CIV100: Mechanics Lecture Notes Module 1: Force & Moment in 2D By: Tamer El-Diraby, PhD, PEng. Associate Prof. & Director, I2C University of Toronto Acknowledgment: Hesham Osman, PhD and Jinyue Zhang, MASc., contributed to this module You Know What to Do! 1
Module Objectives: By the end of this Module you should be able to: Be familiar with the units used in this course Know the requirements regarding significant figures in this course. Know how to resolve a force in any 2 axis. Know how to add 2 or more forces and find their resultant. Graphically Using Algebra Understand the concept of vector, their addition and products. Know how to develop free body diagram Understand the concept of moment in 2D Know how to transfer forces form one point to another in 2D Understand the concept of Equilibrium Make sure to test your knowledge of these objectives before moving to module 2. Module 2 will take these concepts to a higher level The Module in a Nutshell: Forces as Vectors The Nature of Force: A Force has a direction (line of action & sense) and a value. Forces are measured by N (not Kg) Forces are added as geometrical vectors not arithmetic (scalar) values: R = F1 + F2 + F3.. Fn Forces are added by the parallelogram law Forces are added by the triangle rule The Unit Vector is used to express the direction of force: F = Fx i + Fy j + Fz k A Figure to Remember: the Unit Vector Never Use Kg to measure Force. It is measured by N. The Unit Vector is just another way to express X, Y & Z 2
Module in a Nutshell: Forces in 2D In 2D, a force can be resolved in any two dimensions. X & Y are the most common: Fx= F Cos Θ, Fy= F Sin Θ A Force can be represented as a Vector: F = Fx i + Fy j Forces can be added using their components Rx = F1x + F2x + F3x.. Fnx (no need for vectors, as X (i) is the direction) Ry = F1y + F2y + F3y.. Fny (no need for vectors, as Y (j) is the direction) R= Rx 2 + Ry 2 ; tan Θ = Ry / Rx A Figure to Remember Graphical Addition/Resolution of Forces A Figure to Remember: Best Way to Add Forces F Fy= F Sin Θ Fx= F Cos Θ 1. We are adding vectors not values 2. FX and FY are the most used R= Rx2 + Ry2 ; tan Θ = Ry / Rx Always add forces using their X & Y components. This easier and safer! Module in a Nutshell: Moment as Vector The Nature of a Moment A Force could have a moment around an axis and, in 3D, around a point. A Figure to Remember: Moments M1=dx x Fy The moment sense is from dx to Fy Fy F Fx F dy dx M2=dy x Fx The moment sense is from dy to Fx r d M = d x F Note: M1 is positive and M2 is negative; M=dx Fy dy Fx M is a vector coming out of the paper (+ve) or going into the paper (-ve) The General rule is that Moment is a vector product of d and F (in this order). It so happens that in magnitude (only), that the value of the Moment is equal to r.f 3
Quantities, units & significant figures Quantities: Length: Meters (m) & Millimeters (mm). Mass: Kilograms (kg) Force: Newton (N), Kilonewton (kn) & Meganewton (MN). Pressure or stress: Pascal (Pa), Kilopascal (KPa) & Megapascal (MPa) 1 Pa = 1 N/m 2 1 MPa = 1 N/mm 2 Quantities, units & significant figures Significant figures: For all answers use 5 significant figures. What does that mean? 738.73687 -> 738.74 738.73287 -> 738.73 5650910 -> 5650900 0.0025214 -> 0.0025214 (It s already 5 S.F.) 5.1 -> 5.1 Always Use 5 S.F. for final answers as well as during process work (easier). 4
Types of problems encountered Coplaner forces: All forces act in the same plane Non-coplaner forces: Forces in 3-D Scalar vs. Vector Scalar A quantity characterized by a positive or negative number Scalars in statics: mass, volume, and length Vector A quantity that has both a magnitude and a direction Vectors in statics: position, force, and moment 5
Vector Operations Vector Addition We are given two vectors A and B, we need to get R = A + B. We can solve it by parallelogram law. A A R B B We can solve it by triangle construction. A R A R B B Example: Step-by-Step 6
