Post on 19-May-2018
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Ishik University / Sulaimani
Architecture Department
Structure
ARCH 214
Chapter -2-
Force Vectors
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Contents :
1. Scalars and Vectors
2. Vector Operations
3. Vector Addition of Forces
4. Addition of a System of Coplanar Force
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Objectives
To show how to add forces and resolve them into
components using the Parallelogram Law and
Trigonometry Analysis.
To express force and position in Cartesian vector
form and explain how to determine the vector’s
magnitude and direction.
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Scalar :
– A quantity characterized by a positive or negative number.
– Indicated by letters in italic such as A.
Example: mass, volume and length
Vector :
– A quantity that has both magnitude and direction.
Example: position, force and moment
– Represent by a letter with an arrow over it such as or A
– Magnitude is designated as or simply A.
– In this subject, vector is presented as A and its magnitude (positive
quantity) as A.
1. Scalars and Vectors
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A
A
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Vector
– Represented graphically as an arrow.
– Length of arrow = Magnitude of Vector.
– Angle between the reference axis and arrow’s line of action
= Direction of Vector.
– Arrowhead = Sense of Vector
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A. Vector Addition
- Addition of two vectors A and B gives a resultant vector R by the
parallelogram law.
- Result R can be found by triangle construction.
Example: R = A + B = B + A
2. Vector Operations
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B. Vector Subtraction
The resultant of the difference between two vectors 𝐀 and 𝐁 of the same type
may be expressed as:
- Special case of addition
Example: R’ = A – B = A + ( - B )
- Rules of Vector Addition Applies
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C. Multiplication and division of vector by a scalar:
If a vector is multiplied by a positive scalar, its magnitude is
increased by that amount.
When multiplied by a negative scalar it will also change the
directional sense of the vector
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3. Vector Addition of Forces
Experimental evidence has shown that a force is a vector
quantity since it has a specified magnitude, direction, and sense
and it adds according to the parallelogram law.
When two or more forces are added, successive applications of
the parallelogram law is carried out to find the resultant
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Finding a resultant force:
The two component forces 𝐅𝟏 and 𝐅𝟐 acting on the pin in figure,
can be added together to form the resultant force.
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Finding the components of a force:
o Sometimes it is necessary to resolve a force into two components in
order to study its pulling and pushing effect in two specific directions.
o For example, in figure below, F is to be resolved into two components
along two members, defined by u and v.
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Addition of several forces:
o If more than two forces are to be added successive applications of the
parallelogram law can be carried out in order to obtain the resultant
force.
o For example if the three forces 𝐅𝟏, 𝐅𝟐, 𝐅𝟑 act at a point o, the resultant
of any two of the forces is found (𝐅𝟏 + 𝐅𝟐) and then this resultant is
added to the third force yielding the resultant of all three forces.
(𝐅𝐑 = (𝐅𝟏 + 𝐅𝟐) + 𝐅𝟑)
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a) Parallelogram Law
b) Trigonometry 1. sine law 2. cosine law
Procedure of Analysis / Review
Two "component" forces F1 and F2 in Fig.(a) add according to the parallelogram
law yielding, a resultant force FR that forms the diagonal of the parallelogram.
If a force F is to be resolved into components along two axes u and v, Fig.(b).
Then start at the head of force F, and construct lines parallel to the axes, thereby
forming the parallelogram.
The sides of the parallelogram represent the components Fu and Fv.
a) Parallelogram Law
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• Redraw a half portion of the parallelogram to illustrate the triangular
head to tail addition of the components.
• From this triangle, the magnitude of the resultant force can be
determined using the law of cosines, and its direction is determined from
the law of sines.
• The magnitudes of two force components are determined from the law
of sines.
b) Trigonometry Analysis
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Example -1-
The screw eye is subjected to two forces F1 and F2. Determine the
magnitude and the direction of a Resultant force.
Ans:
𝐹𝑅 = 213 N
∅ = 54.7°
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Solution 1: Parallelogram
Unknown: magnitude of FR and angle θ
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N
N
NNNNFR
213
6.212
4226.0300002250010000
115cos150100215010022
Law of Cosines:
Solution 2: Trigonometry
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8.39sin
9063.06.212
150sin
115sin
6.212
sin
150
N
N
NN
Law of Sines
Direction Φ of FR measured from the
horizontal is,
8.54
158.39
Note: The results seem reasonable since it shows FR to have a magnitude larger
than its components and a direction that is between them.
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Example -2-
Determine the magnitude of the component force 𝐅 in figure below
and the magnitude of the resultant force 𝐅𝐑 if 𝐅𝐑 is directed a long
the positive y axis.
Ans:
𝐹 = 245 lb
𝐹𝑅 = 273 lb
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Solution 2: Trigonometry Solution 1: Parallelogram
The magnitudes of FR and F are the two unknowns. They can be
determined by applying the law of sines.
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Example -3-
Determine the magnitude of the resultant force acting on the
screw eye and its direction measured clockwise from the x
axis.
Ans:
𝐹𝑅 = 6.80 kN
θ = 103°
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Example -4-
Resolve the horizontal 600lb force in figure below, into
components action along the u and v axes and determine the
magnitudes of these components.
