Download F - Civil Engineering Department

Chapter 2:
Force Vectors
Engineering Mechanics: Statics
Objectives

To show how to add forces and resolve them
into components using the Parallelogram Law.

To express force and position in Cartesian
vector form and explain how to determine the
vector’s magnitude and direction.
Chapter Outline
 Scalars
and Vectors
 Vector
Operations
 Vector
Addition of Forces
 Addition
of System of Coplanar Forces
Vector operation
 Cartesian
 Addition
Vectors
and Subtraction of
Cartesian Vectors
 Position
 Force
 Dot
Vectors
Vector Directed along a Line
Product
Review of Scalars and Vectors
Number
Any positive or negative value.
e.g.: 5, -3.07, π...
Scalar
A quantity characterized by a positive or
negative number and has a unit.
e.g.: mass (5.1 kg), volume (0.028 m3)...
Scalars and Vectors
Vector
A quantity that has both magnitude and direction
measured from a given axis.
e.g.: Position (3.4 km North), force (8.73 N
towards right side), ...
– Represented by a letter with an arrow over it

such as A or bold such as A

– Magnitude is shown by A or simply A
Scalars and Vectors
Scalars
Characteristics
Application
VALUE
(POSITIVE
OR
NEGATIVE)
&
UNIT
(both togather called
MAGNITUDE)
mass, speed, volume, time
Vectors
VALUE
(POSITIVE
OR
NEGATIVE)
UNIT
(both togather called
MAGNITUDE)
&
DIRECTION
velocity, acceleration force
Scalars and Vectors
Example
Magnitude of Vector = 4 units
Direction of Vector = 20° measured
counterclockwise from the horizontal axis
Sense of Vector = Upward and to the right
The point O is called tail
of the vector and the point
P is called the tip or head
Vector Operations

Multiplication and Division of a Vector by a Scalar
- Product of vector A and scalar a = aA
- Magnitude = aA
- If a is positive (+), sense of aA is the same as
sense of A
- If a is negative (-), sense of
aA, it is opposite to the
sense of A
Vector Operations
Multiplication and Division of a Vector
by a Scalar
- Negative of a vector is found by multiplying the
vector by ( -1 )
- Law of multiplication applies
e.g.: A/a = ( 1/a ) A, if a≠0
Vector Operations
Vector Addition
Addition of two vectors A and B gives a
resultant vector R by the parallelogram law
special case of parallelogram law is triangle
construction (see the next slide)
Commutative rule
e.g.: R = A + B = B + A
Vector Operations
Vector Addition
Parallelogram Law
Triangular Construction
Vector Addition of Forces
Parallelogram Law description
- To resolve a force into components along
two axes directed from the tail of the force
- Start at the head, constructing lines parallel
to the axes
Vector Operations
 Vector
Addition
- Special case: Vectors A and B are
collinear (both have the same line of
action). Triangular shape does not form.
Vector Operations
Vector Subtraction
Special case of addition
e.g.: R’ = A – B = A + ( - B )
Rules of Vector Addition Applies
Vector Operations
Resolution of Vector
Any vector can be resolved into two
components by the parallelogram law
The two components A and B are drawn
such that they extend from the tail of R
Vector Operations
Resolution of a vector is breaking up a vector
into its components.
It is the reverse application of the parallelogram law.
Vector Addition of Forces
When two or more forces are added,
successive applications of the
parallelogram law is carried out to find
the resultant
e.g.: Forces F1, F2 and F3 acts at a point O
- First, find resultant of
F1 + F2
- Resultant,
FR = ( F1 + F2 ) + F3
Vector Addition of Forces
Method 1: (graphical)
parallelogram Law OR
Establishing triangles
Method 2:
Trigonometry
Procedure for Analysis:
Parallelogram Law
- Make a sketch showing the vector addition using the
parallelogram law
- Two “component” forces add according to the
parallelogram law yielding a resultant force.
- Resultant force is shown by the diagonal of the
parallelogram
- The components are the sides of the parallelogram
- Label all the known and unknown force
magnitudes and angles.
- Identify the two unknowns.
Procedure for Analysis:
Trigonometry
- Redraw half portion of the parallelogram
- Magnitude of the resultant force can be determined
by the law of COSINE
- Direction of the resultant force can be determined
by the law of SINE
Procedure for Analysis:
Magnitude of the two components can be
determined by the law of COSINE.
Direction of the two components can be
determined by the law of SINE.
Vector Addition of Forces
Example:
The screw eye is subjected to two forces F1
and F2. Determine the
magnitude and direction
of the resultant force.
Vector Addition of Forces
Solution
 Parallelogram Law
Unknown:
 magnitude of FR and
 angle θ
Vector Addition of Forces
Half portion of the parallelogram
FR= Magnitude of the resultant force
(determined by the law of cosine)
θ= angle (determined by the law of sine)
ϕ = direction of the resultant force = θ +15°
Vector Addition of Forces
Solution
Trigonometry
Law of Cosine
FR 
100 N 2  150 N 2 -
2 100 N 150 N  cos 115
 10000  22500  30000  0.4226
 212.6 N
Vector Addition of Forces
Solution
Trigonometry
Law of Sines
150 N 212.6 N

