CE 201 - Statics Lecture 12. MOMENT OF A FORCE ABOUT A SPECIFIED AXIS Scalar Analysis Mo = (20 N) (0.5 m) = 10 N.m (F tends to turn about the Ob axis)

Slides:



Advertisements
Similar presentations
Forces and Moments MET 2214 Ok. Lets get started.
Advertisements

STATICS OF RIGID BODIES
CE Statics Lecture 3.
CE Statics Lecture 15. Moment of a Force on a Rigid Body If a force is to be moved from one point to another, then the external effects of the force.
Exam 1, Fall 2014 CE 2200.
Forces and Moments MET 2214 Ok. Lets get started.
MOMENT OF A FORCE (SCALAR FORMULATION), CROSS PRODUCT, MOMENT OF A FORCE (VECTOR FORMULATION), & PRINCIPLE OF MOMENTS Today’s Objectives : Students will.
CE Statics Lecture 6.
CE Statics Lecture 10. FORCE SYSTEM RESULTANTS So far, we know that for a particle to be in equilibrium, the resultant of the force system acting.
READING QUIZ The moment of a force about a specified axis can be determined using ___. A) a scalar analysis only B) a vector analysis only C) either a.
MOMENT ABOUT AN AXIS Today’s Objectives:
MOMENT AND COUPLES.
Lecture 1eee3401 Chapter 2. Vector Analysis 2-2, 2-3, Vector Algebra (pp ) Scalar: has only magnitude (time, mass, distance) A,B Vector: has both.
4.6 Moment due to Force Couples
Engineering Mechanics: Statics
MOMENT ABOUT AN AXIS In-Class Activities: Check Homework Reading Quiz Applications Scalar Analysis Vector Analysis Concept Quiz Group Problem Solving Attention.
Engineering Mechanics: Statics
Moment of a force The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point.
MOMENT ABOUT AN AXIS Today’s Objectives: Students will be able to determine the moment of a force about an axis using a) scalar analysis, and b) vector.
CHAPTER TWO Force Vectors.
Engineering Fundamentals Session 9. Equilibrium A body is in Equilibrium if it moves with constant velocity. A body at rest is a special case of constant.
4.4 Principles of Moments Also known as Varignon’s Theorem
Principle of Engineering ENG2301 F Mechanics Section F Textbook: F A Foundation Course in Statics and Dynamics F Addison Wesley Longman 1997.
MOMENT ABOUT AN AXIS Today’s Objectives:
CE Statics Lecture 11. MOMENT OF A FORCE-VECTOR FORMULATION In scalar formulation, we have seen that the moment of a force F about point O or a.
Cont. ERT 146 Engineering Mechanics STATIC. 4.4 Principles of Moments Also known as Varignon ’ s Theorem “ Moment of a force about a point is equal to.
MOMENT ABOUT AN AXIS Today’s Objectives: Students will be able to determine the moment of a force about an axis using a) scalar analysis, and b) vector.
CE Statics Chapter 5 – Lectures 2 and 3. EQUATIONS OF EQUILIBRIUM The body is subjected to a system of forces which lies in the x-y plane. From.
Vectors for Calculus-Based Physics AP Physics C. A Vector …  … is a quantity that has a magnitude (size) AND a direction.  …can be in one-dimension,
Force System Resultants 4 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson Education South Asia Pte Ltd.
MEC 0011 Statics Lecture 4 Prof. Sanghee Kim Fall_ 2012.
Copyright © 2010 Pearson Education South Asia Pte Ltd
Vectors for Calculus-Based Physics
Forces and Moments Mo = F x d What is a moment?
MOMENT OF A FORCE (SCALAR FORMULATION), CROSS PRODUCT, MOMENT OF A FORCE (VECTOR FORMULATION), & PRINCIPLE OF MOMENTS Objectives : a) understand and define.
Rigid Bodies: Equivalent Systems of Forces
Forces Classification of Forces Force System
MOMENT ABOUT AN AXIS (Section 4.5)
MOMENT ABOUT AN AXIS Today’s Objectives:
MOMENT ABOUT AN AXIS Today’s Objectives:
4.5 MOMENT ABOUT AN AXIS - Scalar analysis
MOMENT ABOUT AN AXIS (Section 4.5)
MOMENT ABOUT AN AXIS Today’s Objectives:
Moments of the forces Mo = F x d A moment is a turning force.
VECTORS APPLICATIONS NHAA/IMK/UNIMAP.
Lecture 4.
Engineering Mechanics : STATICS
Vectors for Calculus-Based Physics
MOMENT OF A COUPLE Today’s Objectives: Students will be able to
Copyright © 2010 Pearson Education South Asia Pte Ltd
MOMENT OF A FORCE ABOUT A POINT
The moment of F about O is defined as
MOMENT ABOUT AN AXIS Today’s Objectives:
MOMENT ABOUT AN AXIS Today’s Objectives:
MOMENT OF A FORCE SCALAR FORMULATION, CROSS PRODUCT, MOMENT OF A FORCE VECTOR FORMULATION, & PRINCIPLE OF MOMENTS Today’s Objectives : Students will be.
MOMENT OF A FORCE SCALAR FORMULATION, CROSS PRODUCT, MOMENT OF A FORCE VECTOR FORMULATION, & PRINCIPLE OF MOMENTS Today’s Objectives : Students will be.
Statics Course Code: CIVL211 Dr. Aeid A. Abdulrazeg
Chapter Objectives Concept of moment of a force in two and three dimensions Method for finding the moment of a force about a specified axis. Define the.
KNUS&T Kumasi-Ghana Instructor: Dr. Joshua Ampofo
CE Statics Lecture 9.
ENGINEERING MECHANICS
ENGR 107 – Introduction to Engineering
MOMENT OF A FORCE (SCALAR FORMULATION), CROSS PRODUCT, MOMENT OF A FORCE (VECTOR FORMULATION), & PRINCIPLE OF MOMENTS Today’s Objectives : Students will.
Angular Kinetics: Torques
MOMENT OF A FORCE (Section 4.1)
CE Statics Lecture 13.
CE Statics Lecture 8.
Su-Jin Kim Gyeongsang National University Fall 2017
Presentation transcript:

