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UNIVERSITY OF SARAJEVO MECHANICAL ENGINEERING FACULTY DEFENSE TECHNOLOGIESDEPARTMENT www.dtd.ba

Klasifikacija vektora Slobodni vektor. To je vektor koji se moe pomjerati slobodno u

prostoru i nije vezan za jedinstvenu liniju.

Klizni vektor. To je vektor koji je vaan za jednu liniju pri

djelovanju, ali nije vezan za jedinstvenu taku pri djelovanju. On

.

. .

UNIVERSITY OF SARAJEVO MECHANICAL ENGINEERING FACULTY DEFENSE TECHNOLOGIESDEPARTMENT www.dtd.ba 1 - 2


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Zakon paralograma


Zakon paralelograma i sinusni i kosinusni

zakoni



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Primjer


Rezultatna tri i vie sueljnih slia



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Rezultatna tri i vie sueljnih sila, primjer


UNIVERSITY OF SARAJEVO MECHANICAL ENGINEERING FACULTY DEFENSE TECHNOLOGIESDEPARTMENT www.dtd.ba 1 -10


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Introduction

The objective for the current chapter is to investigate the effects of forces

on particles:

- replacing multiple forces acting on a particle with a single

equivalent or resultant force,

- relations between forces actin on a article that is in a

state of equilibrium.

The focus on particles does not imply a restriction to miniscule bodies.

Rather, the study is restricted to analyses in which the size and shape of

the bodies is not significant so that all forces may be assumed to be

applied at a single point.

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Resultant of Two Forces

force: action of one body on another;

characterized by its point of application,

magnitude, line of action, and sense.

Experimental evidence shows that the

combined effect of two forces may be

.

The resultant is equivalent to the diagonal of

in adjacent legs.

Force is a vectorquantity.



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Vectors Vector: parameters possessing magnitude and direction

which add according to the parallelogram law. Examples:

displacements, velocities, accelerations.

Scalar: parameters possessing magnitude but not

direction. Examples: mass, volume, temperature

- Fixed or bound vectors have well defined points of

application that cannot be changed without affecting

an analysis.

- Free vectors may be freely moved in space without

changing their effect on an analysis.

- Sliding vectors may be applied anywhere along their

.

Equal vectors have the same magnitude and direction.


Negative vector of a given vector has the same magnitude

and the opposite direction.

Addition of Vectors Trapezoid rule for vector addition

Triangle rule for vector addition

BC

BPQQPR += cos2222

Law of cosines,

C QPRrrr

+=

Law of sines,

B A

C

R

B

Q

A sinsinsin==

,

PQQPrrrr

+=+



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Addition of Vectors

Addition of three or more vectors through

repeated application of the triangle rule

more vectors.

Vector addition is associative,

( ) ( )SQPSQPSQPrrrrrrrrr

++=++=++

Multiplication of a vector by a scalar


Resultant of Several Concurrent Forces Concurrent forces: set of forces which all

pass through the same point.

A set of concurrent forces applied to a

particle may be replaced by a single

resultant force which is the vector sum of the

.

vectors which, together, have the same effect

as a single force vector.



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Sample Problem 2.1 Trigonometric solution - Apply the triangle rule.

From the Law of Cosines,

222

( ) ( ) ( )( ) +=

=

155cosN60N402N60N40 22

N73.97=R

BA sinsin

From the Law of Sines,

R

QBA

RQ

=

=

sinsin

A =

=

04.15

N73.97

N60155sin


+=

= 04.35

Sample Problem 2.2SOLUTION:

Find a graphical solution by applying the

Parallelogram Rule for vector addition. The

parallelogram has sides in the directions of

the two ropes and a diagonal in the direction

of the barge axis and length proportional to

5000 lbf.

A barge is pulled by two tugboats.

