Question

Para el mecanismo plano de la Figura, 2 - 28 krad/s constante, 82-35 grados, r2- 45 cm y 83-90 grados en el momento mostrado. Cual es la componente horizontal (en el eje x) de la aceleracion de coriolis en el eslabon deslizante B en cm/s? HINT: Destaca en tu procedimiento el valor de r3, velocidad de deslizamiento By w3. Figura F E B r2 cm r3 cm 03 O2 02 A D

Answer 1

AB= r2=0.45m W₂ = 28 rad/s. Configuration diag om V₂ = Wzx AB Scule : locm 1cm- = 28 x 0.45 = 12.6.ml F B 8 from tue olingomy] r₂ = 26 cm /26cm a 45cm 350 90 A B' the re point on link DE B { slicer } VBBI 20 mm x Scale - Velocity diagram - 20 X0.5 10 m/s = ithm obtained from velocity diagom. Scale: 0.5 m/ b Voi= 14 mm x scale =14 40.5 = -7 mis Vgb e ad ad. • mark the fined ponts as from a drow a line à line perpendicular to link AB direction of we that vector ab (v.) , in the is . . a in а. ED from b doaw line parallel to link ED (that represent motion of slider B links ED. from d draw line perpendicular to link (that represents velocity of B' in link ED with respect to fined point o). by direct measurement and convert scale to obtaine the velocities of points. CS Scanned with CamScanner

C A BB' E { coriolis com sponent of acceleration] ago = 2 Veri w VBB' B (Slider) < a Bel "B' (ponton DE) Ver' = velocitay & Slider B relative to B' 4 w Ws = angular velocity of link De Direction of Coriolis component g acceleration m VBB' = 10 m/s obhuined from velocity diagram] Ws 7 m/s NI = 26.92 rools DB-= 26 cms. DB 0.26 m (obtained from configuration clragom a Bel 2 VBBI W = 2x10 X (26.92) = 538.4 mle CS Scanned with CamScanner