sin θ sin115 
150 N
0.9063
sin θ 
212.6 N
θ  39.8 
Vector Addition of Forces
Solution
Trigonometry
Direction Φ of FR measured from the horizontal
ϕ
= 39.7613 +15
= 54.7613
Vector Operations
Example:
The p and q axes are not perpendicular to each other. If the component of 365 N force along pp
axis is 237.5 N, determine the other component along qq axis and the unknown angle β.
q
p
365 N
38.5°
β
p
q
Vector Operations
Solution:
The p and q axes are not perpendicular to each other. If the component of 365 N force along pp
axis is 237.5 N, determine the other component along qq axis and the unknown angle β.
Establish a paralelogram or a triangle. Given are: resultant and component along pp axis.
q
p
38.5°
β
p
q
Vector Operations
Solution:
The p and q axes are not perpendicular to each other. If the component of 365 N force along pp
axis is 237.5 N, determine the other component along qq axis and the unknown angle β.
Establish a paralelogram or a triangle. Given are: resultant and component along pp axis.
q
p
38.5°
β
38.5°
p
q
Vector Operations
Solution:
The p and q axes are not perpendicular to each other. If the component of 365 N force along pp
axis is 237.5 N, determine the other component along qq axis and the unknown angle β.
Establish a paralelogram or a triangle. Given are: resultant and component along pp axis.
q
p
38.5°
β
β
38.5°
p
q
Addition of a System of
Concurrent Coplanar (2D) Forces
Concurrent at a point:
FR = ∑F
Addition of a System of
Concurrent Coplanar (2D) 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
magnitude of unity ( = 1 )
Addition of a System of
Concurrent Coplanar (2D) Forces
 Cartesian
Vector Notation
F = F x i + Fy j
F’ = F’x i + F’y (-j)
F’ = F’x i – F’y j
Addition of a System of
Concurrent Coplanar (2D) Forces
Resolve vectors into components using x and y
axes system.
Each component of the vector will have a
magnitude and a direction.
The directions are based on the x and y axes. The
“unit vectors” i and j are used to designate x and y
axes.
x and y axes are always perpendicular to each
other. Together, they can be
directed at any inclination.
F = Fx i + Fy j
Cartesian Vector Notation
Vector components may be expressed as
products of the unit vectors with the scalar
magnitudes of the vector components.