If the resultant of the forces

exerted b the tu boats is 5000 lbf

Find a trigonometric solution by applying the

Triangle Rule for vector addition. With thedirected along the axis of the

barge, determine

magnitude and direction of the resultant

known and the directions of the other two

sides parallel to the ropes given, apply the

a) the tension in each of the ropes

for = 45o, The angle for minimum tension in rope 2 is

determined b a l in the Trian le Rule and

.


tension in rope 2 is a minimum. observing the effect of variations in .


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Sample Problem 2.2

Graphical solution - Parallelogram Rulewith known resultant direction and

magn u e, nown rec ons or s es.

lbf2600lbf3700 21 == TT

Trigonometric solution - Triangle Rule

with Law of Sines

=

=

105sin

lbf5000

30sin45sin

21 TT


21 ==

Sample Problem 2.2 The angle for minimum tension in rope 2 is

determined by applying the Triangle Rule and

observing the effect of variations in .

The minimum tension in rope 2 occurs when

.

( ) = 30sinlbf50002T lbf25002 =T

( ) = 30coslbf50001T lbf43301=T

= =



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Rectangular Components of a Force: Unit

Vectors

May resolve a force vector into perpendicular

components so that the resulting parallelogram is a

rectangle. are referred to as rectangularFFrr

andvector components an

yx FFFrrr

+=

Define perpendicular unit vectors which are

parallel to the x and y axes.jirr

and

Vector components may be expressed as products of

the unit vectors with the scalar magnitudes of the

vector com onents..

Fx and Fy are referred to as the scalar components of

jFiFF yxrrr

+=r


Addition of Forces by Summing Components

SPRrrrr

++=

Wish to find the resultant of 3 or more

concurrent forces,

rrrrrrrr

Resolve each force into rectangular components

( ) ( )jSQPiSQP yyyxxxyxyxyxyxrr

+++++=

The scalar components of the resultant are equal

++= xxxx SQPR

to the sum of the corresponding scalar

components of the given forces.

++= SQPR

= xF = yF

To find the resultant magnitude and direction,


x

yyx

RRRR

122 tan=+=


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Sample Problem 2.3

SOLUTION:

Resolve each force into rectangular

components.

Determine the components of the

Calculate the magnitude and direction

force components.

Four forces act on boltA as shown.

Determine the resultant of the force

on the bolt.

of the resultant.



Resolve each force into rectangular

components.

2.754.2780

0.759.129150

2

1

+

++

F

F

compycompxmagforce

r

r

r

9.256.96100

0.1100110

4

3

+

F

Fr

= =

Determine the components of the resultant by

adding the corresponding force components.

.x .y

22 3.141.199 +=R N6.199=R

Calculate the magnitude and direction.


N1.199

N3.14tan = = 1.4


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Equilibrium of a Particle When the resultant of all forces acting on a particle is zero, the particle is

in equilibrium.

Newtons First Law: If the resultant force on a particle is zero, the particle will

remain at rest or will continue at constant speed in a straight line.

Particle acted upon by Particle acted upon by three or more forces:

two forces:

- equal magnitude

- same line of action

- graphical solution yields a closed polygon

- algebraic solution

rr


- opposite sense

00 ==

==

yx FF

Free-Body Diagrams

Space Diagram: A sketch showing

the physical conditions of the

Free-Body Diagram: A sketch showing

only the forces on the selected particle.

.



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Sample Problem 2.4

SOLUTION:

Construct a free-body diagram for the

part c e at t e unct on o t e rope an

cable.

creating a closed polygon from the forces

applied to the particle.

In a ship-unloading operation, a

3500-lb automobile is su orted b

pp y tr gonometr c re at ons to

determine the unknown force

magnitudes.

a cable. A rope is tied to the cable

and pulled to center the automobile

over its intended position. What is


e ens on n e rope


Construct a free-body diagram for the

particle atA.

Apply the conditions for equilibrium.

Solve for the unknown force magnitudes.

== lb3500

ACAB TT

s ns ns n

lb3570=ABT

=AC



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Sample Problem 2.6

SOLUTION:

Choosing the hull as the free body,

draw a free-body diagram.

