园 上活我人峰 Type of Rigid-Body Plane Motion Example a Rectilinear translation Rocket test sled 6) Parallel-link swinging plate (e) Compound pendulum d Seaan Connecting rod in a reciprocating engine 圈上清庆大坐 Scalars Vectors Examples: mass,volume,speed force,velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Parallelogram law Special Notation: None Bold font,a line,an arrow or a“carrot
Scalars Vectors Examples: mass, volume, speed force, velocity Characteristics: It has a magnitude It has a magnitude (positive or negative) and direction Addition rule: Simple arithmetic Parallelogram law Special Notation: None Bold font, a line, an arrow or a “carrot
国上活大峰 Vector manipulations Vector Magnitude Head Magnitude (terminus) Direction Tail (origin) Vector can be represented Graphically rR=Ri+Ryj R=R,+jRy Analytically R=[RR,T R=Reio=R(cos0+jsin) 圈上清庆大坐 Vector manipulations Vector addition: A=Aeio B=Bei0s C=A+B C=A+B=Aeio+Beia =B+A =(Acos0+Bcoseg)+ j(Asin0+Bsin0g) AC-B Vector substraction: A=C-B=Cei@c-Beia =(Ccos0-Bcose)+ j(Csinc-Bsineg) Figure (a)Vector addition and (vector subtraction
Vector manipulations Vector Magnitude Direction Vector can be represented Graphically Analytically Head Magnitude (terminus) Tail (origin) Rx y R jR R e (cos sin ) j R j R [ ]T R R x y R Rx y R i R j Vector manipulations Figure (a) Vector addition and (b) vector subtraction. B A B C A j Ae A B j Be B =( cos cos )+ j( sin sin ) A B j j A B A B Ae Be A B A B CAB =( cos cos )+ j( sin sin ) C B j j C B C B Ce Be C B C B A CB Vector addition: Vector substraction:
国上活大峰 0 B (a) Figure Graphical solution of case 2:(a)given C.A.and B. Figure Graphical solution of case 3 (a)given C.A'.and B': c的solution for A,B°andA'.B". 句solution for A and B. √√o√√o C=A+B 6=A+B 圈上清庆大坐 Vector rotation: R=Reio R R'=Reio=Rei(0+o) R
Figure Graphical solution of case 2: (a) given C, Aˆ , and B; (b) solution for A, Bˆ and A’, Bˆ’. CAB √ √ O √ √ O √ √ O √ O √ Figure Graphical solution of case 3: (a) given C, Aˆ, and Bˆ; (b) solution for A and B. CAB Vector rotation: j R e R ( ) 'e j j R e R R X Y O R R '
国上清大峰 o=i2+j(-1)+k4(rad1s). Vector cross product: r=i(-1)+j10+k2(mm), A=iAs jA,+kA. B=iB +jBy+kB. =i(-2-40)+j-4-4)+k(20-1) i jk =i(-42)+j-8)+k19)mm/s A×B=AA,A BB B. AxB=i(A,B.-A.B)+j(A.B,-AB.)+k(A,B-A,B, A×B=ABsin0 圆上清夫大学 Vector dot product: A[B=ABcos0 Angle 0 is the smallest angle between the two vectors and is always in a range of0°to180°
ω i2 j(1) k4 (rad / s). r i(1) j10 k2 (mm), mm s B B B A A A x y z x y z ( 42) ( 8) (19) / ( 2 40) ( 4 4) (20 1) 1 10 2 2 1 4 i j k i j k i j k i j k υ ω r ω r v Ax Ay A z A i j k Bx By B z B i j k Vector cross product: x y z x y z B B B A A A i j k AB ( ) ( ) ( ) AyBz AzBy AzBx AxBz AxBy AyBx A B i j k A B ABsin Vector dot product: A B ABcos Angle is the smallest angle between the two vectors and is always in a range of 0º to 180º
国上活天大峰 -1- ixj=k jxi=-kj×k=ikxi=-i kxi=j ixk=-j i×i=j×j=k×k=0 i.i=jj=k.k=1 i.j=j.k=k.i=0 圆上泽夫大峰 commutative BxA=-AxB. A.B=B.A Distributive AX(B+C)=A×B+A×C A.(B+C)=A.B+A.C Triple Product 44,4 A(B×C)=B.(C×A)=C.(A×B)=BB,B c.c,c. A×(B·C)=(A.C)B-(A.B)C (A×B)×C=(A.C)B-(B.C)A
0 k i j i k j i i j j k k i j k j i k j k i k j i i.i j.j k.k 1 i.j j.k k.i 0 BA AB. A.B B.A A (B C) A B AC A.(B C) A.B A.C commutative Distributive ( )( )( ) ; xyz xyz x y z A A A B B B CCC A. B C B. C A C. A B ( )( ) ( ) ( ) ( )() A B C A.C B A.B C A B C A.C B B.C A Triple Product