Download - 6161103 2.4 addition of a system
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
For resultant of two or more forces:� Find the components of the forces in the
specified axes
� Add them algebraically
� Form the resultant
In this subject, we resolve each force into rectangular forces along the x and y axes.
yx FFF +=
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Scalar Notation- x and y axes are designated positive and negative
- Components of forces expressed as algebraic - Components of forces expressed as algebraic scalars
Eg:
Sense of direction
along positive x and
y axes
yx FFF +=
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
�Scalar NotationEg:
Sense of direction
yx FFF ''' +=
Sense of direction
along positive x and
negative y axes
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
�Scalar Notation- Head of a vector arrow = sense of the vector graphically (algebraic signs not vector graphically (algebraic signs not used)
- Vectors are designated using boldface notations
- Magnitudes (always a positive quantity) are designated using italic symbols
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Cartesian Vector Notation- Cartesian unit vectors i and j are used to designate the x and y directions
- Unit vectors i and j have dimensionless - Unit vectors i and j have dimensionless magnitude of unity ( = 1 )
- Their sense are indicated by a positive or negative sign (pointing in the positive or negative x or y axis)
- Magnitude is always a positive quantity, represented by scalars Fx and Fy
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Cartesian Vector Notation F = Fxi + Fyj F’ = F’xi + F’y(-j)
F’ = F’xi – F’yj
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Coplanar Force ResultantsTo determine resultant of several coplanar forces:
- Resolve force into x and y - Resolve force into x and y components
- Addition of the respective components using scalar algebra
- Resultant force is found using the parallelogram law
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Coplanar Force ResultantsExample: Consider three coplanar forces
Cartesian vector notation
F1 = F1xi + F1yj
F2 = - F2xi + F2yj
F3 = F3xi – F3yj
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� Coplanar Force ResultantsVector resultant is therefore
FR = F1 + F2 + F3
= F i + F j - F i + F j + F i – F j
2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
= F1xi + F1yj - F2xi + F2yj + F3xi – F3yj
= (F1x - F2x + F3x)i + (F1y + F2y – F3y)j
= (FRx)i + (FRy)j
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Coplanar Force ResultantsIf scalar notation are usedFRx = (F1x - F2x + F3x)
FRy = (F1y + F2y – F3y)FRy = (F1y + F2y – F3y)
In all cases,
FRx = ∑Fx
FRy = ∑Fy
* Take note of sign conventions
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Coplanar Force Resultants- Positive scalars = sense of direction along the positive coordinate axes
- Negative scalars = sense of direction - Negative scalars = sense of direction along the negative coordinate axes
- Magnitude of FR can be found by Pythagorean Theorem
RyRxR FFF 22 +=
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
� Coplanar Force Resultants- Direction angle θ (orientation of the force) can be found by trigonometry
Rx
Ry
F
F1tan−=θ
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Example 2.5
Determine x and y components of F1 and F2
acting on the boom. Express each force as a
Cartesian vector Cartesian vector
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Solution
Scalar Notation
←=−=−= NNNF x 10010030sin2001o
Hence, from the slope triangle
↑=== NNNF y
x
17317330cos2001
1
o
= −
12
5tan1θ
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Solution
Alt, by similar triangles
N
F x
13
12
2602 =
Similarly,
NNF
N
x 24013
12260
13260
2 =
=
=
NNF y 10013
52602 =
=
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Solution
Scalar Notation
→== NNF x 2402402
Cartesian Vector Notation
F1 = {-100i +173j }N
F2 = {240i -100j }N
↓=−=
→==
NNF
NNF
y
x
100100
240240
2
2
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Example 2.6
The link is subjected to two forces F1 and
F2. Determine the magnitude and F2. Determine the magnitude and orientation of the resultant force.
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Solution
Scalar Notation
Σ= FF xRx :
↑=
+=
Σ=
→=
−=
N
NNF
FF
N
NNF
Ry
yRy
Rx
xRx
8.582
45cos40030sin600
:
8.236
45sin40030cos600
oo
oo
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Solution
Resultant Force
( ) ( )N
NNFR
629
8.5828.236 22
=
+=
From vector addition,
Direction angle θ is
N629=
o9.67
8.236
8.582tan 1
=
= −
N
Nθ
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Solution
Cartesian Vector Notation
F1 = { 600cos30°i + 600sin30°j } N
F2 = { -400sin45°i + 400cos45°j } NF2 = { -400sin45°i + 400cos45°j } N
Thus,
FR = F1 + F2
= (600cos30°N - 400sin45°N)i + (600sin30°N + 400cos45°N)j
= {236.8i + 582.8j}N
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Example 2.7
The end of the boom O is subjected to three
concurrent and coplanar forces. Determine
the magnitude and orientation of the the magnitude and orientation of the
resultant force.
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2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
Solution
Scalar Notation
Σ= FF xRx
4
:
↑=
+=
Σ=
←=−=
−+−=
N
NNF
FF
NN
NNNF
Ry
yRy
Rx
8.296
5
320045cos250
:
2.3832.383
5
420045sin250400
o
o
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Solution
Resultant Force
2.4 Addition of a System of Coplanar Forces
2.4 Addition of a System of Coplanar Forces
( ) ( )NNFR 8.2962.383 22 +−=
From vector addition,
Direction angle θ is
N485=
o8.37
2.383
8.296tan 1
=
= −
N
Nθ