1 2force systems force, moment, couple and resultants

2 Force Systems

Force, Moment, Couple and Resultants

Force Definition

• Force is an action that tends to cause acceleration of an object. [Dynamics]

• The SI unit of force magnitude is the newton (N). • One Newton is equivalent to one kilogram-meter per second

squared (kg·m/s2 or kg·m · s – 2)

Examples of mechanical force include the thrust of a rocket engine, the impetus that causes a car to speed up when you step on the accelerator, and the pull of gravity on your body.

• Force is a vector quantity (why?)

Force can result from the action of electric fields, magnetic fields, and various other phenomena.

• Force is the action of one body on another. [Statics]

FORCE SYSTEMSFORCE SYSTEMS

Force is a vector F

Line of action is a straight line colinear with the force

coplanar if the lines of action lie on the same plane

Force System:

concurrent if the lines of action intersect at a point

parallel if the lines of action are parallel

pararell coplaner ?

Hand Print

Scalar

Vector

Unit Vector

Magnitude of Vector

Writing Convention

ˆ i i

same symbol

Recommended Style

In this course, you have to write in this convention.

ˆ i i

2-D Force Systems2-D Force Systems 3-D Force Systems3-D Force Systems

Moment

Couple

Resultants

Moment

Couple

Resultants

Vector (2D&3D)

Basic Concept

Free Vectors: associated with “Magnitude” and “Direction”

: Direction

or V| |V

Magnitude:

Vector :

1 2 V V V

= ( 1)

| | | |W a V

parallelogram

( ) ( )

a bA ab A

a b A aA bA

a A B aA aB

Representation 1 2( )V V V

( ) ( )

A B B A

A B C A B C

triangle( 0)a M

| | | |M b V

+ˆ ˆA A e Ae

(unit vector of )

Operation Addition #5

R A B B ACommutative

Vector

Operation Addition #6Vector

( ) ( )R A B C A B C A B CAssociative

Operation Scalar Multiplication #2

( ) ( )

a bA ab A

a b A aA bA

a A B aA aB

ˆ ˆA A e Ae

ˆ unit vector of e A

associative

distributive wrt scalar addition

distributive wrt vector addition

wrt = with respect to

Component Resolution of a Vector

A vector may be resolved into two components.

Vector

| | ? | | ?A B

Basic relations of Triangle

(C/6, law of cosine, sine)

2 2 2 2 cosc a b ab

sin sin sin

Law of cosine

Law of sine

1 ___V

2 ___V

(Law of sine)

Given V, and , find 1 2,V V

2 2 2 2 cosc a b ab

sin sin sin

Law of cosine

Law of sine

special case: projection vectors are orthogonal to each other

Vector Component and Projection

: vector components of (along axis a and b)

1 2F F R

21, FF

: projections of (onto axis a and b)

(generally)a bF F R

, : orthogonal projections & vector components

Rectangular Components• Most commonly used

22yx FFF

1tan | / |y xF F

yx FFF

ˆ ˆ x yF F i F j

)directions and (in

of components vector ,

FFF yx

)directions and (in

of componentsscalar ,

FFF yx

ˆ ˆ i j vector component = vector projection

jcosF sinF

Fx=? Fy=?

cosxF F

sinyF F

cosxF F

sinyF F

sin( )xF F

cos( )yF F

sin( )xF F

cos( )yF F

= sinF

( cos )

(b>90)

ˆ ˆ7.7 4.6i j

EXAMPLE 2-1EXAMPLE 2-1

Given the magnitude of the tensionin the cable, T = 9 kN, express T interms of unit vector i and j

j)sin9(i)cos9(T

j662.11

662.11

kN ANS

2 210 6 11.662 mAB

Correct?3 S.F. ˆ ˆ7.72 4.63i j

We are using robot arm to put the cylindrical part into a hole.Determine the components of the force which the cylindrical part exerts on the robot along axes (a) parallel and perpendicular to arm AB

