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Chapter 2

STATICS OF PARTICLE

Objectives

•  To show how to add forces and resolve them into components using the parallelogram law.

•  To introduce the concept of the free-body diagram for a particle.

•  To show how to solve particle equilibrium problems using the equations of equilibrium.

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Introduction

•  Study the effect of forces acting on particles

–  How to replace two or more forces acting on a particle by a single force having the same effect resultant force.

–  Relations which exist among the various forces acting on a particle in a state of equilibrium to determine some of the forces acting on the particle.

–  Resolution of a force into components

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Definitions

Scalar - A quantity characterized by a positive or negative number is called a scalar. Examples of scalars used in Statics are mass, volume or length.

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Definitions

Vector - A quantity that has both magnitude and a direction. Examples of vectors used in Statics are position, force, and moment.

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Forces on a particle. Resultant of Two Forces

•  Force action of one body on another; characterized by its point of application, magnitude and direction

•  Direction of a force defines by line of action and sense of the force.

•  Line of action infinite straight line along which the force act; characterized by the angle it forms with some fixed axis

•  Sense of the force indicated by an arrowhead

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•  Force represented by a segment of that line.

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•  Forces P and Q acting on a particle A may be replaces by a single force R; has the same of the particle resultant force

•  By constructing a parallelogram, the diagonal that passes through A represents the resultant parallelogram law for the additional of two forces

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Vector

•  Vectors mathematical expression possessing magnitude and direction; which add according to parallelogram law.

•  Two vectors; have the same magnitude and direction are said to be equal (a)

•  Two vectors; have the same magnitude, parallel lines of action and opposite sense are equal and opposite. (b)

a) b)

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Addition of Vectors •  Vectors add according to the parallelogram law (PL) •  The sum of two vectors P and Q attach the two vectors

to the same point A and using PL; the diagonal passes through A represents the sum of vectors P and Q; denoted by P+Q

•  The sum does not depend upon the order of the vectors; the addition of two vectors is commutative

P + Q = Q + P

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Vector Addition

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•  Alternate method for determining the sum of two vectors triangle rule

•  Draw only half of PL •  The sum of the two

vectors; by arranging P and Q in tip-to-tail fashion and connecting the tail of P with the tip of Q

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Vector Addition

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Vector Substraction CHAPTER 2

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For vector V1 and V2 shown in figure: a: determine the magnitude S of their vector sum S = V1 + V2 b: determine the angle α between S and the positive x-axis c: determine the vector difference D = V1 – V2

V1= 4 kN

V2= 3 kN

450

300

Example 2.1

2.5  Resultant  of  several  concurrent  forces  

Par/cle  A  acted  by  several  upon  coplanar  forces.  

A

P

Q

S

θ  

β  γ  A

P

Q

S

θ  

β  

γ  

α  

R  

end  

start  

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A

P Q

S

θ  

β  

γ  

R  

end  

start  

OR   S  +  Q  +  P  =  R  

Vector  addi;on  is  associated  

P+Q+S  =  (P+Q)+S  =  S+(P+Q)  =  S+(Q+P)  =  S+Q+P  

Applica;on  is  beAer  illustrated  by  the  following  example:  

Example 2.2

Determine  the  resultant  of  the  two  forces.  

100N  

150N  

20°  

30°  

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Solu;on  1  :  Graphical  Method  

using  a  scale  1cm  =  50N  

The  procedure  is  to  start  at  a  point  and  arrange  the  force  vectors  ;p-­‐to-­‐tail.  Resultant  is  the  vector  beginning  from  ‘start’  and  finishing  at  ‘end’.  

100N  20°  

150N  30°  

R  

θ  

start  

end  

R  =  3.3  cm  =  165N  -­‐  using  ruler  Δθ  =  83.5°      -­‐  using  protractor  

Solu;on  2:  Trigonometry  Method  

R  

x  

y  

A  

B    

α  

β  

θ  

1. LAW OF COSINES: R2 = (A)2 + (B)2 ! 2AB cos"

2. LAW OF SINES:

Rsin!

=Bsin"

=Asin#

Defining  ques/on  for  usage:  Do  you  know  2  sides  and  the  angle  them?  IF  answer  is  “YES”,  then  we  use  the  law  of  cosines  IF  answer  is  “NO”,  then  we  must  use  the  law  of  sines  

The  procedure  is  iden;cal  to  the  graphical  method,  but  trigonometry  is  used  to  calculate  the  magnitude  and  direc;on  (angle).  These  is  no  need  to  draw  the  polygon  to  scale.  

Prerequisites:  

3.  Must  be  ‘good’  in  determining  angles  

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! = 20! + 60! = 80!

Cosine law:R2 = 1002 +1502 " 2(100)(150)cos80!

R = 165.2 NSine Law:sin#150

=sin80!

165.2# = 63.4!

$ = # + 20! = 83.4!

