Presentation on theme: "1 2Force Systems Force, Moment, Couple and Resultants."— Presentation transcript:
1 2Force Systems Force, Moment, Couple and Resultants
2 Objectives Students must be able to #1 Course Objective Describe the characteristics and properties of forces and moments, analyse the force system, obtain the resultant and equivalent force systems. Chapter Objectives Use mathematical formulae to manipulate physical quantities Obtain position vectors with appropriate representation. Use and manipulate force vectors Use and manipulate moment vectors Analyse the force system resultants Describe and obtain equivalent systems
3 Force Definition Force is an action that tends to cause acceleration of an object. [Dynamics] The SI unit of force magnitude is the newton (N). One Newton is equivalent to one kilogram-meter per second squared (kg·m/s 2 or kg·m · s – 2 ) Examples of mechanical force include the thrust of a rocket engine, the impetus that causes a car to speed up when you step on the accelerator, and the pull of gravity on your body. Force is a vector quantity (why?) Force can result from the action of electric fields, magnetic fields, and various other phenomena. Force is the action of one body on another. [Statics]
4 FORCE SYSTEMS Force is a vector Line of action is a straight line colinear with the force coplanar if the lines of action lie on the same plane Force System: concurrent if the lines of action intersect at a point parallel if the lines of action are parallel y x
HandPrint Scalar Vector Unit Vector Magnitude of Vector Writing Convention same symbol Recommended Style In this course, you have to write in this convention.
7 FORCE SYSTEMS 2-D Force Systems 3-D Force Systems Moment Couple Resultants Moment Couple Resultants Vector (2D&3D) Basic Concept
8 Free Vectors: associated with “Magnitude” and “Direction” : Direction or V Magnitude: or V Vector : parallelogram Representation triangle +
9 Operation Addition #5 Commutative Vector
10 Operation Addition #6 Vector Associative
11 Operation Scalar Multiplication #2 associative distributive wrt scalar addition distributive wrt vector addition wrt = with respect to
13 Component Resolution of a Vector A vector may be resolved into two components. Vector
14 Basic relations of Triangle (C/6, law of cosine, sine) a b c Law of cosine Law of sine
1 2 15 (Law of sine) Given V, and , find a b c Law of cosine Law of sine Hint
16 special case: projection vectors are orthogonal to each other Vector Component and Projection a : vector components of (along axis a and b) : projections of (onto axis a and b) b a b : orthogonal projections & vector components =
17 Rectangular Components Most commonly used x y vector component = vector projection
18 x y F x =? F y =? x y y x x y F minus >90)
19 EXAMPLE 2-1 Given the magnitude of the tension in the cable, T = 9 kN, express T in terms of unit vector i and j kN ANS T x y Correct? 3 S.F.
20 We are using robot arm to put the cylindrical part into a hole. Determine the components of the force which the cylindrical part exerts on the robot along axes (a) parallel and perpendicular to arm AB (b) parallel and perpendicular to arm BC arm AB ANS arm BC ANS P = 90 N par per par per Defining direction
21 2/2 Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R P=800 N (8cm) T=600 N (6cm) R Graphics Geometric P T R Vector Component (Algebraic) Point of application is B Correct?
22 Example Hibbeler Ex 2-1 #1 Determine the magnitude and direction of the resultant force. Geometric Two forces is not acting at the same point.
23 Geometric Vector Component ( Algebraic) Good? (get full score?) - more explanation - mark answer - 5S.F. Then 3S.F.
24 Good Answer Sheet O Geometric
25 Point of Application
26 Example Hibbeler Ex 2-6 #1 Vector
27 Example Hibbeler Ex 2-6 #2 Vector
28 F1yF1y F1xF1x F2yF2y F2xF2x RyRy RxRx y x o Reference axis (very very important) –Many problems do not come with ref. axis. –Assignment based on convenience/experience The calculations do not reveal the point of application of the resultant force. 1. Graphically Vector summation (addition) –Three ways to be mastered 2. Geometrically 3. Vector component (algebraically) Originally pass through O In case where forces do not apply at the same point of application, you have to find it too!