(b) parallel and perpendicular to arm BC

arm AB

sin 45 90sin 45 63.64 ( )parP P N

cos 45 90cos 45 63.64 ( )perP P N

arm BC

sin 30 9sin 30 4.5 ( )perP P N

cos30 9cos30 7.794 ( )parP P N

P = 90 N

parper

par per

Defining direction

2/2 Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R

P=800 N (8cm)

1 6sin 60tan 40.9

3 6cos60

T=600 N (6cm)

525 N (5.25cm) R

Graphics

Geometric

2 2 2 2 cosR P T PT 524 NR

sin sin

Vector Component (Algebraic)

ˆ800P i

ˆ ˆ600(cos sin )T i j

ˆ ˆ346 393R P T i j

2 2346 ( 393) 524 NR

1 393tan 48.6

Point of application is B

Correct?

Example Hibbeler Ex 2-1 #1

2 21 2 1 2 2 cos

213 NRF F F FF

Determine the magnitude and direction of the resultant force.

Geometric

2 150 212.55 sin sin sin sin115

39.101 39.1

Two forces is not acting at the same point.

2 21 2 1 22 cos

213 NRF F F FF

Geometric

2 150 212.55

sin sin sin sin11539.101 39.1

ˆ ˆ100cos15 100sin15 N

ˆ ˆ150sin10 150cos10 No

Vector Component (Algebraic)

ˆ ˆ122.64 173.60 N

173.60tan tan 54.761

122.64

F F F F i j

Good? (get full score?)

- more explanation

- mark answer

- 5S.F. Then 3S.F. ˆ ˆdirection of , ?i j

Good Answer Sheet

2 21 2 1 2

Using the law of cosine:

100 150 2(100)(150)cos115

212.55 213 N #

Applying the law of sine:

150 212.55

sin sin sin sin11539.101 39.1 #

F F F FF

Geometric

Point of Application

ˆ ˆ(600cos30 600sin30 ) N

ˆ ˆ( 400sin45 400cos 45 ) N

Vector

ˆ ˆ ˆ ˆ(600cos30 600sin30 ) ( 400sin45 400cos45 )

ˆ ˆ236.77 582.84 N

236.77 582.84

629.10 629 N #

tan ( ) tan (582.84 236.77)

67.891 67.9 #

i j i j

Vector

• Reference axis (very very important)– Many problems do not come with ref. axis.– Assignment based on convenience/experience

)ˆˆ()ˆˆ( 221121 jFiFjFiFFFR yxyx

jFFiFFjRiR yyxxyxˆ)(ˆ)(ˆˆ

xxxx FFFR 21

yyyy FFFR 21

The calculations do not reveal the point of application of the resultant force.

R1. Graphically

• Vector summation (addition)– Three ways to be mastered

2. Geometrically

3. Vector component (algebraically)

2 2 21 2 1 22 cosR F F F F

sin sin( )

Originally pass through O

In case where forces do not apply at the same point of application, you have to find it too!

Recommended Problems

Operation Products

1. Dot Products

2. Cross Products

3. Mixed Triple Products

( )A B C

Vector

scalar product cosP Q PQ

ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( )x y z x y zA B A i A j A k B i B j B k

x x y y z zA B A B A B A B

(unit vector) ( three orthogonal vector )

ˆ ˆ ˆ ˆ 0

ˆ ˆˆ ˆ 0

i j j i

i k k i

j k k j

ˆ ˆˆ ˆ ˆ ˆ 1i i j j k k

ˆ ˆ ˆ ˆ ˆ ˆ

x x x y x z

y x y y y z

z x z y z z

A i B i A i B j A i B k

A j B i A j B j A j B k

A k B i A k B j A k B k

ˆ ˆ ˆ P 2 3 4

ˆ ˆ ˆ 4 2 5

Example

Q i j k

Application of Dot Operation

• Angle between two vectors

| || |

P Q• Component’izing Vector

e/ /U U U

/ /ˆˆ ˆ ( ) U e U e

ˆˆ ˆ= ( ) U U e U e

ˆ ˆ( )T e e

ˆ ˆ( )T T e e

which direction?

