Ch.02 Force Vectors

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02. Force Vectors

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics Statics 2.01 Force Vectors

Chapter Objectives

To show how to add forces and resolve them into components using the Parallelogram Law

To express force and position in Cartesian vector form and explain how to determine the vectors magnitude and direction

To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto

another



1. Scalars and Vectors

- Scalar: a scalar is any positive or negative physical quantity

that can be completely specified by its magnitude

Examples: length, mass, and time

- Vector: a vector is any physical quantity that requires both a

magnitude and a direction for its complete description

Examples: force, position, and moment

A vector is shown graphically by an arrow

the length of the arrow: the magnitude of the vector

the angle between the vector and a fixed axis: direction of its line of action

the head or tip of the arrow: the sense of direction of the vector

- Notation: , | |



2. Vector Operations

- Multiplication and division of a vector by a scalar =

- Vector addition = +

Parallelogram law

Triangle rule



2. Vector Operations

- Vector subtraction = = + ()

Parallelogram law

Triangle construction



3. Vector Addition Forces

- Experimental evidence has shown that a force is a vector

quantity since it has a specified magnitude, direction, and

sense and it adds according to the parallelogram law

- Finding a resultant force

The two component forces 1 and 2 acting on the pin can be

added together to form the resultant force = 1 + 2



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- Finding the components of a force

To resolve a force into two components in order to study its

pulling or pushing effect in two specific directions




- Addition of several forces

To resolve a force into two components in order to study its

pulling or pushing effect in two specific directions

= 1 + 2 + 3 = ( 1 + 2 12

) + 3




- Example 2.1 The screw eye is subjected to two forces, 1 and

2. Determine the magnitude and direction of the resultant force

Solution

= 1002 +1502 2100 150 1150 = 212.6 150

=

212.6

1150 = 39.80, = 39.80 + 15.00 = 54.80




- Example 2.2 Resolve the horizontal 600 force into components acting along the and axes and determine the magnitudes of these components

Solution

1200

=600

300 = 600

1200

300= 1039

300

=600

300 = 600

300

300= 600




- Example 2.3 Determine the magnitude of the component

force and the magnitude of the resultant force if is directed along the positive axis

Solution

600=

200

450 = 200

600

450= 245

750

=200

450 = 200

750

450= 273




- Example 2.4 It is required that the resultant force acting on

the eyebolt be directed along the positive axis and that

2 have a minimum magnitude. Determine this magnitude, the angle , and the corresponding resultant force

Solution

Magnitude of 2 is a minimum its line of action is perpendicular to the line of action

of , that is, when = 900

and = 800 600 = 400

2 = 800 600 = 693



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Fundamental Problems

- F2.1: Determine the magnitude of the resultant force acting on the

screw eye and its direction measured clockwise from the axis




- F2.2: Two forces act on the hook. Determine the magnitude of

the resultant force




- F2.3: Determine the magnitude of the resultant force and its

direction measured counterclockwise from the positive axis




- F2.4: Resolve the 30 force into components along the and axes, and determine the magnitude of each of these

components




- F2.5: The force acts on the frame. Resolve this force into

components acting along members and , and determine the magnitude of each component




- F2.6: If force is to have a component along the axis of = 6, determine the magnitude of and the magnitude of its component along the axis



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4. Addition of a System of Coplanar Forces

- Scalar Notation

= +

=

=

- Cartesian Vector Notation

= +

= +



=

=


- Coplanar Force Resultants

= 1 + 2 + 3

= 1 + 1 + (2 + 2 ) + (3 3 )

= 1 2 + 3 + (1 + 2 3)

+ =

+ =

= 2 +

2 , = 1




- Example 2.5 Determine the and components of 1 and

2 acting on the boom. Express each force as a Cartesian vector

Solution




Scalar notation

1 = 200300 = 100 = 100

1 = 200300 = 173 = 173

2 = (12/13) 260 = 240 = 240N

2 = (5/13) 260 = 100 = 100




Cartesian vector notation

1 = 100 + 173

2 = 240 100




- Example 2.6 The link is subjected to two forces 1 and 2. Determine the magnitude and direction of the resultant force

Solution 1

Scalar Notation

+ = : = 600300 400450 = 236.8

+ = : = 600300 + 400450 = 582.8

= (236.8)2+(582.8)2= 629

= 1(582.8/236.8) = 67.90 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien


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Solution 2

Cartesian vector notation

1 = 600300 + 600300

2 = 400450 + 400450

= 1 + 2

= (600300 400450) +(600300 + 400450)

