Question

The 3-lb uniform rod BD is connected to crank AB and to a collar of negligible weight. A couple (not shown) is applied to crank AB causing it to rotate with an angular velocity of 16 rad/s counterclockwise and an angular acceleration of 84 rad/s2 clockwise at the instant shown. Neglecting the effect of friction, determine the reaction at D. А 3 in. B 6 in. D Sin The reaction at Dis ib.

Answer 1

3 in Ist OBES Answer (98)n = YAB WAB (2/2) (16) ft/s = 64 Atis? (ab) =VAB TAB =(22)(84) ft/5 = 214+/- V8 = YAB WAB = (2) (16) (AB)n = 4 ft/s Also, VB = FBC WBO 4 = (2) WBD 280 = 8 rad's consider BD c= instanteneous Center, B VBD *VB = råc trão TBD 6²+8² PBD = 10 in Gin sin → Acceleration diagram for BD Qen (fixed) B 12BD + G 21445 ODS (90)n = 64 21/8 (20/B)n

= AB+ = abtas/B {(apien tapps) Here, (a D/Bn = YBD WED = (123) = 53.333 ft/s C01B)t = Crep dep) = () 480) = 0-8333 LED Acceleration diagram (adet @bt-) Dish 3) aple)t = 0 →-21 - $) 53.333) +*%0.8333 den) 28 → 28D = 127.337 rad's SO Оо (OB (AB)t -> ag = AbtaG/B = agt £ ao/B = agt EcodentDIB)+) → @qx=QB)n + $(@08h+) 2016/) =-21 + + ($63.333) + 3( 49(127.33+))) = -10.501 ft/s = 10.501 4+15? (towards left)

@y (AB)++ Ļ( ² (aden + 4 Qofet) = 64 + £ ( }(53.333) + (49(127.534)) = 122.445 ft/s (upwards) moment of inertia I = the (roo) te (22)(10 = = 0.0054 lb-ft.5" → kinetic diagram of BD By B B Be macly maase Gin K 4in ktin 米 EMB= 2 (MB)eff → R. (6)-W(4) = I dep + m@ay(+)- max(3) → Role)- 34) =(0.0054)127.334) +2.2X122.445) -(S.2) 10.501) (3) Ro= 10.491 eb Good Luck