29 Recommended Problem 2/9, H2-17, 2/12, 2/26, H2-28
31 Three Dimensional Coordinate System y x z Real-life Coordinate System is 3D. Introduce rule for defining the 3 rd axis - “right-hand rule”: x-y-z - for consistency in math calculation (cross vector) How does 2D differs from 3D? y x z 2D
32 - If you known the magnitude and all directional cosines, you can write force in the form of Rectangular Components (3D) y x z - cos( x ), cos( y ), cos( z ) : “directional cosines” of - cos 2 ( x )+cos 2 ( y )+cos 2 ( z ) = 1 is a unit vector in the direction of (maybe +/-) directional cosine Method projection & component
33 Example Hibbeler Ex 2-8 Find Cartesian components of F x y z
34 Given the cable tension T = 2 kN. Write the vector expression of x y z A B A B 1) directional cosine method directionl cosine = Real directional cosine
35 A B A B x y z A B x y z A B Thus, ANS
36 Directional Cosines by Graphics cos 2 ( x )+cos 2 ( y )+cos 2 ( z ) = 1
37 - Usually, the direction of force is not given using the directional cosines. Need some calculation. - Two examples (a) Two points on the line of action of force is given (F also given). x y z A (x 1, y 1, z 1 ) B (x 2, y 2, z 2 ) Two-Point Method Position vector
38 Ans x y z B A kN 2) 2-point construction
39 Write vector expression of. Also determine angle x, y, z, of T with respect to positive x, y and z axes where = unit vector from B to A Thus ANS Consider: T as force of tension acting on the bar
40 Example Hibbeler Ex 2-9 #1 Determine the magnitude and the coordinate direction angles of the resultant force acting on the ring Vector
41 Example Hibbeler Ex 2-9 #2 Vector
42 Example Hibbeler Ex 2-11 #1 Specify the coordinate direction angles of F 2 so that the resultant F R acts along the positive y axis and has a magnitude of 800 N. Vector
43 Example Hibbeler Ex 2-11 #2 Vector
44 Example Hibbeler Ex 2-11 #3 Vector
45 Example Hibbeler Ex 2-15 #1 The roof is supported by cables as shown. If the cables exert forces F AB = 100 N and F AC = 120 N on the wall hook at A as shown, determine the magnitude of the resultant force acting at A. Force
46 Example Hibbeler Ex 2-15 #2 Force
47 Example Hibbeler Ex 2-15 #3 Force
48 Example Hibbeler Ex 2-15 #4 Force
49 (b) Two Angles orienting the line of action of force are given ( , ) y x z Resolve into two components at a time F x = F xy cos( ) = F cos( ) cos( ) F y = F xy sin( ) = F cos( ) sin( ) F z = F sin( ) F xy = F cos( ) Othorgonal projection Method
50 y x z F F xy FxFx FyFy FzFz 65 o 50 o A ns
y x z T A C B 15 o 51 x Ans TZTZ T AB
52 2/110 A force F is applied to the surface of the sphere as shown. The 2 angles (zeta, phi) locate Point P, and point M is the midpoint of ON. Express F in vector form, using the given x-,y- z-coordinates.
72 Homepage URLs Statics official HP (User: Prince Password: Caspian) Session 1 HP (after the end of registration period)
73 FORCE SYSTEMS 2-D Force Systems 3-D Force Systems Moment Couple Resultants Moment Couple Resultants Vector Basic Concept
74 Force Definition Force is an action that tends to cause acceleration of an object. [Dynamics] The SI unit of force magnitude is the newton (N). One newton is equivalent to one kilogram-meter per second squared (kg·m/s 2 or kg·m · s – 2 ) Examples of mechanical force include the thrust of a rocket engine, the impetus that causes a car to speed up when you step on the accelerator, and the pull of gravity on your body. Force is a vector quantity (why?) Force can result from the action of electric fields, magnetic fields, and various other phenomena. Force is the action of one body on another. [Statics]
75 Force Representation Vector quantity –Magnitude –Direction –Point of application 10 N Force Use different colours in diagrams Body outline blue Load red Miscellaneous black (dimension, angle, etc.)