Example: Step-by-Step Vector Operations Vector Subtraction We are given two vectors A and B, we need to get R = A - B. We solve it by parallelogram law. We also solve it by triangular construction. To Remember F A B A R A R -BB -BB 7
Addition of forces: Graphical Method Because force is a vector, they can not be added using simple arithmetic. They have to be added using geometric rules. The parallelogram law The triangle rule A Figure to Remember Graphical Addition/Resolution of Forces R A B F Fy= F Sin Θ Fx= F Cos Θ D R C 1. We are adding vectors not values 2. FX and FY are the most used The action of Force R is equivalent to that of A and B, Then R=A+B Note: if A=4 and B=3, then R=5 (not 7) Vector Operations: The axis do not have to be parallel Resolution of Vector We have a vector R. We want to resolve R into two components, and we know lines of action of the two components is a and b. a A R Resultant B b Components 8
Review Questions For a static s problem your calculations show the final answer as 12345.6 N. What will you write as your final answer? A) 12345.6 N B) 12346 N C) 12 kn D) 12.3 kn E) 12345 N Which one of the following is a scalar quantity? A) Force B) Position C) Mass D) Velocity For vector addition you have to use law. A) Newton s Second B) the arithmetic C) Pascal s D) the parallelogram Can you resolve a 2-D vector along two directions, which are not at 90º to each other? A) Yes, but not uniquely. B) No. C) Yes, uniquely. What is the unit of forces? A) kg B) N Think ahead: Multiplying Vectors Think Ahead Q: If vectors are added geometrically, can we multiply vectors? How? What is the result of multiplying vectors? A: Wait till the Moment section!! 9
An Example The screw eye shown in the figure is subjected to two forces, F 1 and F 2. Determine the magnitude and direction of the resultant force. Adding Multiple Vectors The resultant vector of multiple vectors can be obtained by applying parallelogram law successively. F 2 F 1 Using the parallelogram law to add more than two forces, as shown here, often requires extensive geometric and trigonometric calculation to determine resultant vector. F 3 10
Algebraic Method Adding multiple coplanar forces is easier if the x and y components of each force are found first. F 3 y F 1 -x x We call them algebraic scalars because they have only the scale and sense (positive/negative) -y F 2 Adding component forces in x and y direction like collinear forces, only the magnitude matters. Addition of forces: Algebraic Method To add a set of forces: Resolve each force into its X and Y components Find the resultant force in the X direction: R x Find the resultant force in the Y direction: R y Find the overall resultant R: R = (R x2 + R y2 ) This is the method we will use throughout the course! 11
Example: Step-by-Step Example 2: The formal way 12
Example 2: KISS Fundamentals Revision of Newton s 2nd law: Force = Mass x Acceleration F = mxg(where g = 9.81 m/s 2 ) 1 kg = 9.81 N Engrave in memory! Always! Concept of Pressure (Stress): 1 Pa is the pressure that is exerted by 1N on an area of 1m 2 1 Pa = 1 N/m 2 13
Spot the Trick! 120 Kg 300 N 310 N Think it is trivial? Just ask NASA http://www.ukmetrication.com/appendix3.htm http://www.tysknews.com/depts/metrication/mystery_of_orbiter_crash_solved.htm http://news.bbc.co.uk/1/hi/sci/tech/514763.stm 14
Links Vector arithmetic http://www.pa.uky.edu/~phy211/vecarith/index.html Resultant of a set of concurrent forces http://lectureonline.cl.msu.edu/~mmp/kap4/cd082.htm FBD: Objectives By the end of this section you should be able to: Understand what is a FBD. Know how to draw a FBD for several examples Know how to draw a FBD for several examples Know how to solve problems involving 2D equilibrium of bodies 15
