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

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

Force System Resultant

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) M M P T Moment of a Force Scalar Formulation

2. Vector quantity d o Lime of action (sliding vector) (1) Magnitude ( N-m or lb-ft) M o = 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

3.Resultant Moment of Coplanar Force System d o1 d o2 d o3 d on 1. Definition (1) magnitude of 4.2 Cross Product (2)Direction of perpendicular to the plane containing A & B

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

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

4.3 Moment of a Force – Vector Formulation F d o (1) Magnitude M o =|M o |=| 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. 1. Moment of a force F about pt. O M o = r x F where r = A position vector from pt. O to any pt. on the line of action of force F.

4.4 Principle of moments 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. r o F1F1 F2F2 M o =r x F F = F 1 +F 2 Mo= r x (F 1 +F 2 ) = r x F 1 + r x F 2 = M O1 +M O2

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 4.5 Moment of a force about a specified Axis Moment axis b’b’ b Ө a’a’ a A MaMa M o = r x F O Axis of projection Point O on axis aa’

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

4.Method of Finding Moment about a specific axis (1) Find the moment of the force about point O M o = r x F (2) Resolving the moment along the specific axis M a = M a u a = (M ou a ) u a =[u a ( r x F )]u a If then

d 4.6 moment of a couple 3. Vector Formulation F θ r d F M= r x F |M|=M=|r x F |=r F sinθ =F d 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

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

4. Equivalent Couples The forces of equal couples lie either in the same plane or in planes parallel to one another. 5. Resultant couple moment Apply couple moment at any point p on a body and add them vectorially. A B M1M1 M2M2 M1M1 M2M2 M R =ΣM=Σ r x F A B F -F F plane A // plane B F -F d d

4.7 Equivalent system 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. P F P F A line of action Same external effect Internal effect ? Internal stresses are different.

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

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

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 F R =F 1 +F 2 +……+ΣF (2)Moment summations M R0 = ΣM C +r 1o *F 1 +r 2o *F 2 = ΣM C + ΣM 0 M C :Couple moment in the system M o :Couple moment about pt.O of the force in the system.

4.9 Further Reduction of a force & couple system 1. Simplification to a single Resultant Force (1)Condition F R M R0 or F R *M R0 = 0 (2)Force system A. Concurrent Force system B.Coplanar Force System F 2 F 1 F R Equivalent P = System F n no couple moment y F 1,F 2,F 3 on xy plane F 3 M 1 &M 2 :z direction M R0 =ΣM C + Σr * F P M R0 F 2 x => => d= F R F 1 F R =ΣF

z z F R z F 1 r 2 F 2 M R0 F R = ΣF r 1 y = y = y M 1 p o r 3 F 3 x x M R0 x M 2 d = |F R |d=|M R0 | F R C. Parallel Force System 1. F 1 // F 2 //……// F n 2. M R0 perpendicular to F R, MR0=ΣM+ Σr*F 2. Reduction to a wrench (1) Condition: F R M R0 M R0 =M +M // M = moment component F R M // = moment component // F R

(2) Wrench (or Screw) An equivalent system reduces a simple resultant force F R and couple moment M R0 at pt.0 to a collinear force F R and couple moment M // at pt. o o o a a a a a a b b b b b b FRFR M Ro FRFR FRFR p p M //