= 236.8 + 582.8




- Example 2.7 The end of the boom is subjected to three concurrent and coplanar forces. Determine the magnitude and

direction of the resultant force

Solution

+ = : = 400 + 250450 200

4

5= 383.2

+ = : = 250450 + 200

3

5= 296.8

= (383.2)2+(296.8)2= 485

= 1(296.8/383.2) = 37.80 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



- F2.7: Resolve each force acting on the post into its and components




- F2.8: Determine the magnitude and direction of the resultant force




- F2.9: Determine the magnitude of the resultant force acting

on the corbel and its direction measured counterclockwise from the axis




- F2.10: If the resultant force acting on the bracket is to be 750 directed along the positive axis, determine the magnitude of and its direction



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- F2.11: If the magnitude of the resultant force acting on the

bracket is to be 80 directed along the axis, determine the magnitude of and its direction




- F2.12: Determine the magnitude of the resultant force and its

direction measured counterclockwise from the positive axis



5. Cartesian Vectors

- Right-Handed Coordinate System




- Rectangular Components of a Vector

= + +

- Cartesian Unit Vectors

, ,

- Cartesian Vector Representation

= + +

- Magnitude of a Cartesian Vector

= 2 +

2 + 2




- Direction of a Cartesian Vector: , ,

=

, =

, =

Unit vector

=

+

+

= + +

Relation between the direction cosines

2 + 2 + 2 = 1

and

= = + + = + +




- The direction of can be specified using two angles, and

= + +

=

=

=

= + +



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6. Addition of Cartesian Vectors

= + +

= + +

= +

= ( +) + ( +) +( +)




- Example 2.8 Express the force as a Cartesian vector

Solution

2 + 2 + 2 = 1

2 + 2600 + 2450 = 1

= 1 0.52 0.7072

= 0.5

= 600

1200

must be in the + direction: = 600

= + +

= 200600 + 200600 + 200450

= 100.0 + 100.0 + 141.4




- Example 2.9 Determine the magnitude and the coordinate

direction angles of the resultant

force acting on the ring

Solution

= 1 + 2

= 50 100 + 100+ 60 + 80

= 50 40 + 180

The magnitude of the sum vector

= 502 + 40 2 +1802

= 191.0




The coordinate direction angles

, , are determined from the components of the unit vector

acting in the direction of

=

=50

191

40

191 +

180

191

= 0.2617 0.2094 + 0.9422

So that

= 0.2617 = 74.80

= 0.2094 = 1020

= 0.9422 = 19.60




- Example 2.10 Express the force as a Cartesian vector

Solution

The magnitudes of the components of

= 100600 = 86.6

= 100600 = 50

= 450 = 50 2/2 = 35.4

= 450 = 50 2/2 = 35.4

Realizing that < 0, we have

= 35.4 35.4 + 86.6

= 35.42 + 35.4 2 +86.62

= 100




| | = 100, = 35.4, = 35.4, = 86.6

The coordinate direction angles of can be determined from

the components of the unit vector acting in the direction of

=

=

+

+

=35.4

100

35.4

100 +

86.6

100

= 0.354 0.354 + 0.866

So that

= 0.354 = 69.30

= 0.354 = 1110

= 0.866 = 30.00



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- Example 2.11 Two forces act on the hook. Specify the

magnitude of and its coordinate

direction angles of that the resultant

force acts along the positive axis and has a magnitude of 800

Solution

1 = 11 + 11 + 11

= 300450 + 300600

+3001200

= 212.1 + 150 150

2 = 2 + 2 + 2




We require

= 1 + 2

where

1 = 212.1 + 150 150

2 = 2 + 2 + 2

= 800

800 = 212.1 + 2 + 150 + 2 + (150 + 2)

To satisfy this equation the , , components of must be

equal to the corresponding , , components of 1 + 2

0 = 212.1 + 2 2 = 212.1

800 = 150 + 2 2 = 650

0 = 150 + 2 2 = 150




= 800

1 = 212.1 + 150 150

2 = 212.1 + 650 + 150

The magnitude of 2

2 = (212.1)2+6502 +1502= 700

The coordinate direction angles of 2

2 = 212.1/700 2 = 1080

2 = 650/700 2 = 21.80

2 = 150/700 2 = 77.60




- F2.13: Determine its coordinate direction angles of the force




- F2.14: Express the force as a Cartesian vector







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- F2.18: Determine the resultant force acting on the hook



7. Position Vectors

- , , Coordinates

A right handed coordinate system

to reference the location of points

in space

- Position Vectors

= + +



7. Position Vectors

- The position vector may be directed from point to point in space

+ =

=

= + + + +

= + +



7. Position Vectors

- Example 2.12

An elastic rubber band is attached to points

and . Determine its length and its direction measured from toward

Solution

Establish a position vector from to

= (21) + (2 0) + (3 3)