Express the condition for equilibriumfor the hull b writin that the sum of all

It is desired to determine the drag force

at a given speed on a prototype sailboat

forces must be zero.

Resolve the vector equilibrium

hull. A model is placed in a test channel

and three cables are used to align its

bow on the channel centerline. For a

iven s eed the tension is 40 lb in

equations. Solve for the two unknown

cable tensions.

cableAB and 60 lb in cableAE.

Determine the drag force exerted on thehull and the tension in cableAC.



Choosing the hull as the free body, draw a

free-body diagram.

== 75.1ft7

tan == 375.0ft1.5

tan

= 25.60t

= 56.20

Express the condition for equilibrium

for the hull by writing that the sum of all

forces must be zero.

0=+++= DAEACAB FTTTRrrrrr



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Sample Problem 2.6

Resolve the vector equilibrium equation into

two component equations. Solve for the two

unknown cable tensions.

( ) ( )

( ) ( )ji

jiTAB rr

rrr

lb84.19lb73.34

26.60coslb4026.60sinlb40

+=

+=

jTiT

jTiTT

ACAC

ACACAC

rr

rr

rrr

9363.03512.0

56.20cos56.20sin

+=

+=

iFF

i

DD

rr

=

=

( ) iFT

R

DAC

r

r

3512.073.34

0

++=

=


( ) jTACr

609363.084.19 ++

Sample Problem 2.6

( ) iFT

R

DAC

r

r

3512.073.34

0

++=

=

( ) jTACr

609363.084.19 ++

This equation is satisfied only if each component

( ) 3512.073.3400 ++==

DACx FTF

609363.084.1900 +== ACy TF

lb9.42+=ACT

lb66.19+=DF



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Rectangular Components in Space

The vector is

contained in the

plane OBAC.

Fr

Resolve into

horizontal and vertical

components.

Fr

Resolve into

rectangular componentshF

yh FF sin=

yy FF cos=

sin

cossin

h

y

hx

FF

F

=

=

=


sinsin yF=


r

With the angles between and the axes,F

FFFFFF zzyyxx

rrrr

coscoscos ===

( )

F

kjiF zyx

zyx

r

rrr

coscoscos

=

++=

kji zyxrrrr

coscoscos ++=

is a unit vector along the line of action of Fr

r


and are the direction

cosines for Fr zyx

cosand,cos,cos


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Direction of the force is defined b

the location of two points,

( ) ( )222111 ,,and,, zyxNzyxM

NMd=r

andjoiningvector

zzdyydxxd

kdjdid

zyx

zyx

===

++=

rr

rrr

121212

( )kdjdidd zyx ++=

=rrrr 1


d

FdF

d

FdF

d

FdF zz

yy

xx ===


Based on the relative locations of the

,

pointing fromA towards B.

A l the unit vector to determine the

components of the force acting onA.

Noting that the components of the unit

The tension in the guy wire is 2500 N.

Determine:

vector are t e rect on cos nes or t e

vector, calculate the corresponding

angles.

a) components Fx, Fy, Fz of the force

acting on the bolt atA,

b) the angles , , defining the


direction of the force


20/20


Determine the unit vector pointing fromA towardsB.

( ) ( ) ( )

( ) ( ) ( )m30m80m40

m30m80m40

222 ++=

++=

AB

kjiABrrr

m3.94=

kjirrrr 308040

+

+

=

Determine the com onents of the force.

kjirrr

318.0848.0424.0

...

++=

( )( )kjiFF

rrr

rr

318.0848.0424.0N2500 ++=

=


( ) ( ) ( )kjirrr

N795N2120N1060 ++=

Sample Problem 2.7 Noting that the components of the unit vector are

the direction cosines for the vector, calculate the

corresponding angles.

kji zyxrrr

rrrr

coscoscos

=

++=

...

o1.115=x

o

o

5.71

0.32

=

=

z

y

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