F = Fx i +Fy j
Fx and Fy are referred
to
as
the
scalar

components of F
Addition of a System of
Concurrent Coplanar (2D) 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
Addition of a System of
Concurrent Coplanar (2D) Forces
Coplanar Force Resultants
- Positive scalars = sense of direction
along the positive coordinate axes
- Negative scalars = sense of direction
along the negative coordinate axes
- Magnitude of FR can be found by
Pythagorean Theorem
FR = F
2
Rx
+F
2
Ry
Addition of a System of
Concurrent Coplanar (2D) Forces
Coplanar Force Resultants
- Direction angle θ (orientation of the force)
can be found by trigonometry
θ = tan
1
FRy
FRx
Addition of a System of
Concurrent Coplanar (2D) Forces
Addition of a System of
Concurrent Coplanar (2D) Forces
Example: Add (sum) the given three
coplanar forces
Addition of a System of
Concurrent Coplanar (2D) Forces
Cartesian Vector Method
(solution by three steps)
Step 1: to resolve each force into
its x-y components.
Step 2: to add all the x components
together and add all the y
components together.
These two totals become the resultant
vector.
Step 3: find the magnitude and
the angle of the resultant vector.
Addition of a System of
Concurrent Coplanar (2D) Forces
Example: Add (sum) the given three
coplanar forces
Step 1
Cartesian vector notation
F1 = F1x i + F1y j
F2 = - F2x i + F2y j
F3 = F3x i – F3y j
Addition of a System of
Concurrent Coplanar (2D) Forces
Step 2
Vector resultant is therefore
FR = F1 + F2 + F3
= F1x i + F1y j - F2x i + F2y j + F3x i – F3yj
= (F1x - F2x + F3x)i + (F1y + F2y – F3y)j
= (FRx)i + (FRy)j
Addition of a System of
Concurrent Coplanar (2D) Forces
Coplanar Force Resultants
(→ ) FRx = (F1x - F2x + F3x)
(+↑) F = (F + F – F )
+
Ry
1y
2y
3y
In all cases,
FRx = ∑Fx
FRy = ∑Fy
* Take note of the sign conventions
Addition of a System of
Concurrent Coplanar (2D) Forces
Step 3
The magnitude and the angle of the resultant
vector
Scalar Notation:
2
Rx
2
Ry
FR = F + F
θ = tan
1
FRy
FRx
Addition of a System of
Concurrent Coplanar (2D) Forces
WAYS OF FINDING THE COMPONENTS
OF FORCES IN 2D
By an ANGLE measured from any axis
(Cosine of this angle gives the component
of that axis)
1-
Addition of a System of
Concurrent Coplanar (2D) Forces
WAYS OF FINDING THE COMPONENTS
OF FORCES IN 2D
2- By right angle TRIANGLE of known sides
Addition of a System of
Concurrent Coplanar (2D) Forces
WAYS OF FINDING THE COMPONENTS
OF FORCES IN 2D
3- By COORDINATES or known lengths.
(18, 10)
10 cm
(0, 0)
18 cm
Addition of a System of
Concurrent Coplanar (2D) Forces
Example:
The end of the boom O is subjected to three
concurrent and coplanar forces. Determine
the magnitude and orientation of the
resultant force.
3
35˚
Addition of a System of
Concurrent Coplanar (2D) Forces
Solution
35˚
FRx  Σ Fx :
4
FRx   400 N  250 sin 35 N  200  N
5
  416.606 N
FRy  Σ Fy :
3
FRy  250 cos 35 N  200  N
5
 324.788 N
324.788 N
416.606 N
Addition of a System of
Concurrent Coplanar (2D) Forces
Solution
Resultant Force
2
2
FR   416.606 N   324.788 N 
 528.250 N
From vector addition,
Direction angle θ is
 324.788 N 
θ  tan 1 

  416.606 N 
 37.9402

180  37.9402  142.0598
324.788 N
142.0598°
416.616 N
Addition of a System of
Concurrent Coplanar (2D) Forces
Example:
Addition of a System of
Concurrent Coplanar (2D) Forces
Example:
R
R