 # ENGINEERING MECHANICS STATICS & DYNAMICS

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ENGINEERING MECHANICS STATICS & DYNAMICS
University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS Instructor: Eng. Eman Al.Swaity Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

Students will be able to : a) Resolve a 2-D vector into components
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 Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

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. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

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” Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

VECTOR OPERATIONS Scalar Multiplication and Division Lecture 2
Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE
Triangle method (always ‘tip to tail’): How do you subtract a vector? How can you add more than two concurrent vectors graphically ? Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

RESOLUTION OF A VECTOR “Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

CARTESIAN VECTOR NOTATION
We ‘ resolve’ vectors into components using the x and y axes system Each component of the vector is shown as a magnitude and a direction. 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. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

F = Fx i + Fy j or F' = F'x i + F'y j
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. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

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. Step 3 is to find the magnitude and angle of the resultant vector. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

Example of this process,
Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

You can also represent a 2-D vector with a magnitude and angle.
Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

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 F3 F3 F1 F4 F2 F2 F1 Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

There are four concurrent cable forces acting on the bracket. How do you determine the resultant force acting on the bracket ? Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

Resultant of Colinear forces Forces in same direction
RESULTANT FORCE 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 forces Forces in same direction The resultant force (R) would be R= = 350 N Forces in deferent direction R= = 50 N 200 N 150 N R=350N 200 N 150 N R=50N Lecture 2 Chapter 2 : Force Vector

Resultant of Colinear forces
RESULTANT FORCE Resultant of Colinear forces Fined the resultant force R= = -100N the minus means R direction is at –x direction Fined the resultant force Rx and Ry Rx = = 50N Ry = = 350N Summery Rx = F1x + F2x + F3x = Fx Ry = F1y + F2y + F3y = Fy 100N 150N 270N 220N + x + y 100N 270 N 200 N 120 N 160 N 100 N 250 N 360N Ry = 350N + x + y Rx = 50N Lecture 2 Chapter 2 : Force Vector

Resultant of two concurrent forces
RESULTANT FORCE 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 R F1=40N Lecture 2 Chapter 2 : Force Vector

Resultant of two concurrent forces
RESULTANT FORCE 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 R=114 75N 75N 50N 50N Lecture 2 Chapter 2 : Force Vector

Given: Three concurrent forces acting on a bracket.
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. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

EXAMPLE (continued) F1 = { 15 sin 40° i + 15 cos 40° j } kN
= { i j } kN F2 = { -(12/13)26 i + (5/13)26 j } kN = { -24 i + 10 j } kN F3 = { 36 cos 30° i – 36 sin 30° j } kN = { i – 18 j } kN Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

Summing up all the i and j components respectively, we get,
EXAMPLE (continued) Summing up all the i and j components respectively, we get, FR = { (9.642 – ) i + ( – 18) j } kN = { i j } kN x y FR FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN  = tan-1(3.49/16.82) = 11.7° Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

Given: A hoist trolley is subjected to the three forces shown
GROUP PROBLEM SOLVING 1000 N P 2000 N 40° Given: A hoist trolley is subjected to the three forces shown Find: (a) the magnitude of force, P for which the resultant of the three forces is vertical (b) the corresponding magnitude of the resultant. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

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: Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

GROUP PROBLEM SOLVING (continued)
40° a) ∑ Fx = sin 40° +P – 2000 cos 40° = 0 P = 2000 cos 40° sin 40° = 889 N a) ∑ Fy = sin 40°– 1000 cos 40° = = N = 2052 N Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

Given: Three concurrent forces acting on a bracket
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. Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

GROUP PROBLEM SOLVING (continued)
F1 = { (4/5) 850 i - (3/5) 850 j } N = { 680 i j } N F2 = { -625 sin(30°) i cos(30°) j } N = { i j } N F3 = { -750 sin(45°) i cos(45°) j } N { i j } N Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

GROUP PROBLEM SOLVING (continued)
Summing up all the i and j components respectively, we get, FR = { (680 – – 530.3) i + (-510 – ) j }N = { i j } N y x FR FR = ((162.8)2 + (521)2) ½ = 546 N = tan–1(521/162.8) = 72.64° or From Positive x axis  = = 253 ° Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

ATTENTION QUIZ 1. Resolve F along x and y axes and write it in vector form. F = { ___________ } N A) 80 cos (30°) i sin (30°) j B) 80 sin (30°) i cos (30°) j C) 80 sin (30°) i cos (30°) j D) 80 cos (30°) i sin (30°) j 30° x y F = 80 N 2. Determine the magnitude of the resultant (F1 + F2) force in N when F1 = { 10 i j } N and F2 = { 20 i j } N . A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N Answers: 1. C 2. C Lecture 2 Chapter 2 : Force Vector Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections 2.1,2.2,2.4

ENGINEERING MECHANICS STATICS & DYNAMICS
University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS End of the Lecture Let Learning Continue Lecture 2 Chapter 2 : Force Vector