What is a FBD? A free body diagram is a graphic, symbolic representation of the body (structure, element or piece of an element) with all connecting "parts" removed. The "parts" which have been removed are all of the real physical aspects of the structure. The body is represented by a simple line. Every "part" which has been removed is replaced by the external forces representing the internal forces present where that "part" connects with the other member in the FBD. Loads, distributed or concentrated, are removed and replaced with representational force systemsstems FBD: We have been doing that 16
FBD: Finding the line that will break the system loose Example: A body suspended from cables Where are the Forces: Cable AC, Cable BC -- cut them (they are the two cables holding the structure) 2 3 1 FBD is a tool to help us identify the forces that holds things together. One of the best tricks to do this is to cut the system loose. i.e. find a line that will break the whole structure and expose the forces. A line that will undermine the whole structure and cut it into loose parts FBD: A Beam A beam with loads 17
FBD: a Truss Spot the Mistake! FBD. 18
Cables and Pulleys NO SPRINGS! Cables 0 weight Do not stretch Has only tension ( pulling force) Force always acts in the direction of the cable Pulleys The size could be negligible, check the statement carefully Frictionless Continuous cable passing over a pulley has a constant tension For any angle ɵ Examples of FBD A simple pulley Pulleys are usually hinged at their center to allow rotation 2 reaction components will be generated at the hinge The tension in a cable going around a pulley is always constant T 35 B 120 kg 19
Example: Step-by-Step Example: Step-by-Step 20
Example: Step-by-Step Examples of FBD 21
Spot the Mistake! FBD. Examples of FBD Where should we cut this structural system? What would expose all forces (break the structure loose)? 22
Problem: Which FBD do I draw? This depends on the problem requirement For example: What are the reaction components acting at point D? The FBD is the member AD What is the tensions in cable AC and BC? The FBD is the 510kg object (or the ring C) D 35 Solved Example: Concurrent forces=equilibrium of a particle Find the tension in cables AC and BC if alpha=30 First and foremost, 23
Example If the sack at A has a weight of 20N, determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown. To Remember Solution: Exam Style 24
Example: Exam Style (cont). Spot the Trick! 25
Tips How to draw a FBD: You must have your mind set on which body you need to draw in order to solve the problem. This may sometimes be tricky. You may have to split up the object into several bodies and draw a FBD for each. Identify all forces that are directly applied (Point load, distributed load, tension force, etc..). Transform any distributed load into a single force. Identify any kind of support that is holding the body (a surface, pin, roller, fixed support, etc..) Draw all forces & supports as single forces acting on the body. Identify exactly where they act and the direction they are acting. Nice links Pulley configurations http://www.engr.umd.edu/hamlet/pulley/index.htm 26
Review Questions 1. Using this FBD of Point C (assume in equilibrium), the sum of forces in the x-direction (Σ F X ) is. (Use a sign convention of + ) F A) F 2 sin 50 20 = 0 2 B) F 2 cos 50 20 = 0 20 N 50 C) F 2 sin 50 F 1 = 0 D) F 2 cos 50 + 20 = 0 C F 1 2. Which statement is correct? D is a 20kg block, and F is a 30kg block. A) T BD =T BA =T AC =T CF B) E should has a mass of 50kg C) T BD =T BA and T AC =T CF, but T BA T AC D) This system can not be hold in equilibrium. Moment: The simple case r M=r x F: Moment is the product of the distance (from the point to the force) by the force F 27