100N  20°  

150N  30°  

R  

θ  

start  

end  

80°  

α  

R  =  165.2N   83.4°  

Resolution of a Force into Components CHAPTER 2

MOHD FAUZI HAMID

•  A single force F acting on a particle may be replaced by two or more forces; which have the same effect. –  The forces are called components; the process of

substituting them is called resolving the force F into components

•  For each force F; exist an infinite number of possible set of components –  Numbers of ways in which a given force F maybe resolved

into two components

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“Resolu;on”  of  a  vector  is  breaking  up  a  vector  into  components.    It  is  kind  of  like  using  the  parallelogram  law  in  reverse.                                                                                                                                                                                    

•  One of the two components; P is known –  The second component; Q is obtained by applying the

triangle rule; joining the tip P to the tip of F; the magnitude and direction of Q are determined graphically or by trigonometry

–  Once Q has been determined, both components of P and Q should apply at A

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•  The line of action of each component is known –  The magnitude and sense of the components are

obtained by applying the parallelogram law and drawing lines; through the tip of F parallel to the given lines of action

–  Two well defined components of P and Q; determined graphically or by trigonometry.

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•  The direction of one component may be known while the magnitude of the other component is to be as small as possible (minimum)

•  The appropriate triangle is drawn

P and Qmin is perpendicular

F

Qmin P

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Rectangular Components of a Force •  Force F resolved into a component Fx along the x axis

and Fy along the y axis; the parallelogram drawn to obtain the two component is a rectangular

•  The X and y axis may be chosen in any two perpendicular directions

•  The magnitude of the components Fx and Fy; »  Fx = F cos Fy = F sin

•  Scalar component Fx is positive when the vector component Fx has the same sense as the x axis; and negative when Fx has the opposite sense. Same goes to Fy component.

•  The angle, , defined by; •  tan = Fy/ Fx •  The magnitude F of the force can be obtained using

Pythagorean theorem; •  F = Fx2 + Fy2

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Addition of Forces by Summing x and y Components

•  When three or more forces are to be added; practical trigonometry solution may be difficult.

•  An analytical solution; resolving each force into two rectangular components

•  The horizontal comps.; added into a single force Rx; and the vertical comps. into a single force Ry.

•  The forces Rx and Ry then be added vectorially into the resultant R of the given system

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CHAPTER 2

R = (Rx)2 + (R

y)2

! = tan"1R

y

Rx

#

$%%

&

'((

Rx= P

x+Q

x+ S

x

Ry= P

y+Q

y+ S

y

R

x= F

x! Ry= Fy!

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Solu;on  3:  Rectangular  Component  

100N  

150N  20°  

30°  

100sin20°  150cos30°  

100cos20°  

150sin30°  

! +ve: Rx= F

x" R

x= 100cos20! #150sin30!

Rx= 18.97 N ( ! )

$ +ve: Ry= F

y" R

y= 100sin20! +150cos30!

Ry= 164.1 N ( $ )

Rx  

Ry  

θ  

R = (18.97)2 + (164.1)2 = 165.3 N

!=tan-1 164.118.97

"

#$

%

&' = 83.4!

Example 2.3

Determine  components  of  the  100N  force  along  the  a-­‐a  and  b-­‐b  axes  

60°  

100  N  

25°  

a  

a  

b   b  

Ans:  Fa-­‐a  =  204.9  N,  Fb-­‐b  =  235.7  N  

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Example 2.4

Determine  the  resultant  R  and  the  minimum  force  P,  if  the  resultant  of  the  forces  is  ver/cally  downwards.  

30°  θ  

100  N  

P  

Ans:  P  =  50  N  ,  R  =  86.6N    

Example 2.5

Three  forces,  F1,  F2  and  F3  act  on  point  A,  determine  the  resultant,  R.  

40°  

20°  

45°  

F1  =  200N   F2  =  100N  

F3  =  300N  

Ans:  R  =  207.8N,  θ  =  11.13°  

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Example 2.6

Three forces, F1, F2 and F3 act on point A. If the resultant is known to be on the a-a axis, determine the magnitude of F3 and the resultant R.

A

40°

F2 = 100N F1 = 200 N

20°

30°

F3

45°

a

a

Ans: R = 178.6 N↖,  F3  =  230.4  N  

Solution 1

A

40°

F3 = 100N F1 = 200 N

20°

30°

F3

45°

a

a

R Rcos30  

Rsin30 F3sin45

F3cos45

200sin20 200cos20

The assumption in direction of R must be stated

! +ve: Rx= F

x"R cos30! = 100cos 40! # 200sin20! # F

3cos 45!

0.866R = 8.2 # 0.707F3 ......(1)

$ +ve: Ry= F

y"#R sin30! = 100sin40! + 200cos20! # F#0.5R = 252.2 # 0.707F

3 ......(2)

Solving for R and F3

equ.(1) - equ.(2):1.366R = !244R = !178.6 N replace in equ.(1)0.866(!178.6) = 8.2 ! 0.707F

3

F3= 230.4 N

x

y

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Solution 2

A

40°

F3 = 100N F1 = 200 N

20°

30°

F3

45°

a

a x

y

30°

Use slanting axes where the forces are resolved through 90° component are not vertical and harizontal

Solution 3

polygon

A

40° F3 = 100N

F1 = 200 N 20°

30°

F3 45°

a

a

R1

θ

β

start

end2

end1