4 - Force System Resultant 49

System of Forces and Couple Moments:

Example: In the figure, point O is not on the line of action

of F1 and so this force can be moved to point O provided a

couple moment is added to the body.

Similarly, the couple moment should be added

to the body when we move F2 to point O.

Finally, since the couple moment M is a free vector, it can

just be moved to point O.

50 Chapter 4 - Force System Resultant

111 FrM

222 FrM

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Chapter 4 - Force System Resultants 26

System of Forces and Couple Moments:

If we sum the forces and couple moments, we obtain the

resultant force and the resultant couple moment:

Chapter 4 - Force System Resultant 51

21 FFF R 21 MMMM

OR

System of Forces and Couple Moments:

We can generalize this method of reducing a force and

couple system to an equivalent resultant forces FR acting

at point O and a resultant couple moment (MR)O by

using the following equations:

If the force system lies in the x-y plane and any couple

moments are perpendicular to this plan, then the above

equations reduce to the three scalar equations:

Chapter 4 - Force System Resultant 52

FFR

MMM OOR

xxR FF yyR FF

MMM OOR

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Chapter 4 - Force System Resultants 27

Example 7

Replace the force and couple system by an equivalent

resultant force and couple moment acting at point O.

Chapter 4 - Force System Resultant 53

Solution 7

Force Summation: The 3 kN and 5 kN forces are resolved

into their x and y components as shown below. We have:

54 Chapter 4 - Force System Resultant

kN 598.5kN 55

330coskN 3

xxR FF

kN 50.6kN 4kN 55

430sinkN 3

yyR FF

kN 58.850.6598.52222 RyRxR FFF

o

xR

yR

F

F3.49

598.5

50.6tantan 11

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Chapter 4 - Force System Resultants 28

Solution 7

Moment Summation: The moments of 3 kN and 5 kN

about point O will be determined using their x and y

components. We have:

55 Chapter 4 - Force System Resultant

kN 46.2kN.m 46.2m 2.0kN 4

m 5.0kN 55

4m 1.0kN 5

5

3

m 1.030coskN 3m 2.030sinkN 3

OOR MM

Example 8

Replace the force and couple system acting on the

member by an equivalent resultant force and couple

moment acting at point O.

Chapter 4 - Force System Resultant 56

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Chapter 4 - Force System Resultants 29

Solution 8

Force Summation: Since the couple forces of 200 N are

equal but opposite, they produce zero resultant force.

The 500 N force is resolved into its x and y components:

Chapter 4 - Force System Resultant 57

N 300kN 5005

3

xxR FF

N 3507505

4N 500

yyRFF

N 462N 350N 3002222 yRxRR FFF

4.49

N 300

N 350tantan 11

xR

yR

F

F

Solution 8

Moment Summation: Since the couple moment is a free

vector, it can act at any point on the member, then:

Chapter 4 - Force System Resultant 58

N.m 5.37N.m 5.37

m 1N 200m 25.1N 750

m 15

3N 500N 5.2

5

4N 500

cOOR MMM

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Chapter 4 - Force System Resultants 30

4.8 Reduction of a Simple Distributed Loading

Sometimes, a body may be subjected to a loading that is

distributed over its surface.

For example:

The pressure of the wind on the face of a sign.

The pressure of water within a tank.

The weight of sand on the floor of a storage container.

The pressure exerted at each point on the surface

indicates the intensity of the loading. It is measured

using Pascals Pa (N/m2).

Chapter 4 - Force System Resultant 59

Uniform Loadings Along a Single Axis:

The most common type of distributed

loading encountered in engineering practice

is generally uniform along a single axis.

Example: The beam (or plate) that has a

constant width and is subjected to a pressure

loading that varies only along the x axis.

This loading can be described by the

function:

We can replace this coplanar parallel force

system with a single equivalent resultant

force FR acting at a specific location on the

beam.

60 Chapter 4 - Force System Resultant

2N/m )(xpp

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Chapter 4 - Force System Resultants 31

Magnitude of Resultant Force:

The magnitude of FR is equivalent to the

sum of all the forces in the system:

Integration must be used since there is an

infinite number of parallel forces dF acting

on the beam.

Since dF is acting on an element of

length dx, and w(x) is a force per unit

length, Then:

Therefore, the magnitude of the

resultant force is equal to the total area A

under the loading diagram.

AL

R AdAdxxwF

61 Chapter 4 - Force System Resultant

FFR

dAdxxwdF

Location of Resultant Force:

The location of the line of action of FR

can be determined by equating the moments

of the force resultant and the parallel force

distribution about point O.

dF produces a moment of:

Then for the entire length:

x

L

R dxxxwFx

A

A

L

L

dA

xdA

dxxw

dxxxwx

62 Chapter 4 - Force System Resultant

dxxxwxdF

OOR MM

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Chapter 4 - Force System Resultants 32

Example 9

Determine the magnitude and location of the equivalent

resultant force acting on the shaft.

Chapter 4 - Force System Resultant 63

Solution 9

Since w = w(x) is given, this problem will be solved by

integration.

The differential element has an area:

Chapter 4 - Force System Resultant 64

N 1603

0

3

260

0

m2

36060x

33m2

0

2

xdxdAF

A

R

m 5.1

N 160

4

0

4

260

N 160

0

m 2

460

N 160

60444m 2

0

2

x

dxxx

dA

xdA

x

A

A

dxxwdxdA 260

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Chapter 4 - Force System Resultants 33

Example 10

The granular material exerts the distributed loading on

the beam. Determine the magnitude and location of the

equivalent resultant of this load.

Chapter 4 - Force System Resultant 65

Solution 10

The area of the loading diagram is a trapezoid. It can be

divided into a rectangular and triangular loading.

Chapter 4 - Force System Resultant 66

kN 225kN/m 50m 92

11 F

kN 450kN/m 50m 92 F

m 3m 93

11 x

m 5.4m 92

12 x

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Chapter 4 - Force System Resultants 34

Solution 10

The two parallel forces F1 and F2 can be reduced to a

single resultant force FR.

The magnitude of FR is:

We can find the location of F with reference to point A:

Chapter 4 - Force System Resultant 67

kN 675450225 RF

FFR

AAR MM

4505.42253675 x

m 4x


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