/ /U U

cosP Q PQ

/ /, ?U U

ˆ ˆ ˆ U 2 3 4

1 ˆ ˆ ˆˆ 3

Example

e i j k

65o50o

cos50 3.21 kNoxyF F

cos65 1.36 kNox xyF F

sin50 3.83 kNozF F

sin 65 2.91 kNoy xyF F

ˆ ˆ ˆ

ˆ ˆ ˆ 1.36 2.91 3.83 kN

x y zF F F

F i j k

i j k Ans

ˆ ˆ ˆ

ˆ ˆ ˆ764 115 207 N

x y zT T T

T i j k

2 2 791.55 Nxz x zT T T Ans

1 10tan 81.5

ˆ ˆ( )T e e

ˆ ˆ( )T T e e

ˆˆ764 207xzT i k

which direction??

= C A B

right-hand rule(A then B)

“Cross Product” of Vectors

line which are perpendicular with both vectors

magnitude

21 ( )A B

= ˆ ( | | | |sin ) C

• Commutative Law is not valid

Operation Cross Product

A B B A

( ) ( ) ( ) ( )a A B aA B A aB A B a

( ) ( ) ( )A B C A B A C

Laws of Operations

• Associative wrt scalar multiplication

• Distributive wrt vector addition

? A B B A

? A B B A C A B

ˆ ˆˆ ˆ ˆ ˆ 0i i j j k k

ˆˆ ˆi j k i

x-y-z complies with right-hand rule

ˆˆ ˆj k i ˆ ˆ ˆk i j

ˆˆ ˆ U U i U j U kzx y

ˆ ˆˆ ˆ ˆ ˆ( ) ( )U V U i U j U k V i V j V kz zx y x y

ˆˆ ˆ ˆ ˆ ˆ( ) ( ) ( )

ˆ ˆ ˆ ˆˆ ˆ( ) ( ) ( )

U V i i U V i j U V i kx x x y x z

U V j i U V j j U V j ky x y y y z

U V k i U V k j U V k kz x z y z z

ˆ ˆ( ) ( )

U V U V U V i U V U V jz z z zy y x xU V U V kx y y x

This term can be written in a determinant form

How to calculate cross product

ˆˆ ˆ V V i V j V kzx y

ˆˆ ˆ ˆ ˆ

x y z x y

i j k i j

U U U U U

V V V V V

ˆˆ ˆ

U V U U U

Cross Product

ˆˆ ˆ( ) ( ) ( )U V U V U V i U V U V j U V U V kz z z zy y x x x y y x

Why cross product?

• Mathematical Representation of Moments, Torque

• Perpendicular Direction• Area Calculation

ˆOABCnB A

Area = ?A B

Mixed Triple Product

ˆ ˆ ˆ

ˆ ˆ ˆ( ) ( )

ˆ ˆ ˆ( )

ˆ ˆ ˆ( ) ( ) ( )

x y z x y z

y z z y x z z x x y y z

U V W U i U j U k V V V

U i U j U k

V W V W i V W V W j V W V W k

( ) ( ) ( ) ( )x y z z y y x z z x z x y y zU V W U V W V W U V W V W U V W V W

( )x y z

U V W V V V

( )V W U

x y z x y z

U U U W W W

W W W U U U

V V V V V V

( ) ( )U V W W U V

Why mixed triple product?

• Mathematical Representation of Moments along the axis.

• Volume Calculation

, ,o FM

ˆOABCn

: ( )Volume C B A

( ) C B A

Volume must always +

Mixed Triple Product Scalar

Operation Product Summary

Cross Product Vector

Dot Product Scalar

Vector

Homepage URLs

Statics official HP http://www.lecturer.eng.chula.ac.th/fmekmn/ (User: Prince Password: Caspian)

Session 1 HPhttp://pioneer.netserv.chula.ac.th/~lsawat/course/statics/

http://blackboard.it.chula.ac.th/ (after the end of registration period)

2-D Force Systems2-D Force Systems 3-D Force Systems3-D Force Systems

Moment

Couple

Resultants

Moment

Couple

Resultants

VectorBasic Concept

Force Definition

• Force is an action that tends to cause acceleration of an object. [Dynamics]

• The SI unit of force magnitude is the newton (N). One newton is equivalent to one kilogram-meter per second squared

(kg·m/s2 or kg·m · s – 2)

Examples of mechanical force include the thrust of a rocket engine, the impetus that causes a car to speed up when you step on the accelerator, and the pull of gravity on your body.