What is Moment? The moment of a force about a point/axis provides a measurement of the tendency for rotation. The larger force or the distance, the greater the turning effect 2D-Moment: Multiplying Vectors M1=dx. Fy The moment sense is from dx to Fy Fy F Fx F dy dx M2=dy. Fx The moment sense is from dy to Fx M= M1 + M2 =dx. Fy -dy. Fx d M = d x F We know how to add Vectors! How to multiply them? 28
The Arm of a Moment? To Remember Fx To make life easy, let us see this in 2D. z M O O y M O =0 O y Fy dy Fx dy x Example: Step-by-Step 29
Where is the Moment Vector For a given coplanar force system (all forces are in the x-y plane) Each force generates a moment about O Actually it is a moment about z-axis The resultant moment Mo is the sum of the moments of all forces (algebraic sum) M O = ( Fd ) = F1d 1 + F 2d 2 + F 3d 3 If the resultant force F R is known, the resultant moment can also be obtained by M = F d = ( F ) d O R O but usually d o is hard to obtain. O F 2 F 3 x O F 1 y F R The Right Hand Rule Moment is a vector has a magnitude M O = Fd M O is the magnitude of moment M O F is the magnitude of force F d is the perpendicular distance from point O to the line of action of the force F Has direction, determined by right-hand-rule M O O y d F x 30
But, Where is the Moment Vector But, Where is the M Vector? M1=dx. Fy Going in the paper; negative Fy Θ F Fx M F Θ M2=dy. Fx Coming out of the paper Positive d Mz Mz= M1 + M2 =dx. Fy - dy. Fx = d. F. Sin Θ How to find the direction of the moment: 1. Draw a vector (d) from the point to the point of action of the force 2. Point your right hand from d to F, your thumb points to M 3. Mz= d. F Sin Θ 31
Resultant Moment Given the force F=20N, determine the moment of F about point O. To Remember F Fy=Fsin30º 3 30º 7 Fx=Fcos30º d 7tan30º 3+7tan30º 30º (3+7tan30º)/tan30º O [(3+7tan30º)/tan30º] sin30º Example Given the force system shown, determine moment of these forces around the z axis. F1 F2 40kN (0, 1) O (1.1, 0) (3.4, 2) 20kN 30º F3 15kN 32
Review Questions If a force of magnitude 20kN can be applied in four different 2-D configurations (P,Q,R, & S), select the cases resulting in the maximum and minimum moment values on the nut. (Max, Min). A) (Q=Max, P=Min) B) (R=Max, S=Min) C) (P=Max, R=Min) P Q R D) (Q=Max, S=Min) S Using the CCW (counterclockwise) direction as positive, the net moment of the two forces about point P is A) 10 N m B) -20 N m 10 N 5 N C) 20 N m 3 m P 2 m D) 20 kn m Cross Product Also known as the vector product The vector product V of 2 vectors P and Q is written as V = P x Q V V O O Q Q P P The magnitude of V = P Q sin θ The line of action of V is perpendicular to the plane containing P and Q The direction of V is found by the right hand rule. Your fingers point from P to Q, your thumb points to the direction of V. 33
Cross Product: Example Vector Product V x = P y Q z -P z Q y V y = P z Q x -P x Q z V z = P x Q y -P y Q x y Example: P = ( 5,0,8) Q = (10,0,0) Vx = 0 Vy=80 Vz= 0 V = (0,80,0) z V P Q x What are the units of moment? Units depend on what units of force and distance you are using. For example: Force Distance Moment N m N.m KN m KN.m N mm KN.mm KN mm KN.mm 34
Moment Example Find the moment produced by the shown set of forces about point O. Assume distances are given in meters. ΣMo = - (5 x 12) clockwise + (3 x 5) counterclockwise + (7.5 x 8) counterclockwise = -60 + 15 + 60 = 15 KN.m y (0,12) 5 kn (-8,5) 3 kn (17,5) 7.5 kn O x Varignon s Theorem Simply stated means that: Moment created by a force = Moment created by the components of a force Very helpful in problem solving. Example: Let us study different direction for F M1=dx x Fy The moment sense is from dx to Fy A Figure to Remember: Moments Fy F Fx F dy dx M2=dy x Fx The moment sense is from dy to Fx r d M = d x F Note: M1 is positive and M2 is negative; M=dx Fy dy Fx M is a vector coming out of the paper (+ve) or going into the paper (-ve) 35