= 3 + 2 + 6

These components of can also be determined directly by realizing that they

represent the direction and distance one

must travel along each axis in order to move

from to



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7. Position Vectors

The length of the rubber band

= (3)2+22 +62= 7

Formulating a unit vector in the direction of

=

=

3

7 +

2

7 +

6

7

The components of this unit vector give the

coordinate direction angles

= 3/7 = 1150

= 2/7 = 73.40

= 6/7 = 31.00



8. Force Vector Directed Along A Line

Quite often, the direction of a force is specified by two points

through which its line of action passes

Formulate as a Cartesian vector by realizing that it has the same direction and sense as the position vector directed from point to point on the cord

= =

=

+ + ( )

2 + 2 + ( )2




- Example 2.13

The man pulls on the cord with

a force of 70. Represent this force acting on the support as a Cartesian vector and

determine its direction

Solution

Force is shown in the figure

The direction of this vector, , is determined from the position

vector ( {24} {8} {12} )

= 12 8 24




The magnitude of , which represents the length of cord

| | = 122+(8)2+(24)2= 28

Forming the unit vector that

defines the direction and sense

of both and

=

=

12

28

8

28

24

28

Since , has a magnitude of 70 and a direction specified by

= = 7012

28

8

28

24

28

= 30 20 60




=

=

12

28

8

28

24

28

From the components of the unit

vector

= 12/28 = 64.60

= 8/28 = 1070

= 24/28 = 1490




- Example 2.14 Express the force acts on the hook as a Cartesian

vector

Solution

The coordinates for points and

(2, 0, 2)

(2,3.464,3)

To go from to : {4} {3.464} {1} , thus

=

=4 +3.464 +

(4)2+3.4642+0.18572=0.7428 +0.6433 +0.1857

Force expressed as a Cartesian vector

= = 750(0.7428 + 0.6433 + 0.1857)

= 577 + 482 + 139



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- Example 2.15

The cables exert forces = 100, = 120 on the wall hook at , determine the resultant force acting at as a Cartesian vector

Solution = +

= 4 4

= 42 + 42 = 5.66

= ( /)

= 70.7 70.7

= 4 + 2 4

= 42+22+(4)2=6

= ( /)

= 80 + 40 80

= 150.7 + 40 150.7 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



- F2.19: Express tile position vector in Cartesian vector form, then determine its magnitude and coordinate direction angles




- F2.20: Determine the length of the rod and the position vector

directed from to . What is the angle ?












- F2.23: Determine the resultant force at



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- F2.24: Determine the magnitude of the resultant force at



9. Dot Product

- dot product = scalar product of vectors

=

- Laws of Operation

Commutative law: =

Multiplication by a scalar: = = ()

Distributive law: + = ( ) + ( )

- Cartesian Vector Formulation

= + +

= + +



= + +

9. Dot Product

- Applications

The angle formed between two vectors or intersecting lines

= 1

, 00 1800

The components of a vector parallel and perpendicular to a line

= =

= 2 2



9. Dot Product

- Example 2.16 Determine the magnitudes of the projection of

the force onto the and axes

Solution

The projection forces

()= 100450 = 70.7

()= 100150 = 96.6



9. Dot Product

- Example 2.17 The frame is subjected to a horizontal force

= 300 . Determine the magnitude of the components of

this force parallel and

perpendicular to member



9. Dot Product

Solution

=

=2 + 6 + 3

22 + 62 + 32

= 0.286 + 0.857 + 0.429

=

= (300 )(0.286 +0.857 +0.429 )

= 257.1

=

= (257.1) 0.286 +0.857 +0.429

= 73.5 + 220 + 110

=

= 73.5 + 80 110



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9. Dot Product

- Example 2.18 The pipe is subjected to the force of = 80. Determine the angle

between and the pipe segment and the projection

of along this segment

Solution

Angle

= 2 2 + , = 3

= 3 + , = 10

=

=2 0 + 2 3 +(1)(1)

3 10=0.7379



= 42.50

9. Dot Product

Components of

The component of along is shown in the figure

=

=2 2 +

(2)2+(2)2+12

= 2

3

2

3 +

1

3

= 80

= 803 +

10= 75.89 +25.30

=

= (75.89 +25.30 ) 2

3

2

3 +

1

3

= 59.0




- F2.25: Determine the angle between the force and the line












- F2.28: Determine the component of projection of the force along the line



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- F2.29: Find the magnitude of the projected component of the force along the pipe




- F2.30: Determine the components of the force acting parallel and perpendicular to the axis of the pole



Ch.02 Force Vectors

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