chapter 2: concurrent force systemsqiw4/academic/engr0135/chapter2.pdf · 2010. 9. 14. · a force...

Department of Mechanical Engineering

Chapter 2: Concurrent force systems


Objectives To understand the basic characteristics of forces To understand the classification of force systems To understand some force principles

To know how to obtain the resultant of forces in 2D and 3D systems

To know how to obtain the components of forces in 2D and 3D systems


Characteristics of forces Force: Vector with magnitude and direction Magnitude – a positive numerical value representing

the size or amount of the force

Directions – the slope and the sense of a line segment used to represent the force– Described by angles or dimensions– A negative sign usually represents opposite

direction Point of application

– A point where the force is applied– A line of action = a straight line extending through

the point of application in the direction of the force

The force is a physical quantity that needs to be represented using a mathematical quantity


Example

α

1000 N

i

j

Line of action

Point of application

magnitude

direction


Vector to represent Force

A vector is the mathematical representation that best describes a force

A vector is characterized by its magnitude and direction/sense

Math operations and manipulations of vectors can be used in the force analysis


Free, sliding, and fixed vectors Vectors have magnitudes, slopes, and senses, and lines of

applications

A free vector– The application line does not pass a certain point in space

A sliding vector– The application line passes a certain point in space

A fixed vector– The application line passes a certain point in space– The application point of the vector is fixed


Vector/force notationThe symbol representing the force bold face

or underlined letters

The magnitude of the force lightface (in the text book, + italic)

AAorA == A


Classification of forces Based on the characteristic of the interacting bodies:

– Contacting vs. Non-contacting forces Surface force (contacting force)

– Examples: » Pushing/pulling force» Frictions

Body force (non-contacting force)– Examples:

» Gravitational force» Electromagnetic force


Classification of forces Based on the area (or volume) over

which the force is acting– Distributed vs. Concentrated forces

Distributed force– The application area is relatively large

compare to the whole loaded body– Uniform vs. Non-uniform

Concentrated force– The application area is relatively small

compare to the whole loaded body


What is a force system? A number of forces (in 2D or 3D system)

that is treated as a group: A concurrent force system

– All of the action lines intersect at a common point

A coplanar force system– All of the forces lie in the same plane

A parallel force system– All of the action lines are parallel

A collinear force system– All of the forces share a common line of

action


The external and internal effects A force exerted on the body has two effects:

– External effects» Change of motion» Resisting forces (reactions)

– Internal effects» The tendency of the body to deform develop

strain, stresses– If the force system does not produce change of

motion » The forces are said to be in balance» The body is said to be in (mechanical)

equilibrium


External and internal effectsExample 1: The body changes in motion

Example 2: The body deforms and produces (support) reactions The forces must be in balance

Not fixed, no (horizontal) support

Fa

F

Support Reactions

Fixed support


Principle for force systems Two or more force systems are equivalent when their

applications to a body produce the same external effect Transmissibility Reduction =

– A process to create a simpler equivalent system– to reduce the number of forces by obtaining the

“resultant” of the forces Resolution =

– The opposite of reduction– to find “the components” of a force vector

“breaking up” the resultant forces


Principle of Transmissibility Many times, the rigid body assumption is taken only the

external effects are the interest The external effect of a force on a rigid body is the same for

all points of application of the force along its line of action


Resultant of Forces –Review on vector addition

Vector addition

Triangle method (head-to-tail method)– Note: the tail of the first vector

and the head of the last vector become the tail and head of the resultant principle of the force polygon/triangle

Parallelogram method– Note: the resultant is the diagonal

of the parallelogram formed by the vectors being summed

ABBAR +=+=

RA

B

B

AR


Resultant of Forces – Review on geometric laws

Law of Sines

Laws of Cosines

α

β

γ

cos2cos2cos2

222

222

222

accbaaccababbac

−+=

−+=

−+=

A

B

C

c

a

b

βγ

α


Resultant of two concurrent forces

The magnitude of the resultant (R) is given by

The direction (relative to the direction of F1) can be given by the law of sines

φ

γ

cos2

cos2

212

22

12

212

22

12

FFFFRFFFFR

++=

−+=

RF φβ sinsin 2=

Pay attention to the angle and the sign of the last term !!!


Resultant of three concurrent forces and more

Basically it is a repetition of finding resultant of two forces

The sequence of the addition process is arbitrary The “force polygons” may be different The final resultant has to be the same


Resultant of more than two forces The polygon method becomes tedious when dealing

with three and more forces It’s getting worse when we deal with 3D cases It is preferable to use “rectangular-component”

method


Example Problem 2-1 Determine:

– The resultant force (R)– The angle θ between the R and the x-axis

Answer:– The magnitude of R is given by

– The angle α between the R and the 900-lb force is given by

– The angle θ therefore is

lbRR

14133.141340cos)600)(900(2600900 0222

≈=++=

o836.153.1413

)40180sin(600

sin 00

=

−=

α

α

000 8.5035836.15 =+=θ


Example Problem 2-2

Determine– The resultant R– The angle between the R

and the x-axis


Another example

If the resultant of the force system is zero, determine– The force FB

– The angle between the FBand the x-axis


Force components


Resolution of a force into components

The components of a resultant force are not unique !!