76 Type of Forces External force Internal force Reactive force Applied force Force Strain Stress Contact force Body force Force Concentrated Distributed
77 Cables & Springs Cable in tension Force
78 2/2 Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force R P=800 N (8cm) T=600 N (6cm) R Graphical Geometric P T R Algebraic Point of application is B Correct?
79 Not OK. ! How to add sliding vectors (forces)? A A A is applied at point A A Point of Application is wrong Point of application Principle of Transmissibility Still OK.
F F F2F2 F1F1 F2F2 F1F1 F F F2F2 F1F1 R1R1 R2R2 R R Point of application R This graphical method can be used to find Line of action Special case: Addition of Parallel Sliding Force F F F2F2 F1F1 R1R1 R2R2 R1R1 R2R2 line of action The better and efficient way will be discussed later, when we learn the concept of “moment”, “couple”, and “resultant force”
81 T VD1 TyTy TxTx x y 60 Point of application, But no physical meaning Move all forces to that concurrent point Application Point Ans
82 How to add sliding vectors (forces)? A A is applied at point A Point of application There is better way to find the point of application (or line of action), but you have to learn the concept of moment and couples first.
84 moment axis Moment is a vector Moment In addition to the tendency to move a body, force may also tend to rotate a body about an axis From experience (experiment) magnitude depends only on “F” and “d” (magnitude) summation Direction
85 Moment Definition x y z O Moment is a vector quantity. –Magnitude –Direction –Axis of Rotation The unit of moment is N·m The moment-arm d (perpendicular distance) The right-hand rule determined by vector cross product Sign convention: 2D +k or CCW is positive. Moment of a force or torque
86 Moment about point A : Mathematical Definition (3D) A r d -Magnitude: -Point of application: point A -Direction: right-hand rule + (Unit: newton-meters, N-m) d A Fd - 2D, need sign convention and be consistent; e.g. + for counter- clockwise and – for clockwise 2D X from A to point of application of the force d
The moment of a force about any point is equal to the sum of the moments of the components of the force about that point 87 can be used with more than 2 components + O x y d1d1 d2d2 M o = -F x d 2 +F y d 1 Same? Varignon’s Theorem (Principle of Moment) sum of moment (of each force) = moment of sum (of all force) Useful with rectangular components
88 Principle of Transmissibility & Moment A O O r d X Y Z M = Fr sin = Fd Sliding force has the same moment convenient - direction: same - magnitude: Principle of Transmissibility is based on the fact that “moving force along the line of action causes no effect in changing moment” position vector: from A to any point on line of action of the force.
89 d Sample 2/5 Calculate the magnitude of the moment about the base point O of 600N force in five different ways. 600N m 2m A O Solution I: 2D Scalar Approach 600N m 2m A O Solution II: 3D Vector Approach CW x y CW or CCW? CW Correct?
90 600N m 2m A O F2F2 F1F1 Solution III: Varignon’s theorem 600N m 2m A O F2F2 B d1d1 F1F1 F2F2 F1F1 C d2d2 Solution IV: Transmissibility Solution V: Transmissibility +
91 EXAMPLE 2.8 In raising the flagpole, the tension T in the cable must supply a Moment about O of 72 kN-m. Determine T. 15 m ANS o 12 m 30 m d
92 Example Hibbeler Ex 4-7 #1 Determine the moment of the force about point O. Moment Correct?
93 Example Hibbeler Ex 4-7 #2 Moment 3D Vector Approach Scalar Approach (Varignon’s theorem)
94 Couple - Couple is a summed moment produced by two force of equal magnitude but opposite in direction. O a d + M = F(a+d) – Fa = Fd magnitude does not depend on distance a (point O), i.e. any point on the body has the same magnitude. - tendency to rotate the “whole” object. - no effect on moving object as translation. 2D representations: (Couples) C couple is a free vector C C Effect of Pure Rotation
95 Moment Couple Definition #2 A couple moment is a free vector It can act at any point since M depends only upon the position vector r directed between the forces. Moment of a Couple O A B Moment
96 A B A Force-couple systems - Line of action of a force on a body may be changed if a couple is added to compensated for the change in the tendency to rotate of that body. Procedure may be reversed to combine a force with a couple Force-couple system d B A B Principle of transmissibility No changes in net external effect The direction and magnitude of Force can not be changed, only line of action (i.e. only change to other pararell line)
97 Principle of Transmissibility is based on the fact that moving force along the line of action causes no effect in changing moment F A B F A B C A B F C A B F from new location (B) to old location (A) F A B F A B No Moment: Principle of Transmissibility
98 In the viewpoint of Mechanics, Result of force to these systems are equal Why using equivalent system? A Force-couple system B A B Principle of transmissibility real (physical) system equivalent system All force systems are equal.
