Question

A straight rack rests on a gear of radius r = 3 in. and is attached to a block B as shown. Knowing that at the instant shown = 20, the angular velocity of gear Dis 42 rad's clockwise, and it is speeding up at a rate of 2.6 rad/s?, determine the angular acceleration of AB and the acceleration of block B. A B The angular acceleration of AB is The acceleration of block B is rad/s2 clockwise. (Round the final answer to three decimal places.) (Round the final answer to one decimal place)

Answer 1

Given data! Bin Crear nadius, r. 4.2 vual sec Angular velocity, Wp = 4.2 rua Anguloi Accele mtion, dp. 2.b raals² 0:20 Solution let, P= Point on crear At Contact point. es 0 : point on rack gr K PQ Y Y/tono ruose B + Position of P with respelt to ö rple Tplo - ( sine i + cose) YpID: 3 ( sin 20 € + cos203) YpID=1.020 i + 2.819; Position coith respect to B, ( cosei - sinoj) tano YOIB Point a 3 roBa ( cosaoi- sinzo ;) tanzo Y/B = 7.7451 - 2.819 => Grear o Wowgear - 4.2 rad/sec 2 Vp = Wor - 4.2 X 3 Vo = 12.6 in/sec g VP= 12.6( cos o i – Sinoj) vp = 11.841 – 4.309j in sec

at Point P Acceleration Alelehatroni Normal Tangential Acceleration: Capt OXY = 2.6x3(cosedi - Sin203) (ap)n = woord 4.2²x 3 52.926-605205 - Sinzoi) Cap)n = -18.09 i – 49.73; in (apt = 7.329 i – 2.667; ine Sect Set Contact Point to rack => Gear Vo= VP= 11.841 - 4.309 j in Isec VB- Vot + Vela WAB = WAB I & AB AB • VQ + WABK X YBlo VB VB → VBI 11.841 – 4.309j + WAB KX(7-7451 - 2.813) OLB ab - VBI = (11.84 + 2.81) i +(-4.309 +7.74594) j Equating į components, -4.309 +7.745 WAB=0 WAB - 0.556 rad Isec Acceleration Anglysis: Gear to rack Contact Point Coolt Capt = 7.329 1 - 2.6613 inlse doon = (ap) ntr word Å Wrot = WAB- WD = 0.556-642)

Wrd = 4.756 vad) sec D 2 2 r word 3x (4.756) ( Sin 20 i + cos20 %) 2 Y wrd' = 23.21 + 6.3.766 ☺ (@o)na -18.09 i – 49.73; +23.211 +63.7663 Qen: 5:12 i + 14.04) in /s? ao= (90) taon 7.329 i - 2.667) +5.12; +14.04; ao 12.449 +11.373) at Acceleration point B aba aQ+ able ao + LAB K X rolo WAB Noia (12.4491 +11-373)) + CABKXC 7.745-2.815) 2 -(0.5567" xl 7.7451 – 2.813) - abi (121449 +2.81 LAB – 2:394) i + C11.373 +7.745 LAB +0.8686); Equate 'j' components 11.373 + 7.745 CAB + 0.8686 XAB = -17:24 7.745 LAB =-1.58 rad/s2

P Ze Equate į Components AB = 12.449 + 2.81*(-1.58) – 2.39 4 5.615 in /s² AB Result LAB 1. 58 rad 52 IN a) Angular Acceleration Of AB, ав 5.615 in s? block, B b) Acceleration