• Force is a vector quantity (why?)

Force can result from the action of electric fields, magnetic fields, and various other phenomena.

• Force is the action of one body on another. [Statics]

Force Representation

• Vector quantity– Magnitude– Direction– Point of application

Use different colours in diagrams• Body outline blue• Load red• Miscellaneous black

(dimension, angle, etc.)

Type of ForcesType of Forces

External force

Internal force

Reactive force

Applied force

Strain

Stress

Contact force

Body force

Concentrated

Distributed

Cables & Springs

Cable in tension

spring constantk

2/2 Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R

P=800 N (8cm)

1 6sin 60tan 40.9

3 6cos60

T=600 N (6cm)

525 N (5.25cm) R

Graphical

Geometric

2 2 2 2 cosR P T PT 524 NR

sin sin

Algebraic

ˆ800P i

ˆ ˆ600(cos sin )T i j

ˆ ˆ346 393R P T i j 2 2346 ( 393) 524 NR

1 393tan 48.6

Point of application is BCorrect?

Not OK. !

How to add sliding vectors (forces)?

21 FFR

is applied at point AR

Point of Application is wrong

Point of application

Principle ofTransmissibility

Still OK.

This graphical method can be used to find Line of action

Special case: Addition of Parallel Sliding Force

line of action

The better and efficient way will be discussed later, when we learn the concept of “moment”, “couple”, and “resultant force”

1 1 cos60 200 NoxT T

1 1 sin 60 346 NoyT T

2 2 693 Nx yR R R

1ˆ ˆ200 346 NT i j

2ˆ400 NT i

1 2ˆ ˆ600 346 NR T T i j

Point of application,But no physical meaning

Move all forces to that concurrent point

1 o346tan =29.97

Application Point Ans

How to add sliding vectors (forces)?

21 FFR

is applied at point AR

There is better way to find the point of application(or line of action), but you have to learn the concept ofmoment and couples first.

moment axis

Moment is a vector

MomentIn addition to the tendency to move a body, force may also tend to rotate a body about an axis

From experience (experiment)magnitude depends only on “F” and “d”

(magnitude)summation

Direction

M Fd i i

Moment Definition

( )O zM• Moment is a vector quantity.

– Magnitude– Direction– Axis of Rotation

• The unit of moment is N·m• The moment-arm d (perpendicular distance)• The right-hand rule• determined by vector cross product• Sign convention: 2D +k or CCW is positive.• Moment of a force or torque

Moment about point A :

Mathematical Definition (3D)

,A FM r F

-Magnitude:

-Point of application: point A

-Direction: right-hand rule

(Unit: newton-meters, N-m)

- 2D, need sign convention and be consistent; e.g. + for counter- clockwise and – for clockwise

from A to point of application of the force

| || | sinr F

The moment of a force about any point is equal to the sum of the moments of the components of the force about that point

can be used with

more than

2 components

Mo = -Fxd2+Fyd1

** OM r F

*1 2OM r F r F

1 2( )F F

Varignon’s Theorem (Principle of Moment)

1 2 1 2!

Yes r F r F r F F

sum of moment (of each force) = moment of sum (of all force)

Useful with rectangular components

Principle of Transmissibility & Moment

M = Fr sin a = FdSliding force has the same moment

convenient

AX AYr F r F

- direction: same

- magnitude:

Principle of Transmissibility is based on the fact that“moving force along the line of action causes no effect in changing moment”

position vector: from A to any point on line of action of the force.

Sample 2/5 Calculate the magnitude of the moment about the base point O of 600N force in five different ways.

Solution I: 2D Scalar Approach

ood 40sin240cos4

35.4600 dFM O

2610 N-m Ans

Solution II: 3D Vector Approach

ˆ 2610 N-m k

OM r F

ˆ ˆ ˆ ˆ2 4 600 cos40 sin 40o o i j i j

CW or CCW? CWCorrect?