99 Understanding Force-Couple system A B A B D D Moment about point B of force F = tendency of force F to rotate the object at point B couple occurs when moving Force F from A to B ( couple occurs when moving Force F parallel to its line of action to the point B) Equivalent System
100 Vector Diagram F F 12 m 70m P P Ans Be careful of the direction of moment
101 2/11 Replace the force F by an equivalent force-couple system at point O. Ans M 50 kN x y 0.1m 0.25 m Couple occurred when moving F to O = Moment of F about O CCW Correct?
102 o ANS Got the meaning? Moving all 3 forces to point O Sum of couples Engine number 3 fails. Determine the force-couple system on the body about point o. couples occuring when moving forces. (direction: left) (CW) sum of moments? x y +
103 Example Hibbeler Ex 4-14 #1 Replace the current system by an equivalent resultant force and couple moment acting A. Resultant
104 Example Hibbeler Ex 4-14 #2 Resultant
105 Ans b 300N 20 o 60 N-m D exactly cancelled b cos20
107 Resultant of many forces-couple is the simplest force-couple combination which can replace the original forces/couples without changing the external effects on the body they act on 2/6 Simplest Resultant -Add two at a time get line of action of -Add many do not get line of action of y x Point of application
108 Easier way to get a resultant + its location any forces + couples system 1) Pick a point (easy to find moment arms) d1d1 d2d2 d3d3 O F1d1F1d1 2) Replace each force with a force at point O + a couple F2d2F2d2 F3d3F3d3 O 4) Replace force-couple system with a single force d=M o /R 3) Add forces and moments O Mo=(Fidi)Mo=(Fidi) M o =Rd O arbitrary (forces + couples : same procedures) any forces + couples system single-force + special single-couple (wrench) 2D single-force system (no-couple) or single-couple system 3D resultant
109 2/87 Determine the resultant and its line of action of the following three loads. R = ( 2.4cos sin cos20 ) i +( -2.4sin cos cos20 ) j kN M = -2.4*0.2cos *0.12cos *0.3cos20 kN-m But we want to find the line of action of the “pure” resultant force (the one which has no additional couple) M (force-couple system) O R move the R to point X where Resultant Moment is zero find the point X where the Resultant Moment is zero R O X M M + N = 0 Move 3 forces to point O, Sums their force and couples Note: R is the same regardless with the location point we move the force to Note: M depends on the location where we move the force to + why?
110 M = kN-m At point O (0,0) M O R At point X (x,y) N M O R R P ( line of action ) couples cancelled + P O (0,0) P (x,y) Two equivalent systems Moment at any point must be the same on both system Sys 1 Sys 2 Pick Point O Correct?
111 M = kN-m At point O (0,0) M O R How to locate Point P O (0,0) + d d d Manually Canceling Couples How to find line of action ? or P P P O
112 Equivalent System Definition Two force-couple systems are equivalent Equivalent System M O R O R P Sys 1 Sys 2
113 A car stuck in the snow. Three students attempt to free the car by exert forces on the car at point A, B and C while the driver’s actions result in a forward thrust of 200 N as shown in picture. Determine 1) the equivalent force-couple system at the car center of mass G 2) locate the point on x-axis where the resultant passes. x y G
114 For line of action of resultant At y = 0; x = m. ANS x y G x y G b Couple Cancellation If you want to find only b (not line of action itself) + or -, you have to find out manually Sys I Two equivalent systems Moment at point G must be the same on both system Sys II Two equivalent systems (2D)
115 Determine the resultant (vector) and the point on x and y axes which must pass. x y G
116 x y O For line of action of resultant If y = 0; x = 7.42 m. ANS x y O x = 0; y = m.