Chapter 2: Force Vectors - Civil Engineering Department...
Transcript of Chapter 2: Force Vectors - Civil Engineering Department...
Engineering Mechanics: Statics
Chapter 2: Force Vectors
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 Vectors
Addition and Subtraction of
Cartesian Vectors
Position Vectors
Force Vector Directed along a Line
Dot Product
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)...
Review of Scalars and Vectors
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 or bold such as A
– Magnitude is shown by or simply A
A
A
Scalars and Vectors
Scalars
Vectors
Characteristics
VALUE
(POSITIVE
OR
NEGATIVE)
&
UNIT
(both togather called
MAGNITUDE)
VALUE
(POSITIVE
OR
NEGATIVE)
UNIT
(both togather called
MAGNITUDE)
&
DIRECTION
Application
mass, speed, volume, time
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
aA
Multiplication and Division of a Vector by a Scalar
- Product of vector A and scalar a = aA
- Magnitude =
- 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 Addition
Parallelogram Law
Triangular Construction
Vector Operations
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 Addition of Forces
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
Resolution of a vector is breaking up a vector into its components.
It is the reverse application of the parallelogram law.
Vector Operations
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
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.
Procedure for Analysis:
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
Vector Addition of Forces
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°
Half portion of the parallelogram
Solution
Trigonometry Law of Cosine
N
NNNNFR
6.212
4226.0 300002250010000
115cos 150 100 2 150 10022
-
Vector Addition of Forces
Solution
Trigonometry Law of Sines
8.39
9063.0 6.212
150sin
sin115
N 212.6
θ sin
N 150
θ
θN
N
Vector Addition of Forces
Solution
Trigonometry Direction Φ of FR measured from the horizontal
354.761
1539.7613
=
+=ϕ
Vector Addition of Forces
q
q
p
p
365 N
β 38.5°
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.
Example:
Vector Operations
q
q
p
p
β 38.5°
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.
Solution:
Vector Operations
Establish a paralelogram or a triangle. Given are: resultant and component along pp axis.
q
q
p
p
β 38.5°
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.
Solution:
Vector Operations
38.5°
Establish a paralelogram or a triangle. Given are: resultant and component along pp axis.
q
q
p
p
β 38.5°
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.
Solution:
Vector Operations
38.5°
β
Establish a paralelogram or a triangle. Given are: resultant and component along pp axis.
∑FFR =
Addition of a System of Concurrent Coplanar (2D) Forces
Concurrent at a point:
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 )
Cartesian Vector Notation F = Fx 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
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.
F = Fx i + Fy j
x and y axes are always perpendicular to each other. Together, they can be
directed at any inclination.
Cartesian Vector Notation
jFiFF yx
+=
F
Vector components may be expressed as
products of the unit vectors with the scalar
magnitudes of the vector components.
Fx and Fy are referred to as the scalar
components of
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
RyRxR FFF 22 +=
Addition of a System of Concurrent Coplanar (2D) Forces
Coplanar Force Resultants - Direction angle θ (orientation of the force)
can be found by trigonometry
Rx
Ry
F
F1tanθ =
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
Step 1: to resolve each force into
its x-y components.
Step 3: find the magnitude and
the angle of the resultant vector.
Step 2: to add all the x components
together and add all the y
components together.
These two totals become the resultant
vector.
Cartesian Vector Method (solution by three steps)
Addition of a System of Concurrent Coplanar (2D) Forces
Example: Add (sum) the given three
coplanar forces
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 1
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
Step 2
( )( )↑→
+
+
Addition of a System of Concurrent Coplanar (2D) Forces
Coplanar Force Resultants FRx = (F1x - F2x + F3x)
FRy = (F1y + F2y – F3y)
In all cases,
FRx = ∑Fx
FRy = ∑Fy
* Take note of the sign conventions
Step 3
Scalar Notation:
Rx
Ry
RyRxR
F
F
FFF
1
22
tanθ =
+=
Addition of a System of Concurrent Coplanar (2D) Forces
The magnitude and the angle of the resultant vector
WAYS OF FINDING THE COMPONENTS OF FORCES IN 2D
1- By an ANGLE measured from any axis
(Cosine of this angle gives the component of that axis)
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.
Addition of a System of Concurrent Coplanar (2D) Forces
(18, 10)
(0, 0)
10 cm
18 cm
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
N788.324
N5
3200N35cos250F
:FF
N606.416
N5
4200N35sin250N400F
:FF
Ry
yRy
Rx
xRx
Σ
Σ
35˚
324.788 N
416.606 N
Addition of a System of Concurrent Coplanar (2D) Forces
Solution
Resultant Force
From vector addition,
Direction angle θ is
N
NNFR
250.528
788.324 606.41622
0598.1429402.37180
9402.37
N 606.416
N 788.324tan 1
θ
324.788 N
416.616 N
Addition of a System of Concurrent Coplanar (2D) Forces
142.0598°
Example:
Addition of a System of Concurrent Coplanar (2D) Forces
Example:
Addition of a System of Concurrent Coplanar (2D) Forces
R
R