Chapter 2 Statics of Particles

Chapter 2Chapter 2

Statics of ParticlesStatics of Particles

Addition of ForcesAddition of Forces

Parallelogram Rule:

The addition of two forces P and Q :

A

→P

→Q →

P→Q+ =

→R

Draw the diagonal vector represent the resultant force

Draw two parallel lines of vectors to form a parallelogram


Triangle Rule:

A

→P

→Q

Alternative method to Parallelogram Rule

Draw the vector from starting point to the tip of second vector

Arrange two vectors in tip-to-tail fashion

→P

→Q+ =

→R


Triangle Rule:

A

→P

→Q

The order of the vectors does not matter

→P

→Q + =

→Q

→P+

Addition of two forces is commutative

→P

→Q+ =

→R


More than two forces:

A

→P

→Q

Arrange the given vectors in tip-to-tail fashion

→S

Connect the tail of first vector to the tip of last vector→P

→Q+ =

→R

→S+


More than two forces:

A

→P

→Q

→S →

P→Q+ =

→R

→S+

A

→P

→Q

→S

The order of the vectors does not matter

→S

→P+ =

→Q+

→Q

→S+

→P+ =

→P

→Q+

→S+

Rectangular Components of a Rectangular Components of a ForceForce

→F

→Fx=

→Fy

+

→Fx , Rectangular components of

→F

→Fy

θ : Angle between x-axis and F

measured from positive side of x-axis

Rectangular Components of a Rectangular Components of a ForceForceUnit VectorsUnit Vectors

→ i ,

Unit vectors→ j

iFF xx

jFF yy

jFiFF yx

Rectangular Components of a Rectangular Components of a ForceForceUnit VectorsUnit Vectors

→ i ,

Unit vectors→ j

iFF xx

jFF yy

22yx FFF

jFiFF yx

cosFFx sinFFy

Rectangular Components of a Rectangular Components of a ForceForceExampleExample

NFFx 665)145cos(800cos 0

NFFy 459)145sin(800sin 0

iFx

)665(

jFy

)459(

jiF

)459()655(

A Force of 800 N is applied on a bolt A. Determine the horizontal and vertical components of the force

Rectangular Components of a Rectangular Components of a ForceForceExampleExample

NFFx 240)14.323cos(300cos 0

NFFy 180)14.323sin(300sin 0

0

0

14.32386.36360

86.368.010

88cos

AB

jFy

)459(

A man pulls with a force of 300 N on a rope attached to a building.

Determine the horizontal and vertical components of the force applied by the rope at point A

1068 22 AB

Addition of Forces by Summing Addition of Forces by Summing Their componentsTheir components

A

→P

→Q

→S

→P

→Q+ =

→R

→S+

jRiRjSiSjQiQjPiP yxyxyxyx

yyyyyy

xxxxxx

FRSQPR

FRSQPR

)(

)(

Addition of Forces by Summing Addition of Forces by Summing Their components - ExampleTheir components - Example

Problem 2.22 on page 33

7 kN

9 kN5 kN

Determine the resultant force applied on the bolt

ForcForcee

MagnitudMagnitudee

x x ComponentComponent

y y ComponenComponentt

FF11 5 kN5 kN 5* cos 5* cos (40) =(40) =

3.83 kN3.83 kN

5* sin 5* sin (40) =(40) =

3.21 kN3.21 kN

FF22 7 kN7 kN 7*cos(1107*cos(110)=)=

-2.39 kN-2.39 kN

7* sin 7* sin (110)=(110)=

6.57 kN6.57 kN

FF33 9 kN9 kN 9*cos(1609*cos(160)=)=

-8.46 kN-8.46 kN

99* * sin(160)=sin(160)=

3.08 kN3.08 kN

RR -7.01 -7.01 kNkN

12.86 12.86 kNkN

jkNikNR

)86.12()01.7(

Equilibrium of ParticleEquilibrium of Particle

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

is in equilibrium

Equilibrium of Particle - ExampleEquilibrium of Particle - Example

Determine the resultant force applied on point A

ForcForcee

MagnituMagnitudede

x x ComponenComponentt

y y ComponenComponentt

FF11 300300 300300 00

FF22 173.2173.2 00 -173.2-173.2

FF33 200200 -100-100 -173.2-173.2

FF44 400400 -200-200 346.4346.4

RR 00 00

N

N

N

N

200*cos(240) = -100200*cos(240) = -100

200*sin(240) = -173.2200*sin(240) = -173.2

Equilibrium of Particle – Newton’s Equilibrium of Particle – Newton’s First LawFirst Law

If the resultant force acting on a particle is zero,

• the particle remains at rest (if originally at rest)

• the particle moves with a constant speed in a straight line ( if originally in motion)

Equilibrium State

Equilibrium of Particle - ExampleEquilibrium of Particle - Example

Load with mass of 75 kg

Determine the tensions in each ropes of AB and AC

N

N

N

200*cos(240) = -100200*cos(240) = -100

200*sin(240) = -173.2200*sin(240) = -173.2

NkgsmmgW 736)75()/81.9( 2

Since the load is in equilibrium state, The resultant force at A is zero.

