Taylor’s University
Engineering Statics (ENG 60103)
By
Dr. Azadeh Ghadimi
PhD (Engineering)
MEng. Aerospace/Aerodynamic Eng.
BEng. Heat and fluid Mech. Eng.
Office @ D9B10
Email: azadeh.ghadimi@taylors.edu.my
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1
Rules of class attendance!!
• Maximum late entry to the class is 10 minutes.
• Handphones should be silent during the class time.
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Contact hours and evaluation
• Mode of delivery
Lectures: 2 hours per week
Tutorials: 1.5 hours per week (average)
Labs:
0.6 hours per week (average)
• Assessments:
40% Continuous Assessment
60% Final Examination (Hurdle*)
* Hurdle is defined as follows:
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Learning outcomes:
• LO 1. Evaluate the internal forces and external reactions
on a pin-jointed truss.
• LO 2. Calculate the bending and shear stresses and
deflections of statically determinate beams.
• LO 3. Show the likely bending moment and shear force
diagrams of laterally loaded beams.
• LO 4. Utilise the free-body diagram as a means of
identifying loads, and apply correctly the conditions of
equilibrium and the principle of compatibility to a
structure.
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Assessment methods
Distribution
Test 1
Test 2
Lab report
Final Examination
(%)
10
10
20
60
Total
100
LO 1
X
X
LO 2
LO 3
X
X
X
X
LO 4
X
X
X
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Assessment schedule
Assessment
Methods
1 2 3 4 5
Test 1
Test 2
Lab report
Final
Examination
6
Week No.
7 8 9 10 11 12
13 14
X
X
Two weeks after conducting a lab session
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Reference material
1. J.L. Meriam and L.G. Kraige. “Engineering Mechanics,
Vol.1, Statics.” 7th edition. John Wiley and Sons, 2011.
2. F. P. Beer, E. R. Johnston, D. F. Mazurek. "Vector
Mechanics for Engineers", 10th or latest edition, McGraw
Hill, 2013.
3. R. C. Hibbeler. “Engineering Mechanics – Statics.” 13th
Edition. Prentice Hall, 2013
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Engineering Mechanics:
Statics in SI Units, 12e
2
Force Vectors
Part 1
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Chapter Objectives
• Parallelogram Law
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Chapter Outline
1.
2.
3.
4.
5.
Scalars and Vectors
Vector Operations
Vector Addition of Forces
Addition of a System of Coplanar Forces
Appendix (Newton’s laws)
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2.1 Scalars and Vectors
• Scalar
– A scalar quantity can be completely specified by its
magnitude.
– Indicated by letters in italic such as A
e.g. Mass, volume and length
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2.1 Scalars and Vectors
• Vector
– A quantity that has magnitude and direction
e.g. Position, force and moment

– Represent by a letter with an arrow over it, A

– Magnitude is designated as A
– In this subject, vector is presented as A and its
magnitude (positive quantity) is italicized as A.
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2.1 Scalars and Vectors
• Vector
• The length of the arrow represents the magnitude of the
vector.
• The angle θ between the vector and a fixed axis defines the
direction of its line of action.
• The head or tip of the arrow indicates the sense of direction.
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2.2 Vector Operations
• Multiplication and Division of a 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 direction (sense) of the vector.
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2.2 Vector Operations
• Vector Addition
- Addition of two vectors A and B gives a resultant
vector R by the parallelogram law
- Result R can also be found by triangle construction
- Commutative e.g. R = A + B = B + A
- Special case: Vectors A and B are collinear (both
have the same line of action)
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2.2 Vector Operations
• Vector Subtraction
- Special case of addition
e.g. R’ = A – B = A + ( - B )
- Rules of Vector Addition Applies
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2.3 Vector Addition of Forces
Finding a Resultant Force
• Parallelogram law is carried out to find the resultant
force
• Resultant,
FR = ( F1 + F2 )
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2.3 Vector Addition of Forces
Procedure for Analysis
• Parallelogram Law
– Make a sketch using the parallelogram law
– 2 components forces add to form the resultant force
– Resultant force is shown by the diagonal of the
parallelogram
– The components is shown by the sides of the
parallelogram
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2.3 Vector Addition of Forces
Procedure for Analysis
• Trigonometry
– Redraw half portion of the parallelogram
– Magnitude of the resultant force can be determined by
the law of cosines
– Direction of the resultant force can be determined by the
law of sines
– Magnitude of the two components can be determined by the
law of sines
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Example 2.1
The screw eye is subjected to two forces, F1 and F2.
Determine the magnitude and direction of the resultant
force.
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Solution
Parallelogram Law
Unknown: magnitude of FR and angle θ
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Solution
Trigonometry
Law of Cosines
100 N 2  150 N 2  2100 N 150 N cos 115
10000  22500  30000 0.4226  212.6 N  213 N
FR 

