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ENGINEERING MECHANICS STATICS & DYNAMICS
Instructor: Eng. Eman Al.Swaity
University of PalestineCollege of Engineering & Urban Planning
Chapter 2 : Force VectorLecture 2
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2–D VECTOR ADDITION Objective:
Students will be able to :
a) Resolve a 2-D vector into components
b) Add 2-D vectors using Cartesian vector notations.
In-Class activities:• Application of adding forces• Parallelogram law • Resolution of a vector using Cartesian vector notation (CVN) • Addition using CVN
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BASIC TERMS
Chapter 2 : Force VectorLecture 2
Vector: A quantity that has both a magnitude & direction, & obeys the parallelogram law of addition. Force & velocity are vectors
Scalars: A quantity characterized by a positive or negative number. Mass, length are scalars
External force: forces external to a body can be either applied forces or reactive forces.
Internal forces: the resistance to deformation, or change of shape, exerted by the material of a body.
Transmissibility of a Force: A force may be considered as acting at any point on its line of action so long as the direction and magnitude are unchanged.
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SCALARS AND VECTORS (Section 2.1)
Scalars Vectors
Examples: mass, volume force, velocity
Characteristics: It has a magnitude It has a magnitude
and direction
Addition rule: Simple arithmetic Parallelogram law
Special Notation: None Bold font, a line, an
arrow or a “carrot”
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VECTOR OPERATIONS
Scalar Multiplication
and Division
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VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE
Parallelogram Law:
Triangle method (always ‘tip to tail’):
How do you subtract a vector? How can you add more than two concurrent vectors graphically ?
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“Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse.
RESOLUTION OF A VECTOR
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CARTESIAN VECTOR NOTATION
• Each component of the vector is shown as a magnitude and a direction.
• We ‘ resolve’ vectors into components using the x and y axes system
• The directions are based on the x and y axes. We use the “unit vectors” i and j to designate the x and y axes.
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For example,
F = Fx i + Fy j or F' = F'x i + F'y j
The x and y axes are always perpendicular to each other. Together,they can be directed at any inclination.
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ADDITION OF SEVERAL VECTORS
Step 3 is to find the magnitude and angle of the resultant vector.
• Step 1 is to resolve each force into its components
• Step 2 is to add all the x components together and add all the y components together. These two totals become the resultant vector.
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Example of this process,
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You can also represent a 2-D vector with a magnitude and angle.
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TYPE OF FORCE SYSTEMS
Chapter 2 : Force VectorLecture 2
Concurrent forces The lines of action of all forces
pass through the common point
Coplanar forces The line of action of all forces lie
in the same plan
Colinear forces Two or more forces act on body
along the same straight line
F1
F2
F3
F4
F1
F2
F3
F1F2
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APPLICATION OF VECTOR ADDITION
There are four concurrent cable forces acting on the bracket.
How do you determine the resultant force acting on the bracket ?
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Definition The resultant force is a single force which can replace two
or more forces and produce the same effect on the body as the forces.
Resultant of Colinear forcesForces in same direction
The resultant force (R) would be R=200+150 = 350 N
Forces in deferent direction
The resultant force (R) would be R=200-150 = 50 N
R=50N
RESULTANT FORCE
200 N150 N
200 N150 N
R=350N
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Resultant of Colinear forcesFined the resultant force
R=220+100-270-150 = -100N
the minus means R direction is at –x direction
Fined the resultant force Rx and Ry
Rx = 200+120-270 = 50N
Ry = 360+250-160-100 = 350N
Summery
Rx = F1x + F2x + F3x = Fx
Ry = F1y + F2y + F3y = Fy
100N
220N
100N150N
270N
360N
270 N
200 N120 N
160 N
100 N
250 N
Rx = 50N
Ry
= 3
50N
+ x
+ y
+ x
+ y
RESULTANT FORCE
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RESULTANT FORCE
R
F1=40N
Resultant of two concurrent forces Parallelogram Law (graphical
method) From figure below the R is resultant
in magnitude and direction R = 50N & = 37 by measurement
Algebraic method As R forms triangle then; F2=30N
2 240 30 50R N
30tan 0.75
4036.87
2 21 2R F F
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RESULTANT FORCE
75N
50N
Resultant of two concurrent forces
EX. Two concurrent forces 75N & 50N acting on a body 50° angle between the two forces. Magnitude and direction of the resultant are required.
Graphical method
75NR=114
50N
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EXAMPLE
Given: Three concurrent forces acting on a bracket.
Find: The magnitude and angle of the resultant force.
Plan:
a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
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EXAMPLE (continued)
F1 = { 15 sin 40° i + 15 cos 40° j } kN
= { 9.642 i + 11.49 j } kN
F2 = { -(12/13)26 i + (5/13)26 j } kN
= { -24 i + 10 j } kNF3 = { 36 cos 30° i – 36 sin 30° j } kN
= { 31.18 i – 18 j } kN
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EXAMPLE (continued)
Summing up all the i and j components respectively, we get,
FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN
= { 16.82 i + 3.49 j } kN
x
y
FRFR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN
= tan-1(3.49/16.82) = 11.7°
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GROUP PROBLEM SOLVING
Given: A hoist trolley is subjected to the three forces shown
1000 N
P
2000 N
40°40°
Find:
(a) the magnitude of force, P for which the resultant of the three forces is vertical
(b) the corresponding magnitude of the resultant.
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GROUP PROBLEM SOLVING (continued)
Plan:
a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) The resultant being vertical means that the horizontal components sum is zero.
Solution:
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a) ∑ Fx = 1000 sin 40° +P – 2000 cos 40° = 0
P = 2000 cos 40° - 1000 sin 40° = 889 N
GROUP PROBLEM SOLVING (continued)
1000 N
P
2000 N
40°40°
a) ∑ Fy = -2000 sin 40°– 1000 cos 40°
= -1285.6-766 = -2052 N = 2052 N
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GROUP PROBLEM SOLVING
Given: Three concurrent forces acting on a bracket
Find: The magnitude and angle of the resultant force.
Plan:
a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Chapter 2 : Force VectorLecture 2
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F1 = { (4/5) 850 i - (3/5) 850 j } N
= { 680 i - 510 j } N
F2 = { -625 sin(30°) i - 625 cos(30°) j } N
= { -312.5 i - 541.3 j } N F3 = { -750 sin(45°) i + 750 cos(45°) j } N
{ -530.3 i + 530.3 j } N
GROUP PROBLEM SOLVING (continued)
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GROUP PROBLEM SOLVING (continued)
Summing up all the i and j components respectively, we get,
FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N
= { - 162.8 i - 521 j } N
FR = ((162.8)2 + (521)2) ½ = 546 N
= tan–1(521/162.8) = 72.64° or
From Positive x axis = 180 + 72.64 = 253 °
y
x
FR
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ATTENTION QUIZ1. Resolve F along x and y axes and write it in
vector form. F = { ___________ } N
A) 80 cos (30°) i - 80 sin (30°) j
B) 80 sin (30°) i + 80 cos (30°) j
C) 80 sin (30°) i - 80 cos (30°) j
D) 80 cos (30°) i + 80 sin (30°) j
2. Determine the magnitude of the resultant (F1 + F2) force in N when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N .
A) 30 N B) 40 N C) 50 N
D) 60 N E) 70 N
30°
xy
F = 80 N
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Chapter 2 : Force VectorLecture 2
ENGINEERING MECHANICS STATICS & DYNAMICS
University of PalestineCollege of Engineering & Urban Planning