engineering mechanics statics & dynamics instructor: eng. eman al.swaity university of palestine...

ENGINEERING MECHANICS STATICS & DYNAMICS

Instructor: Eng. Eman Al.Swaity

University of PalestineCollege of Engineering & Urban Planning

Chapter 2 : Force VectorLecture 2

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


BASIC TERMS


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.

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”


VECTOR OPERATIONS

Scalar Multiplication

and Division


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 ?


“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


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.


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.


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.


Example of this process,


You can also represent a 2-D vector with a magnitude and angle.


TYPE OF FORCE SYSTEMS


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

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 ?


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


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


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


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


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.


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


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°


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.


GROUP PROBLEM SOLVING (continued)

Plan:



c) The resultant being vertical means that the horizontal components sum is zero.

Solution:


a) ∑ Fx = 1000 sin 40° +P – 2000 cos 40° = 0

P = 2000 cos 40° - 1000 sin 40° = 889 N


1000 N

P

2000 N

40°40°

a) ∑ Fy = -2000 sin 40°– 1000 cos 40°

= -1285.6-766 = -2052 N = 2052 N


GROUP PROBLEM SOLVING

Given: Three concurrent forces acting on a bracket

Find: The magnitude and angle of the resultant force.

Plan:



c) Find magnitude and angle from the resultant components.


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




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


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



ENGINEERING MECHANICS STATICS & DYNAMICS

University of PalestineCollege of Engineering & Urban Planning

engineering mechanics statics & dynamics instructor: eng. eman al.swaity university of palestine...

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