Download - 6161103 2.4 addition of a system
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 +=
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 +=
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
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
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
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
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
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
� 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
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
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 +=
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
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
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
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 =
=
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
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.
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
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
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
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.
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
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