The direction of the components must be fixed (given)

FEDCHIGBAR

+=+=++=+= )(


How to obtain the components of a force (arbitrary component directions)?

Steps:– Draw lines parallel to u and v crossing

the tip of the R– Together with the original u and v

lines, these two lines produce the parallelogram

– The sides of the parallelogram represent the components of R

– Use law of sines to determine the magnitudes of the components

Parallel to v

Parallel to u

oov

ou FF

110sin900

25sin45sin==

NF

NF

ov

o

u

405110sin

25sin900

677110sin

45sin900

0

0

==

==


Example Problem 2-5 Determine the components of F = 100 kN along

the bars AB and AC

Hints:– Construct the force triangle/parallelogram – Determine the angles α, β, γ– Utilize the law of sines


Another example

Determine the magnitude of the components of R in the directions along u and v, when R = 1500 N


Rectangular components of a force What and Why rectangular components?

– Rectangular components all of the components are perpendicular to each other (mutually perpendicular)

– Why? One of the angle is 90o ==> simple Utilization of unit vectors Rectangular components in 2D and 3D Utilization of the Cartesian c.s.

Arbitrary rectangular


The Cartesian coordinate system

The Cartesian coordinate axes are arranged following the right-hand system (shown on the right)

The setting of the system is arbitrary, but the results of the analysis must be independent of the chosen system

x y

z


Unit vectors A dimensionless vector of unit magnitude The very basic coordinate system used to specify coordinates in

the space is the Cartesian c.s. The unit vectors along the Cartesian coordinate axis x, y and z

are i, j, k, respectively The symbol en will be used to indicate a unit vector in some n-

direction (not x, y, nor z) Any vector can be represented as a multiplication of a

magnitude and a unit vector

nn AeeAA ==

nn BeeBB −=−=

A is in the positivedirection along n

B is in the negativedirection along n

AA

AAen ==


The rectangular components of a force in 2D system

While the components must be perpendicular to each other, the directions do not have to be parallel or perpendicular to the horizontal or vertical directions

x

y

Fy = Fy j

Fx = Fx i

i

j

F

θ

jiFFF yxyx FF +=+=

x

y

yx

y

x

FF

FFF

FFFF

1

22

tan

sincos

−=

+=

==

θ

θθ

Department of Mechanical EngineeringFF

FF

FF

FFFF

FFFFFF

zz

yy

xx

zyx

zz

yy

xx

111

222

coscoscos

cos

coscos

−−− ===

++=

=

==

θθθ

θ

θθ

The rectangular components in 3D systems

FFFF

F

FFFF

zyxn

n

zyx

zyx

kjiFe

eFkji

FFFF

++==

=

++=

++=

x

y

z

Fy = Fy jFx = Fx i

Fz = Fz k

F

i

k

j

en

θz

θx θy

kjie zyxn θθθ coscoscos ++=


Dot Products of two vectors

θθ coscos AB==•=• BAABBA

θ

A

B

It’s a scalar !!!Special cosines:

Cos 0o = 1Cos 30o = ½ √3Cos 45o = ½ √2Cos 60o = 0.5Cos 90o = 0


Dot products and rectangular components The dot product can be used to obtain the rectangular

components of a force (a vector in general)

nt

nnn

nnn

nnn

AAA

AAA

eeAA

eAeA

−=

•=

==•=

)(

cosθ (magnitude)

(the vectorial componentin the n direction)

The component along en

The component along et

Remember, en and et are perpendicular


Cartesian rectangular components

The dot product is particularly useful when the unit vectors are of the Cartesian system (the i, j, k)

x

y

Fy = Fy j

Fx = Fx i

i

j

F

θ

kF

jFiF

•=

=

−=•==•=

z

y

x

F

FFFFF

θ

θθ

sin

)90cos(cos

90-θAlso, in 3D,

jjFiiFjiFFF )()( •+•=+=+= yxyx FF


More usage of dot products …

Dot products of two vectors written in Cartesian system

The magnitude of a vector (could be a force vector), here A is the vector magnitude

The angle between two vectors (say between vectors Aand B)

zzyyxx BABABA ++=• BA

zzyyxx AAAAAAAA ++===• 22 0cosAA

++= −

ABBABABA zzyyxx1cosθ


The rectangular components of arbitrary direction

znzynyxnx

nznynx

nzyx

nn

ttnn

zyx

zyx

FFFFFF

FFFF

FFFFF

θθθ coscoscos

)(

++=

•+•+•=

•++=•=

+=

++=

++=

ekejeiekji

eF

eeFkji

FFFF

kjie znynxnn θθθ coscoscos ++=

z

x

y

Fy = Fy jFx = Fx i

Fz = Fz k

F

i

k

j

en

θzn

θxn θyn

Fn

Ft

Can you show the following?