Solved Example: Concurrent forces Find the reaction components at C and E Couples A couple is a special type of moment that occurs with 2 equal and opposite forces. The resultant force of a couple is zero. BUT, the resultant of a couple is not zero; it is a pure moment. The moment of a couple is the product of the magnitude of one of the forces and the perpendicular distance between their lines of action. 36
Couples Example: Step-by-Step 37
Solved Examples Finding the resultant of a set of non-concurrent non-parallel forces y (0,5) 2 kn 10 kn 4 4 kn (5,2) 3 35 (-1,0) O 10 KN.m x Fundamentals Moment = Force x Distance Units can be N.m KN.m N.mm KN.mm etc... 38
Spot the Tricks! F d M = F x d Real world examples of moments Q: where is the tension? 39
Real world examples of couples FAQ: Resultants of a set of forces Forces Resultant A force passing through the point (No Moment). Elsewhere, it is a force and a moment? A force in the same direction and a moment.? A force in some direction and a moment?? 40
Conditions for Equilibrium For a particle Newton s First Law! Resultant of all external forces is ZERO! For a 2D body: 40 N 40 N T+T T+T Conditions for Equilibrium How do we take the moment into account? Sum to ZERO! To Remember New condition of equilibrium F = 0 M O = 0 The sum of all the external forces acting on the body is equal to ZERO! The sum of the moments of the external forces about a point is equal to ZERO! 41
Support Reactions If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction. If rotation is prevented, a moment is exerted on the body. A Figure To Remember 42
A Figure To Remember Let us see some real world examples Put it to the test: Can you move it vertically (basically penetrate the floor)? Can you move it laterally? Can you rotate it? ice Concrete Laminate 43
Some Real World Examples Put it to the test: Can you move it vertically (basically penetrate the floor)? Can you move it laterally? Can you rotate it? Bridge Bearings 44
How do they do it? Free Body Diagram Why FBD? How? Isolate the body being studied, make it free from its surroundings Outline the shape of the body Identify all the forces that the surroundings exert on the body 45
Example: Step-by-Step Example: Exam style Determine the horizontal and vertical components of reaction for the beam loaded as shown. The beam has no weight. To Remember 46
Review Questions How many unknown support reactions exerting on the beam at supporting point A? A) One vertical force B) Two, one vertical force, one moment C) Two, one vertical force, one horizontal force D) Three, one vertical force, one horizontal force, one moment By applying EoE (Equations of Equilibrium) on a 2D rigid body, up to how many unknowns can we solve? A) 1 unknown Fx = 0 Fa = 0 M = 0 A B) 2 unknowns Fy = 0 M A = 0 M B = 0 C) 3 unknowns Mo = 0 M B = 0 M C = 0 D) 4 unknowns Today s Objective Two-Force Member What is a two-force member? What are the features of those two forces? What can a two-force member tell in solving gproblems? Three-Force Member What is a three-force member? What are the features of those three forces? How do we utilize those features to solve problems? An example 47
Two-Force Members The name says it all! Features of the two forces Have same magnitude Be collinear But opposite sense Some 2-force members M O B A B A Three-Force Members It has only three forces! Features of the three forces Concurrent forces A M O or Parallel forces Some 3-force members 48
An Interesting Example 400N 300N 100N 100N By Bx 300N Dx Dy Ay Ax Dx Dy Bx By Ay Ax An Interesting Example Dx D Dy D Dx Dy Ay Ax 49
An Interesting Example By Bx Dx Dy Dx Ax Dy Ay Fundamentals For a set of concurrent forces there exist only 2 equilibrium equations: ΣFx = 0 ΣFy = 0 For a set of non-concurrent forces there exist 3 equilibrium equations: ΣFx = 0 ΣFy = 0 ΣM = 0 50
Think Ahead Think Ahead Q: Why does it help to support his upper back to the pole? Q Why is he tying himself to the pole? 51