engineering mechanic - chapter 2a
DESCRIPTION
Engineering Mechanic - Chapter 2A UTMTRANSCRIPT
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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
CHAPTER 2
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.
CHAPTER 2
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)
CHAPTER 2
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
CHAPTER 2
• 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
CHAPTER 2
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
CHAPTER 2
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