Ans:
𝐹𝑢 = 1039 lb
𝐹𝑣 = 600 lb
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4. Addition of a System of Coplanar Force
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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Scalar Notation
- x and y axes are designated positive and negative
- Components of forces expressed as algebraic scalars
Example: Sense of direction along positive x and y axes
Instead of using the angle 𝜽, the direction of 𝐅 can also be defined using a
small "slope" triangle, shown in figure.
It is also possible to represent the x and y components of a force in terms of
Cartesian unit vectors i and j
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Cartesian Vector Notation
F = Fxi + Fyj
• We can express 𝐅 as a Cartesian vector.
Coplanar Force Resultants
• In coplanar force resultant case, each force is resolved into its x and
y components.
• Then the respective components are added using scalar algebra since
they are collinear.
• For example,
consider the three concurrent forces,
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• Each force is represented as a Cartesian vector.
Cartesian vector notation;
F1 = F1xi + F1yj
F2 = - F2xi + F2yj
F3 = F3xi – F3yj
• Vector resultant is therefore,
FR = F1 + F2 + F3
= F1xi + F1yj - F2xi + F2yj + F3xi – F3yj
= (F1x - F2x + F3x)i + (F1y + F2y – F3y)j
= (FRx)i + (FRy)j
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• We can represent the components of the resultant force of any number of
coplanar forces symbolically by the algebraic sum the x and y
components of all the forces.
• In all cases, FRx = ∑Fx
FRy = ∑Fy
• Take note of sign conventions
• Magnitude of FR can be found by
Pythagorean Theorem, Ry
2Rx
2
RFFF
• Direction angle θ (orientation of the force)
can be found by trigonometry, Rx
Ry
F
Ftan
1
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Example -5-
The link is subjected to two forces F1 and F2. Determine the magnitude and
orientation of the resultant force.
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Solution :
N
NNF
FF
N
NNF
FF
Ry
yRy
Rx
xRx
8.582
45cos40030sin600
:
8.236
45sin40030cos600
:
Solution 1: Scalar Notation;
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Resultant Force
From vector addition, Direction angle θ is
N
NNFR
629
8.5828.23622
9.67
8.236
8.582tan 1
N
N
Solution 2: Cartesian Vector Notation
F1 = { 600cos30°i + 600sin30°j } N F2 = { -400sin45°i + 400cos45°j } N Thus, FR = F1 + F2 = (600cos30°N - 400sin45°N)i + (600sin30°N + 400cos45°N)j = {236.8i + 582.8j}N
*The magnitude and direction of FR are determined in the S.1nle manner as before.
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Example -6-
Determine the x and y components of 𝐅𝟏 and 𝐅𝟐 acting on the boom shown in
figure, express each force as a Cartesian vector.
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Solution;
NNNF
NNNF
y
x
17317330cos200
10010030sin200
1
1
Scalar Notation;
N10013
5260
N24013
12260
2
2
y
x
F
FBy similar triangles we have;
NNF
NF
y
x
100100
240
2
2
Scalar Notation;
Cartesian Vector Notation;
NjiF
NjiF
100240
173100
2
1
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Example -7-
The link is subjected to two forces F1 and F2. Determine the magnitude and
orientation of the resultant force.
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N
NNF
FF
N
NNF
FF
Ry
yRy
Rx
xRx
8.582
45cos40030sin600
:
8.236
45sin40030cos600
:
Scalar Notation:
N629N8.582N8.236F22
R
Resultant Force;
From vector addition, direction angle θ is;
9.67N8.236
N8.582tan 1
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Resultant Force; N629N8.582N8.236F22
R
From vector addition, direction angle θ is; 9.67
N8.236
N8.582tan 1
Cartesian Vector Notation;
F1 = { 600cos30°i + 600sin30°j } N
F2 = { -400sin45°i + 400cos45°j } N
Thus,
FR = F1 + F2
= (600cos30ºN - 400sin45ºN)i
+ (600sin30ºN + 400cos45ºN)j
= {236.8i + 582.8j}N
The magnitude and direction of FR are determined in the same manner as
before.
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Example -8-
Resolve each force acting on the post into its x and y components.
Ans:
F1𝑥 = O N
F1y = 300 N
F2𝑥 = −318 N
F2y = 318 N
𝐹3𝑥 = 360 N
𝐹3𝑦 = 480 N
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Example -9-
Determine the magnitude and direction of the resultant force.
Ans:
𝐹𝑅 = 567 N
𝜃 = 38.1°
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Example -10-
Determine the magnitude of the resultant force acting on the corbel and its
direction θ measured counterclockwise from the x axis.
Ans:
𝐹𝑅 = 1254 lb
𝛷 = 78.68°
𝜃 = 180 + 𝛷 = 259°
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Example -11-
If the resultant force acting on the bracket is to be 750 N directed along the
positive x axis, determine the magnitude of 𝐅 and its direction θ.
Ans:
𝜃 = 31.76°
𝐹 = 236 N
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Example -12-
Determine the magnitude of the resultant force and its direction θ measured
counterclockwise from the positive x axis.
Ans:
𝜃 = 39.8°
𝐹 = 31.2 kN
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References :
1. Engineering Mechanics-Statics by R.C.-Hibbeler, 12th Edition.
2. Lecture Notes and Exercises on Statics, by Dr. Abdulwahab Amrani.