CE Statics Lecture 12

MOMENT OF A FORCE ABOUT A SPECIFIED AXIS Scalar Analysis Mo = (20 N) (0.5 m) = 10 N.m (F tends to turn about the Ob axis) How can we determine the component of Mo about the y-axis (My)? If  is known, then My = Mo cos (  ) It is clear that in order to find My Determine Mo Resolve Mo along the y-axis A F = 20 N 0.5 m O MOMO dy = 0.4 m b  dx = 0.3 m x y z MyMy

Generally, to find My directly, it is necessary to find the perpendicular distance from the line of action of the force to the y-axis (da), then: Ma = F (da) where: a is the aa-axis da is the perpendicular distance from F line of action to aa- axis Note: F will not have a moment about an axis if F is parallel to that axis or passes through it.

Vector Analysis The two steps performed with scalar analysis can be performed with vector analysis, therefore: Mo = rA  F = (0.3 i j)  (-20 k) = (-8 i + 6 j) N.m The component of Mo along the y- axis is determined from the dot product, then: My = Mo. u = (-8 i + 6 j). j = 6 N.m A F = (- 20 k) N r OA O MOMO 0.4 m b  0.3 m x y z MyMy u = j

Generally, if a body is subjected to force F acting at point (A), what is Ma?? A F r OA O M O = r  F b a b a Ma 

(1) Calculate Mo ( Mo = r OA  F ) The moment axis is perpendicular to the plane containing F and r (say bb) (2) Ma is the component along aa, where its magnitude is found from the dot product, ( Ma = Mo cos (  ) = Mo. u a ), where u a is the unit vector defining the direction of aa, then: Ma = (r  F). u a Since the dot product is commutative (A. B = B. A), then: Ma = u a. (r  F)

The combination of the dot and cross products is called the triple scalar product. i j k Ma = (u ax i + u ay j + u az k ). rx ry rz Fx Fy Fz Or simply,

u ax u ay u az Ma = u a. (r  F) = rx ry rz Fx Fy Fz If Ma is +ve, then it has the same direction as ua and if Ma is –ve, then it is in the opposite direction of u a. Whence Ma was calculated, Ma can be expressed in Cartesian vector form by: + Ma = Ma u a = [ u a. (r  F) ] u a The resultant moment of a system of forces about the axis is: Ma =  [ u a. (r  F) ] = u a.  (r  F)