Law of Sines
150 N 212.6 N

sin 
sin 115
150 N
0.9063
sin  
212.6 N
  39.8
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Solution
Trigonometry
Direction Φ of FR measured from the horizontal
  39.8  15
 54.8 
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2.4 Addition of a System of Coplanar Forces
When a force is resolved into two components along
the x and y axes, the components are called
rectangular components.
• Scalar Notation
– The rectangular components of force F are found using
the parallelogram law, so that:
F  Fx  Fy
Fx  F cos  and Fy  F sin 
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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 components.
– Unit vectors i and j have dimensionless magnitude of
unity ( = 1 )
– Magnitude is always a positive quantity, represented by
scalars Fx and Fy
F  Fx i  Fy j
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2.4 Addition of a System of Coplanar Forces
• Coplanar Force Resultants
To determine resultant of several coplanar forces:
– Resolve force into x and y components
– Addition of the respective components using scalar
algebra
– Resultant force is found using the parallelogram law
– Cartesian vector notation:
F1  F1x i  F1 y j
F2   F2 x i  F2 y j
F3  F3 x i  F3 y j
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2.4 Addition of a System of Coplanar Forces
• Coplanar Force Resultants
– Vector resultant is therefore
FR  F1  F2  F3
 FRx i  FRy j
– If scalar notation are used
FRx  F1x  F2 x  F3 x
FRy  F1 y  F2 y  F3 y
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2.4 Addition of a System of Coplanar Forces
• Coplanar Force Resultants
– In all cases we have
FRx   Fx
FRy   Fy
* Take note of sign conventions
– Magnitude of FR can be found by Pythagorean Theorem
FR  F  F
2
Rx
2
Ry
and   tan
-1
FRy
FRx
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Example 2.5
Determine x and y components of F1 and F2 acting on the
boom. Express each force as a Cartesian vector.
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Solution
Scalar Notation
F1x  200 sin 30 N  100 N  100 N 
F1 y  200 cos 30 N  173 N  173 N 
Hence, from the slope triangle, we have
5
  tan 1  
 12 
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Solution
By similar triangles we have
 12 
F2 x  260   240 N
 13 
5
F2 y  260   100 N
 13 
Scalar Notation: F  240 N 
2x
F2 y  100 N  100 N 
Cartesian Vector Notation: F1   100i  173j N
F2  240i  100 j N
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Example 2.6
The link is subjected to two forces F1 and F2. Determine the
magnitude and orientation of the resultant force.
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Solution I
Scalar Notation:
FRx  Fx :
FRx  600 cos 30 N  400 sin 45 N
 236.8N 
FRy  Fy :
FRy  600 sin 30 N  400 cos 45 N
 582.8N 
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Solution I
Resultant Force
FR 
236.8N 2  582.8N 2
 629 N
From vector addition, direction angle θ is
 582.8N 

 236.8N 
  tan 1 
 67.9
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Solution II
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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Acknowledgement
• Dr.Hosseini, School of Engineering Taylor’s university,
lakeside campus, Engineering Static, 2012
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