slide #: 1 chapter 3 rigid bodies : equivalent systems of forces

Slide #: 1

Chapter 3Chapter 3

Rigid Bodies : Rigid Bodies : Equivalent Systems Equivalent Systems of Forcesof Forces

Slide #: 2

IntroductionIntroduction

In previous lessons, the bodies were assumed to be particles.

Actually the bodies are a combination of a large number of particles.

particle

body

Slide #: 3

IntroductionIntroduction

translation

rotation

A force may cause a body to move in a straight path that is called

Another force may cause the body to move differently, for example,

The external forces may cause a motion of translation or rotation or both. The point of application of force determines the effect.

Slide #: 4

Principle of Transmissibility : Principle of Transmissibility : Equivalent ForcesEquivalent Forces

F

F’

According to the principal of transmissibility, the effect of an external force on a rigid body remains unchanged if that force is moved along its line of action. Two forces acting on the rigidbody at two different points have the same effect on that body if they have the same magnitude, same direction, and same line of action. Two such forces are said to be equivalent.

Slide #: 5

• 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 : Equivalent Principle of Transmissibility : Equivalent ForcesForces

Slide #: 6

Vector Product of Two VectorsVector Product of Two Vectors

• Concept of the moment of a force about a point is more easily understood through applications of the 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

Slide #: 7

Vector Product of Two VectorsVector Product of Two Vectors

• Vector products:

- are not commutative,

- are distributive,

- are not associative,

QPPQ 2121 QPQPQQP

SQPSQP

Slide #: 8

Vector Product of Unit VectorsVector Product of Unit Vectors

It follows from the definition of the vector product of two vectors that the vector products of unit vectors i, j, and k are:

i x i = j x j = k x k = 0

0

0

0

kkikjjki

ijkjjkji

jikkijii

Slide #: 9

Vector Products in Terms of Rectangular Vector Products in Terms of Rectangular ComponentsComponents

The rectangular components of the vector product V of twovectors P and Q are determined as follows: Given

P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k

V = (Py Qz - Pz Qy ) i + (Pz Qx - Px Qz ) j + (Px Qy - Py Qx ) k

V = P x Q

V = Vx i + Vy j + Vz k

Slide #: 10

where

Vx = Py Qz - Pz Qy

Vy = Pz Qx - Px Qz

Vz = Px Qy - Py Qx

V = P x Q = Vx i + Vy j + Vz k

V = P x Q =i

Px

Qx

jPy

Qy

kPz

Qz

Expansion ofDeterminant

Vector Products in Terms of Rectangular Vector Products in Terms of Rectangular ComponentsComponents

Slide #: 11

V = P x Q =i j k

P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz kPx Py Pz Qx Qy Qz

Expansion of DeterminantExpansion of Determinant

Slide #: 12

V = P x Q =i j k

P = 4 i + 3 j + 2 k Q = 6 i + 1 j + 5 k4 3 2 6 1 5

Expansion of Determinant - ExampleExpansion of Determinant - Example

Slide #: 13

V = P x Q =i46

j31

k25

Calculation of Determinant : Method -1Calculation of Determinant : Method -1

i46

j31

Slide #: 14

V = P x Q =i46

j31

k25


i46

j31

1 2 3

4 5 6

V = P x Q = + + - - -1 2 3 4 5 6

Slide #: 15

V = P x Q =i46

j31

k25


i46

j31

15 i 12 j 4 k

18 k 2 i 20 j

V = P x Q = 13 i - 8 j -14 k

Slide #: 16


D =a11

a21

a31

a12

a22

a32

a13

a23

a33

a11 a12 a13

a21

a31

a22

a32

a23

a33

= D11

D11 : Minor of a11 in D

a22

a32

a23

a33

D11 = = a22* a33 – a32* a23

Slide #: 17


D =a11

a21

a31

a12

a22

a32

a13

a23

a33

a11 a12 a13

a21

a31

a23

a33

= D12


a22

a32

a21

a31

a23

a33

D12 = = a21* a33 – a31* a23

Slide #: 18


D =a11

a21

a31

a12

a22

a32

a13

a23

a33

a11 a12 a13

= D13


a22

a32

a21

a31

a23

a33

D13 = = a21* a32 – a31* a22

a22

a32

a21

a31

Slide #: 19


D =a11

a21

a31

a12

a22

a32

a13

a23

a33

a11 a12 a13

D = a11 D11 - a12 D12 + a13 D13

a22

a32

a21

a31

a23

a33

D= a11 (a22* a33 – a32* a23) - a12 (a21* a33 – a31* a23)

+ a13 (a21* a32 – a31* a22)

Slide #: 20

V = P x Q =i46

j31

k25


V = P x Q = 31

25

i 46

31

k- 46

25

j +

V = P x Q = i (15-2) – j (20-12) + k (4-18)

V = P x Q = 13 i – 8 j – 14 k

Slide #: 21

Moment of a Force About a Point Moment of a Force About a Point

The moment of force F about point O is defined as the vector product

MO = r x F

where r is the position vector drawn from point O to the point of application of the force F. The angle between the lines of actionof r and F is .

• A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application.

Slide #: 22


The magnitude of the moment of F about O can be expressed as

MO = rF sin = Fd

where d is the perpendicular distance from O to the line of action of F.

• 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.

Slide #: 23


• 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 counter clockwise, 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 counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.

Slide #: 24

Varignon’s Theorem Varignon’s Theorem

• The moment about a given 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.