0)30cos()130cos(

000

ACAB

xACxABxx

TT

TTFR

0736)30sin()130sin(

000

ACAB

yACyAByy

TT

TTFR

Equilibrium of Particle – Example Equilibrium of Particle – Example (continued)(continued)

N

N

N

200*cos(240) = -100200*cos(240) = -100

200*sin(240) = -173.2200*sin(240) = -173.2

0)866.0()643.0(

0)30cos()130cos( 00

ACAB

ACAB

TT

TT

0736)500.0()766.0(

0736)30sin()130sin( 00

ACAB

ACAB

TT

TT

(1)

(2)

NT

NT

AC

AB

480

647

Equilibrium of Particle – Equilibrium of Particle – Example-2Example-2

Two cables are tied together at C and they are loaded as shown

Determine the tensions in cable AC and BC

Problem 2.44 / page 41


C

BCT

ACT

kNW 3

Draw “Free Body” Diagram


C

BCT

ACT

kNW 3

For simplicity

2F

1F

3F


C

2F

1F

kNF 33

Since the system is in equilibrium

The resultant force at C is zero.

2

0321 yyyyy FFFFR

0)sin()sin()sin( 332211 FFF

0)270sin()sin()180sin( 03221 FFF

ofdefinitionthememberRe

1


Since the system is in equilibrium

The resultant force at C is zero.

0)270sin()6.43sin()13.143sin( 03

02

01 FFF

01

0 13.14386.3618086.36400

300tan

Arc

02 6.43

525

500tan

Arc

C

2F

1F

kNF 33

2 1 2


0)270sin()6.43sin()13.143sin( 03

02

01 FFF

03000)69.0()6.0( 21 FF

0321 xxxxx FFFFR

0)cos()cos()cos( 332211 FFF

0)270cos()6.43cos()13.143cos( 03

02

01 FFF

0)724.0()8.0( 21 FF

(1)

(2)

C

2F

1F

kNF 33

2 1


03000)69.0()6.0( 21 FF

0)724.0()8.0( 21 FF

(1)

(2)

NFNF 24332202 21

C

2F

1F

kNF 33

2 1

Forces in Space ( 3 D )Forces in Space ( 3 D )

)cos( xx FF )cos( yy FF


)cos( zz FF


)cos( zz FF )cos( xx FF )cos( yy FF

•The three angle define the direction of the force F

•They are measured from the positive side of the axis to the force F

•They are always between 0 and 180º

zyx ,,


kFjFiFF zyx


kji

F

kjiF

F

kFjFiF

magnitudeits

vector

F

F

zyx

zyx

zyx

coscoscos

)coscos(cos

coscoscos

•The vector is the unit vector along the line of action of F


kji

F

kjiF

zyx

zyx

coscoscos

coscoscos

kFjFiF

kFjFiFF

zyx

zyx

coscoscos

•The vector is the unit vector along the line of action of F


)cos( zz FF )cos( xx FF )cos( yy FF

given are F force a of F and F ,F components When the zyx

2z

2y

2x FFF F

F

F

F

F

F

F zz

yy

xx coscoscos


Direction of the force is defined by the location of two points,

222111 ,, and ,, zyxNzyxM


d

FdF

d

FdF

d

FdF

kd

Fdj

d

Fdi

d

FdF

kdjdidd

FF

FF

zz

yy

xx

zyx

zyx

1

12

12

12

and joining vector

zzd

yyd

xxd

where

kdjdid

NMd

z

y

x

zyx

222zyx dddd

kdjdidd

d

d

zyx

1

Sample Problem 2.7Sample Problem 2.7

The tension in the guy wire is 2500 N. Determine:

a) components Fx, Fy, Fz of the force acting on the bolt at A,

b) the angles x, y, zdefining the direction of the force

mzzd

myyd

mxxd

BA

z

y

x

30)30(0

80080

40400

)0,80,0()30,0,40(

12

12

12

mdddd zyx 3.94222

BAd and joining vector

km) (jm) (im) -(d

308040

kji

kji

d

d

318.0848.0424.0

3080403.94

1

d

kji

318.0848.0424.0

• Determine the components of the force.

kji

kji

FF

N 795N 2120N1060

318.0848.0424.0N 2500


• Noting that the components of the unit vector are the direction cosines for the vector, calculate the corresponding angles.

kji

kji zyx

318.0848.0424.0

coscoscos

5.71

0.32

1.115

z

y

x

or

2500

795cos

2500

2120cos

2500

1060cos

F

F

F

F

N

N

F

F zz

yy

xx


Addition of Concurrent Forces in Addition of Concurrent Forces in SpaceSpace

• The resultant R of two or more vectors in space

R

R

R

R

R

R

RRRR

sFRsFRsFR

ForcesR

zz

yy

xx

zyx

zzyyxx

coscoscos

'''

222


A concrete wall is temporarily held by the cables shown.

The tension is 840 N in cable AB and 1200 N in cable AC.

Determine the magnitude and direction of the resultant vector on stake A m

m

m

m


Equilibrium of a Particle in SpaceEquilibrium of a Particle in Space

0

0

0

z

y

x

F

F

F

When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium

Problem 2.103 on page 60

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Chapter 2 Statics of Particles