topic resultant and resolution of...
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SKAA 1213 ‐ Engineering Mechanics
TOPIC 2
RESULTANT AND RESOLUTIONOF FORCES
Lecturers: Rosli AnangRosli Anang
Dr. Mohd Yunus IshakDr. Tan Cher Siang
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C FConcurrent Forces ‐forces acting at a point
F1
F2
F4
F3
2
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Collinear ForcesCollinear Forces –forces acting in the same line
F3
F4
F2
F3
F1F1
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Coplanar ForcesCoplanar Forces –forces acting in a same plane
F1
F3
F1
F4
F2
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Force is a vector, therefore parallelogram law is applicable., p g pp
Used to;1) To find resultant force.2) Resolving a known force into two components.
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Vector Addition of Forces
If only two forces are added, the resultant the forcesacting at a point can be determined by;g p y;
Apply the sine and cosine laws.Parallelogram law
F2F1
F1
F2
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Example 1Determine the magnitude of the resultant
Example 1
force on the lever shown and its direction measured counterclockwise from the positive x axis. [Answer: R = 0.346 N , β = 275.3˚]
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Example 2A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is 30 kN directed along the axis of the barge ;
p
barge,;(a) determine the tension in each of the ropes when α = 35o,(b) determine α for which the tension in rope A is minimum.
A
(a) TA = 20 kN , TB = 17.8 kN
(b) TA = 19.3 kN , TB = 23.0 kN
α
40o
A
40
B
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Resolution of Forces• A single force can be broken into two separate components. The two components can be determined by using the
ll l l
Resolution of Forces
Force F can be replaced by FA and FBth t d th ff t
parallelogram law. It is the reverse of resultant.
that produce the same effect
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Example 3Determine the components of force F=100N forDetermine the components of force F=100N for the axis system of a and b as shown.
[Answer : FA = 50.8 N , FB = 95.4 N ][ A B ]
aF=100N
b70o
30o30o
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Scalar notation is used to solve problems for coplanar forces 2DScalar notation is used to solve problems for coplanar forces – 2D
200mmPF3 =3KN
20° AB
C
D
600N360mm
θF4 =1KN
Cartesian vector notation is used in three dimensional problems. F1 = 2KN
600N
11
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R lt t f th 2 FIf more than two forces are added;Resultant of more than 2 Forces
It requires extensive geometricIt requires extensive geometric and trigonometric calculation to determine the magnitudeF2 to determine the magnitude and direction of the resultant. F1
2
F3
An easier way is to use Rectangular‐component method.y g p
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The sense of direction is represented graphically by the arrow headThe sense of direction is represented graphically by the arrow head . For analytical work, establish a notation for representing the sense of direction of the rectangular components. This can be done by either Scalar Notation or Cartesian Vector Notation.
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Rectangular Components of Coplanar ForceRectangular Components of Coplanar Force
Method• Sum up the components of each force along specified axes
algebraically, and then form a resultant.
Method
The axes may in general
• Resolved each force into its rectangular components Fx and Fywhich lie along the x and y axes.
FF
yThe axes may in general be directed in any inclination.
Fy
Fx F’ = F’x + F’y
FxF = Fx + Fy
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E l 4Four forces act on an eye bolt. Determine the
Example 4y
resultant of the forces on the bolt.
y[ANSWER FRx= 111.8N , FRy= ‐244.8N , FR = 269 N , Θ = ‐ 65.5o ]
y
xxF1=30N
F4=100N65o
30o
20o
F2=120NF3=90N
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Example 5Resolve the force F acting at C into two components acting along members AC and BC
Example 5
components acting along members AC and BC, if θ = 20o. Determine θ, so that the component F is directed toward C and has a magnitudeFAC is directed toward C and has a magnitude of 150N.
A C
[Answer : FCB = 163.8 N , FAC = 168.4 N , θ = 4.3o ]
35o
F=100Nθ
B
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1.Scalar Notation• Since the x and y axes have designated +ve and –ve directions,
the magnitude and directional sense of the components of a force can be expressed in terms of algebraic scalarsforce can be expressed in terms of algebraic scalars.
• the component is represented by +ve scalar F if the sense of direction is along the +ve axis and vice versa.
y
Fy
F
y
x
F’xF’y θ
Fx
xθ
F’x = F’ cos θF’y = ‐ F’ sin θ (– ve opposite to the direction
f i )
F’Fx = F cos θ
Fy = F sin θ of +ve y‐axis)y
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2. Cartesian Vector Notation• In 2D, the Cartesian unit vectors i and j are used to show the
direction of the x and y axes
• The unit vectors have a dimensionless magnitude, and it’s described analytically by + and – signs, depending whether they are pointing along the +ve or –ve x or y axesy yy p g g yy
Fj
F’x
F’y
y
xFx i
Fyi-j
F = Fxi + Fyj F’ = F’x i +F’y(-j)F’ = F’x i - F’yj
F’ x
Fx = F cos θFy = F sin θ F’x = F’ cos θ
F’y = F’ sin θ
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Example 6pDetermine the magnitude and orientation of the resultant
force using Cartesian vector method. [Answer : FR = 13.3 i + 354.8 j ]
y
F1=250NF2=325N
y
60o
45o
x
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Coplanar Resultant Force [2D]Coplanar Resultant Force [2D] (Rectangular‐Component Method)
• Either of the two methods (Scalar or Cartesian vector) can be• Either of the two methods (Scalar or Cartesian vector) can be used to determined the resultants.
E h f i l d i d d l• Each force is resolved into x and y components and total up all the components using scalar algebra.
• Apply parallelogram law to obtain resultant force by adding the resultant of the x and y components.
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y
F1FUsing Scalar notation
( ) F F F FF1yF
x
1F2 ( +) FRx = F1x ‐ F2x + F3x( +) FRy = F1y + F2y ‐ F3y
F1xF2xF3x
F2y
F3Using Cartesian vector notation
F3y
F
F1yF2y
F1 = F1x i + F1y j
F2 = ‐F2x i + F2y j
yF1F2F1xF2x
F3x
F3 = F3xi ‐ F3yj
F = F + F + F
x
F3y
FR = F1 + F2 + F3= F1xi + F1y j ‐ F2xi + F2y j + F3xi ‐ F3y j
= (F1x- F2x + F3x) i + (F1y + F2y ‐ F3y) jF3
( 1x 2x 3x) ( 1y 2y 3y) j
= FRx i + FRy j
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h l b d b h l b fThe resultant can be represented by the algebraic sum of the x and y components:
y
FRX = ∑ Fx
FRFRy
y
RX ∑ x
FRY = ∑ FyFRxxθ
The magnitude FR can be found from Pythagoras theorem;
The direction angle, θ
theorem;
θ = tan‐1FRYF
g ,FRX
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E l 7Express each of the three forces acting the column in Cartesian vector form and compute the
Example 7in Cartesian vector form and compute the magnitude of the resultant force.
[Answer : FR = 443.5 N , θ = - 84.3o ]
y
F 80 NF1=175 N F2=250N
x
F3=80 N
40o4
3
5