Bellringer

1. Bellringer Compare and explain in complete sentences and formulas how using the Newton’s three laws of motion find the resultant force.

2. HOMEWORK SUBMIT YOUR PROJECT (50%), ALL 3 COMPONENTS

3. Force System Resultant

4. P T M M Moment of a Force - - - Scalar Formulation 1.Moment A measure of the tendency of the force to cause a body to rotate about the point or axis. • Torque (T) • Bending moment (M)

5. 2. Vector quantity d o Lime of action (sliding vector) (1) Magnitude ( N-m or lb-ft) Mo = Fd d = moment arm or perpendicular distance from point O to the line of action of force. (2) Direction Right-Hard rule A. Sense of rotation ( Force rotates about Pt.O) Curled fingers B. Direction and sense of moment Thumb

6. 3.Resultant Moment of Coplanar Force System don do1 do3 do2 4.2 Cross Product 1. Definition (1) magnitude of (2)Direction of perpendicular to the plane containing A & B

7. j i 2. Law of operation (1) (2) (3) 3. Cartesian Vector Formulation (1) Cross product of Cartesian unit vectors.

8. (2) Cross product of vector A & B in Cartesian vector form

9. 4.3 Moment of a Force – Vector Formulation d F o 1. Moment of a force F about pt. O Mo= r x F where r = A position vector from pt. O to any pt. on the line of action of force F . (1) Magnitude Mo=|Mo|=| r x F | =| r|| F | sinθ=F r sinθ =F d (2) Direction Curl the right-hand fingers from r toward F (r cross F ) and the thumb is perpendicular to the plane containing r and F.

10. 4.4 Principle of moments F1 r F2 o Mo=r x F F = F1+F2 Mo= r x (F1+F2) = r x F1+ r x F2 = MO1+MO2 Varignon’s theorem The moment of a force about a point is equal to the sum of the moment of the force’s components about the point .

11. 4.5 Moment of a force about a specified Axis a b Ma Ө Mo= r x F b’ A Moment axis a’ 1. Objective Find the component of this moment along a specified axis passes through the point about which the moment of a force is computed. 2. Scalar analysis (See textbook) 3. Vector analysis Point O on axis aa’ O Axis of projection

12. Moment of a force F about point 0 Mo= r × F Here, we assume that bb’ axis is the moment axis of Mo (2) Component of Mo onto aa´ axis Ma = Ma ua Ma=Mo cosθ =Mo●ua=( r × F ) ●ua =trip scalar product Here Ma=magnitude of Ma ua= unit vector define the direction of aa´ axis

13. If then 4.Method of Finding Moment about a specific axis (1) Find the moment of the force about point O Mo= r x F (2) Resolving the moment along the specific axis Ma = Maua = (Mo•ua) ua =[ua •( r x F )]ua

14. 4.6 moment of a couple d d θ F F r 1. Definition ( couple) 偶力矩 Two parallel forces have the same magnitude, opposite distances, and are separated by a perpendicular distance d. 2. Scalar Formulation (1) Magnitude M=Fd (2) Direction & sense (Right-hand rule) •Thumb indicates the direction •Curled fingers indicates the sense of rotation 3. Vector Formulation M= r x F |M|=M=|r x F |=r F sinθ =F d

15. -F r A rB F rA o B B F -F r o’ A Mo=Mo’= r x F=M o Remark: (1) The couple moment is equivalent to the sum of the moment of both couple forces about any arbitrary point 0 in space. Mo= rAx( -F )+ rB x F =(-rA+rB) x F =r x F= M (2) Couple moment is a free vector which can act at any point in space.

16. 4. Equivalent Couples The forces of equal couples lie either in the same plane or in planes parallel to one another. F -F -F F A plane A // plane B F d -F d B M2 M1 A B M2 M1 5. Resultant couple moment Apply couple moment at any point p on a body and add them vectorially. MR=ΣM=Σ r x F

17. 4.7 Equivalent system A F P P F 1. Equivalent system When the force and couple moment system produce the same “external” effects of translation and rotation of the body as their resultant , these two sets of loadings are said to be equivalent. 2. Principle of transmissibility The external effects on a rigid body remain unchanged,when a force, acting a given point on the body, is applied to another point lying on line of action of the force. line of action Same external effect Internal effect ? Internal stresses are different.

18. F F F -F -F o 3. Point O is on the line of action of the force A A A equivalent equivalent o o o Original system Sliding vector 4. Point O is not on the line of action of the force F Couple moment F Mc= r x F M=r x F A A F r P A F o line of action o Original system Force on Point A =Force on point O + couple moment on any point p.

19. o o A A F F F F o o X P A A d M= F d (Free vector) Mo= F d Example: Point O is on the line of action of the force Point O is not on the line of action of the force

20. 4.8 Resultant of a force & couple system 1. Objective Simplify a system of force and couple moments to their resultants to study the external effects on the body. 2. Procedures for Analysis (1)Force summations FR=F1+F2+……+ΣF (2)Moment summations MR0= ΣMC+r1o*F1+r2o*F2= ΣMC+ ΣM0 MC:Couple moment in the system Mo:Couple moment about pt.O of the force in the system.

21. y F1,F2,F3 on xy plane F3 M1&M2:z direction MR0=ΣMC+Σr * F P MR0 F2 x => => d= FR F1 FR=ΣF F2 F1 FR Equivalent P = System Fn no couple moment 4.9 Further Reduction of a force & couple system 1. Simplification to a single Resultant Force (1)Condition FRMR0 or FR*MR0 = 0 (2)Force system A. Concurrent Force system B.Coplanar Force System

22. z z FR z F1 r2 F2MR0FR= ΣF r1 y = y = y M1 p o r3F3 x x MR0 x M2d = -------------- |FR|d=|MR0|FR C. Parallel Force System 1. F1 // F2 //……// Fn 2. MR0 perpendicular to FR, MR0=ΣM+Σr*F 2. Reduction to a wrench (1) Condition: FRMR0 MR0=M +M// M = moment component FR M// = moment component // FR

23. (2) Wrench (or Screw) An equivalent system reduces a simple resultant force FR and couple moment MR0 at pt.0 to a collinear force FR and couple moment M// at pt. FR MRo a b o b FR FR a a M// a b M// o b p o p b b a a