Question

Shown in figure is a Scotch yoke mechanism. What is the input and output motion of the mechanism? Derive (a) an expression for the position vector, Re, velocity, V=RB and acceleration, AB =Rg of the point B on yoke and (b) an expression for the position vector, RA, velocity, VA=RA and acceleration, AA =RA of the point A on slider in terms of 02, 02, 2 and 02A=a (crank length) Α. y OzA=a 002 I >> B Position, velocity and acceleration of a vector in plane R = a cos(O) i +a sin(0)j, a: length of vector R VER --a o sin(O) i + a cos() A=V=R-a a [-sin(0) it cos(O) jl-a 02 (cos(O) i+ sin(0)j]

Answer 1

o a Given - O2 Aza y Solution А W242 DO2 VITI & cotch yoke mechanism i's used to conuest the rotong motion of inpect crank into motion of the slider. the slicking ( Expression for position vector Ra, relocity Rea and of point a acceleration Position vector Re a cos@i + asincaj for Ra and we are considering .- QE Q2 anti'clockwise tue. - Ra - a cas @2)i + asin (@2) j V - = R = -aw sino i taw cos@); Va = Ŕa = a awsin@z) i ta w cos (02) į Similarly for Occeleration A = v=RDQ * [-sim@i+ cxc@j) a.02 [coscos i t sim@i]

Aa= a * E sin(@2) i + cos (Ⓡa);] a w2 ( casei & sim (02) ;] Similarly l) Expression for position vector RB, Rs and Re of point respect Here therefon B with to 02-0 and R = aces @ è + asinte); RB = cosoit a sinass= a i for velocity Ves cos b) ; VB aw (sino) i 7 aw 8 COM@j (sinco] itaw a wj a Ř =VB = o it and acceleration RE 1 - 2x [-sin@) è + cos@)] -QW2 [cosce)i + sinkosi] x 6 sin (0) i+ COMO)j] a wrcoso it sinoj] + a RE AB- (a wz) (a ); AB = RB