Summarizing …. The components of a force resultant are not unique Graphical methods (triangular or parallelogram methods)

combined with law of sinus and law of cosines can be used to obtain components in arbitrary direction

Rectangular components are components of a force (vector) that perpendicular to each other

The dot product can be used to – obtain rectangular components of a force vector– obtain the magnitude of a force vector (by performing self-

dot-product)– Obtain the angle between two (force) vectors


Example Problem 2-6 Find the x and y scalar components of the

force Find the x’ and y’ scalar components of

the force Express the force F in Cartesian vector

form for the xy- and x’y’- axes


Example Problem 2-6

θ

β

NFNFNFNF

FFFFFFFF

y

x

y

x

o

o

yx

yx

23832sin45038232cos450

39762sin45021162cos450

323062622890

)90cos(cos

)90cos(cos

'

''

====

====

=−=

=−=

−==

−==

β

θ

ββ

θθ

NN yx )238382()397211('' eejiF +=+=

Writing the F in Cartesian vector form:


Example Problem 2-8 Find the angles θx, θy, and θz

(θx is the angle between OB and x axis and so on ..)

The x, y, and x scalar components of the force.

The rectangular component Fn of the force along line OA

The rectangular component of the force perpendicular to line OA (say Ft)

B


Example Problem 2-8

To find the angles:– Find the length of the

diagonal OB, say d– d = 5.831 m– Use cosines to get the

angles

The scalar components in the x, y, and z directions:

B

oz

oy

ox

0.59831.53cos

7.46831.54cos

0.59831.53cos

1

1

1

==

==

==

−

−

−

θ

θ

θ

kNFFkNFFkNFF

zz

yy

xx

862.12cos

150.17cos862.12cos

==

====

θ

θθ

kN)862.12150.17862.12( kjiF ++=


Example Problem 2-8 To find the rectangular component Fn

of the force along line OA:– Needs the unit vector along OA– Method 1 : Follow the method

described in the book– Method 2: utilize the vector

position of A (basically vector OA)

– Remember, that any vector can be represented as a multiplication of its magnitude and a unit vector along its line of application

kjir 313 ++== AOAkjikji

kjirre A

688.0230.0688.036.4

313313

313222

++=++

=

++

++==

AOA


Example Problem 8-2

The scalar component of F along OA

The vector component of F along OA

The vector component of F perpendicular to OA

The scalar component of F perpendicular to OA

OAeF •=OAF

kNFF

OA

OA

643.21688.0862.12230.0150.17688.0862.12)688.0230.0688.0()862.12150.17862.12(

=×+×+×=++•++= kjikji

kjieeFF OAOAOA

86.1497.486.14)688.0230.0688.0(6.21)(

++=++=•= kji

)218.122()86.1497.486.14()862.12150.17862.12(

kjikjikjiFFF OA

++−=++−++=−=t

kNF tt 50.12)2(18.12)2(|)218.122(||| 222 =−++−=−+−== kjiF

Check: kNFFF tOA 2550.12643.21 2222 ≈+=+=


Resultants by rectangular components

The Cartesian rectangular components of forces can be utilized to obtain the resultant of the forces

x

y

F1

F2

F1x

F2x

F2y

F1y

•Adding the x vector components, we obtain the x vector component of the resultant

•Adding the y vector components, we obtain the y vector component of the resultant

•The resultant can be obtained by performing the vector addition of these two vector components

∑ +== xxxx 21 FFFR

∑ +== yyyy 21 FFFR

jiRRR yxyx RR +=+=


Resultants by rectangular components

The scalar components of the resultant

The magnitude of the resultant

The angles formed by the resultant and the Cartesian axes

All of the above results can be easily extended for 3D system

iiFFR 21 xxxxxx RFF =+=+= )( 21

jjFFR 21 yyyyyy RFF =+=+= )( 21

22yx RRR +=

RR

RR y

yx

x11 coscos −− == θθ


Please do example problems 2-9, 2-10, and 2-11


HW Problem 2-20

Determine the non-rectangular components of R


HW Problem 2-37Determine the

components of F1 and F2in x-y and x’-y’ systems


HW Problem 2-44

Express the cable tension in Cartesian form

Determine the magnitude of the rectangular component of the cable force

Determine the angle α between cables AD and BD

Typo in the problem!!!

B(4.9,-7.6,0)C(-7.6,-4.6,0)

Don’t worry if you don’t get the solution in the back of the book


HW Problem 2-46

Determine the scalar components Express the force in Cartesian

vector form Determine the angle α between the

force and line AB


HW problems 2-55

Given: F1 = 500 lb, F2 = 300 lb, F3 = 200 lb

Determine the resultant Express the resultant in the

Cartesian format Find the angles formed by the

resultant and the coordinate axes


HW Problem 2-49

Given T1 and T2 are 650 lb, Determine P so that the resultant of T1, T2 and P is

zero

chapter 2: concurrent force systemsqiw4/academic/engr0135/chapter2.pdf · 2010. 9. 14. · a force...

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