600N400

Solution III: Varignon’s theorem

NF o 46040cos6001

NF o 38640sin6002

460(4) 386(2)OM

2610 N-m (CW)

Solution IV: Transmissibility

od 40tan241 m68.5

1 1 2610 N-m (CW)OM F d

Solution V: Transmissibility

od 40cot422 m77.6

2 2 2610 N-m (CW)OM F d

EXAMPLE 2.8EXAMPLE 2.8

In raising the flagpole, the tension T in the cable must supply a Moment about O of 72 kN-m. Determine T.

30sin 60

25.981m

72 (12)sin 43.898Td T

8.65T kN

1 25.981tan 43.898

60o12 m

Determine the moment of the force about point O.

ˆ ˆ(400sin30 400cos30 ) N

ˆ ˆ(200 346.41 ) N

Moment

Correct? ˆ ˆ , ?i j

ˆ ˆ ˆ

0.4 0.2 0 N-m

200 346.41 0

0.4 0.2 ˆ N-m200 346.41

ˆ ˆ98.564 98.6 N m

k k Ans

Moment

ˆ ˆ(400sin30 400cos30 ) N

ˆ ˆ(200 346.41 ) N

400sin 30(0.2) 400cos 30(0.4)

40 138.56 98.564 N m

ˆ98.6 N m #

ˆ ˆ(0.4 0.2 ) mr i j

3D Vector Approach

Scalar Approach (Varignon’s theorem)

Couple- Couple is a summed moment produced by two force of equal magnitude but opposite in direction.

M = F(a+d) – Fa = Fd

magnitude does not depend on distance a (point O),i.e. any point on the body has the same magnitude.

Couple Fd

- tendency to rotate the “whole” object.

- no effect on moving object as translation.

2D representations: (Couples)

couple is a free vector

Effect of Pure Rotation

Force-couple systems- Line of action of a force on a body may be changed if a couple is added to compensated for the change in the tendency to rotate of that body.

Procedure may be reversed to combine a force with a couple

Force-couple system

Principle of transmissibility

No changes in net external effect

The direction and magnitude of Force can not be changed, only line of action (i.e. only change to other pararell line)

BAC r F

Principle of Transmissibility is based on the fact thatmoving force along the line of action causes no effect in changing moment

B FC r F

from new location (B)to old location (A)

No Moment:Principle of Transmissibility

In the viewpoint of Mechanics,Result of force to these systemsare equal

Why using equivalent system?

Force-couple system

Principle of transmissibility

real (physical) system

equivalent system equivalent system

All force systems are equal.

Understanding Force-Couple system

Moment about point B of force F = tendency of force F to rotate the object at point B

couple occurs when moving Force F from A to B ( couple occurs when moving Force F parallel to its line of action to the point B)

Equivalent System

Vector Diagram

7012 PF

P703600

51.42 kNP

Be careful of the direction of moment

2/11 Replace the force F by an equivalent force-couple system at point O.

250 .mm

++ cos 20 (0.1cos 25 0.25cos10 )F

sin 20 (0.1sin 25 0.25sin10 )o o oF

17.3 N-m

M50 kN

0.25 m

Couple occurred when moving F to O= Moment of F about O

CCWˆ ˆ17.1 46.9 kNF i j

Correct?

o 90 90 90 270R F kN

)21(90)12(90)21(90Mo

mkN1080 ANS

Got the meaning?

Moving all 3 forces to point O

Sum of couples

Engine number 3 fails. Determine the force-couple system on the body about point o.

couples occuring when moving forces.

(direction: left)

sum of moments?

Replace the current system by an equivalent resultant force and couple moment acting A.