2121 FrFrFFr

Slide #: 25

Rectangular Components of the Moment of a Rectangular Components of the Moment of a Force Force

x

y

z

Fx i

Fz k

Fy j

x i

y j

z k

O

A (x , y, z )

r

The moment of F about O,

kFjFiFF

kzjyixrFrM

zyx

O

,

Slide #: 26

The rectangular components of the moment Mo of a force F are determined by expanding thedeterminant of r x F.

Mo = r x F =ixFx

jyFy

kzFz

= Mx i + My j + Mzk

whereMx = y Fz - z Fy My = zFx - x Fz

Mz = x Fy - y Fx


Slide #: 27


The moment of F about B,

FrM BAB

/

kFjFiFF

kzzjyyixx

rrr

zyx

BABABA

BABA

/

zyx

BABABAB

FFF

zzyyxx

kji

M

Slide #: 28

Moment of a Force About a Point Moment of a Force About a Point Sample Problem 3.4Sample Problem 3.4

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

Slide #: 29


SOLUTION:

12896120

08.003.0

kji

M A

kirrCAr ACAC

m 08.0m 3.0

FrM ACA

kji

kji

CD

DCFF

N 128N 69N 120

m 5.0

m 32.0m 0.24m 3.0N 200

N 200

Slide #: 30


SOLUTION:

12896120

08.003.0

kji

M A

096

0.08-128i -MA =

0.3-120

j0.3

-1200

96k+0.08-128

MA = (0- (96)(0.08)) i – ((0.3)(-128) – (-120)(0.08)) j + (96)(0.3) k

MA = - 7.68 i + 28.8 j + 28.8 k

Slide #: 31


SOLUTION:

kjiM A

mN 8.82mN 8.82mN 68.7

Slide #: 32

Rectangular Components of the Moment of a Rectangular Components of the Moment of a ForceForce

For two-dimensional structures,

xy

ZO

xyO

yFxF

MM

kyFxFM

0

0

yx

O

FF

yx

kji

M

FrM AOO

dFM O

Slide #: 33

Rectangular Components of the Moment of a Rectangular Components of the Moment of a ForceForce

For two-dimensional structures,

xBAyBA

ZB

xBAyBAB

FyyFxx

MM

kFyyFxxM

0

0

yx

BABAB

FF

yyxx

kji

M

FrM ABB

dFM B

Slide #: 34


A 100-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 240-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.

Slide #: 35


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.

in. 12lb 100

in. 1260cosin.24

O

O

M

d

FdM

in lb 1200 OM

Slide #: 36


b) Horizontal force at A that produces the same moment,

m 8.20

m N 1200

m. 8.20m 1200

m 8.2060sinm 24

F

FN

FdM

d

O

N 7.57F

Slide #: 37


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 42

. 1200

m 42m 1200

NmF

FN

FdM O

N 50F

Slide #: 38


d) To determine the point of application of vertical 240 N force to produce the same moment,

m 5cos60

m 5N 402

m. 1200

N 240m 1200

OB

Nd

dN

FdM O

m 10OB

Slide #: 39


e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force.

Slide #: 40

Scalar Product of Two VectorsScalar Product of Two Vectors

• The scalar product or dot product between two vectors P and Q is defined as

resultscalarcosPQQP

• Scalar products:

- are commutative,

- are distributive,

- are not associative,

PQQP

2121 QPQPQQP

undefined SQP

• Scalar products with Cartesian unit components,

000111 ikkjjikkjjii

kQjQiQkPjPiPQP zyxzyx

2222 PPPPPP

QPQPQPQP

zyx

zzyyxx

Slide #: 41

Scalar Product of Two Vectors : Scalar Product of Two Vectors : ApplicationsApplications

• 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:

Slide #: 42

Mixed Triple Product of Three Mixed Triple Product of Three VectorsVectors

• 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

Slide #: 43

Mixed Triple Product of Three Mixed Triple Product of Three VectorsVectors

zyx

zyx

zyx

xyyxz

zxxzyyzzyx

QQQ

PPP

SSS

QPQPS

QPQPSQPQPSQPS

• Evaluating the mixed triple product,

Slide #: 44

Moment of a Force About a Given Moment of a Force About a Given AxisAxis

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

FrM O

• Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis,

FrMM OOL

zyx

zyx

OL

FFF

zyxM

OL) axis of cosinesdirection ( force of components , ,

F ofn applicatio ofpoint of scoordinate z ,y x,

F force of components F ,F ,F

where

zyx

zyx

Slide #: 45


• Moments of F about the coordinate axes,

xyz

zxy

yzx

yFxFM

xFzFM

zFyFM

• Moment MOL of the force F about OL measures the tendency of F to make the rigid body rotates about the fixed axis OL

Slide #: 46


If the axis does not pass through the origin

• Moment of a force about an arbitrary axis,

BABA

BA

BBL

rrr

Fr

MM

zyx

BABABA

zyx

BL

FFF

zzyyxxM

Slide #: 47

Sample Problem 3.5Sample Problem 3.5

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.

A cube is acted on by a force P as shown. Determine the moment of P

Slide #: 48


• Moment of P about A,

kjP

jiaM

kjP

kPjPP

jiajaiar

PrM

A

AF

AFA

2

22/2/

kjiaPM A

2/

• Moment of P about AB,

kjiaPi

MiM AAB

2

2aPM AB

Slide #: 49


1116

23

12

3

1

3

aP

kjiaP

kjiM

kjiaP

M

kjia

kajaia

r

r

MM

AG

A

GA

GA

AAG

6

aPM AG

Slide #: 50


zyx

AFAFAF

zyx

BL

FFF

zzyyxxM

Alternative method using determinant

6/

2/2/0

0

3/13/13/1

aP

PP

aaM BL

slide #: 1 chapter 3 rigid bodies : equivalent systems of forces

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