100 400cos 45

382.84 N

600 400sin45

882.84 N

Resultant

( 382.84) ( 882.84)

962.27 962 N #

882.84tan tan

382.84

66.556 66.6 #

R Rx Ry

100(0) 600(0.4) 400sin45 (0.8) 400cos45 (0.3)

551.13 551 N m #

Resultant

6020cos300 ob

0.213 mb

60 N-m

exactly cancelled

b cos20

• Resultant of many forces-couple is the simplest force-couple combination which can replace the original forces/couples without changing the external effects on the body they act on

2/6 Simplest Resultant

-Add two at a time get line of action of R

-Add many do not get line of action of R

...321 , xx FR yy FR

22 )()( yx FFR )/(tan 1xy RR

xF2 xF3

Easier way to get a resultant + its location

any forces + couples system

1) Pick a point (easy to find moment arms)

2) Replace each force with a force at point O + a couple

4) Replace force-couple system with a single force

d=Mo/R

3) Add forces and moments

Mo=(Fidi)

arbitrary

(forces + couples : same procedures)

any forces + couples system single-force + special single-couple (wrench)

single-force system (no-couple)

or single-couple system

resultant

2/87 Determine the resultant and its line of action of the following three loads.

R = ( 2.4cos20 -1.5sin20 -3.6cos20 ) i +( -2.4sin20 -1.5cos20 +3.6cos20 ) j kN

M = -2.4*0.2cos20 -1.5*0.12cos20 -3.6*0.3cos20 kN-m

(force-couple system)

Move 3 forces to point O,Sums their force and couples

Note: R is the same regardless with the location point we move the force to

Note: M depends on the location where we move the force to

M = -1.635 kN-m

At point O (0,0)

ˆ ˆ1.64 0.99 kNR i j

At point X (x,y)

ˆ 1.635 k kNM

ˆ ˆ0 ( )M xi yj R

ˆˆ ˆ ˆ ˆ( ) ( 1.64 0.99 ) 1.635xi yj i j k

0.99 1.64 1.635x y ( line of action )

couples cancelled

O (0,0)

P (x,y)

ˆ ˆ(0 ) (0 )r x i y j

Two equivalent systems Moment at any pointmust be the same on both system

Sys 2 ˆ ˆ( )xi yj R M

Pick Point O

Correct?

M = -1.635 kN-m

At point O (0,0)

ˆ ˆ1.64 0.99 kNR i j

How to locate Point P

O (0,0)

ˆ 1.635 k kNM

| | | 1.635 |0.853 m

| | ( 1.64) ( 0.99)

1 1 0.99tan tan 31.118

1.64y o

( sin , cos )d d

(0.441, 0.73)

( 0.73) 0.99

0.441 1.64

d2 2 2x y d

1( )tan

Manually Canceling Couples

How to find line of action ?

Equivalent System Definition

Two force-couple systems are equivalent

Equivalent System

system I system II

( ) ( )

( ) ( )O O

Sys 1 Sys 2

A car stuck in the snow. Three students attempt to free the car by exert forces on the car at point A, B and C while the driver’s actions result in a forward thrust of 200 N as shown in picture.

Determine

1) the equivalent force-couple system at the car center of mass G

2) locate the point on x-axis where the resultant passes.

jFiFR yx

i30sin250200200400R

j30cos250350

j567i925R

350(1.65) 250sin 30 (0.825)GM

For line of action of resultant

ˆˆ ˆ ˆ ˆ( ) (925 567 ) 690x i y j i j k

ˆ : 567 925 690k x y

At y = 0; x = +1.218 m. ANS

j567i925R

( ) | |y GR b M

567(b)=690

b=1.218 m

ˆ ˆ0 ( )M xi yj R

ˆ ˆ( )xi yj R M

Couple Cancellation

If you want to find only b (not line of action itself)

+ or - , you have to find out manually

Two equivalent systems Moment at point Gmust be the same on both system

, ,G SysII G SysIM M

Sys II

Two equivalent systems(2D)

Determine the resultant (vector) and the point on x and y axes which must pass.

jFiFR yx

ˆ25 20sin 30

ˆ ( 20cos30 30)

j3.47i15

25(5) 30(9)

(20cos30 )(9) (20sin 30 )(5)

mkN351

j3.47i15R

mkN351

For line of action of resultant

k351)j3.47i15()jyix(

351y15x3.47

If y = 0; x = 7.42 m.

x = 0; y = -23.4 m.

1 2force systems force, moment, couple and resultants

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