Question

Vector Mechanics Dynamics

The 6.85 lb disk A has a radius of 4.35 in and is idle. The 12.09 lb disk B has a radius of 7.49 in and an angular velocity of 910 when it contacts the disk A. Neglecting the friction in the bearings, determine the total thrust "Ft" exerted by the friction force on the disco A.

Р A Фо ТА ТВ

Answer 1

Feel free to ask any question but remember to rate the answer.

Moment of Inertia of dist A IS 2 IA= L x (WA XA2 g Th 5 ( -X(6.85X4.2 12 IA= 0.01397 lb. A g2 32.2 M. I of disk B IS, IB 1X (WB) DB2 g X X (7.4972 12 IB=0.07313 lb. Pt. 52 FBD of disk A is IA (WA)o=0 DAWA + M А WAT А

Moment about point A IS ot (Ft) VA - TAWA ft= 0.01397.WA. (4.356127 Ft= 0.03855WA FBD of disk B is as follows. IB WD fowe +Nt B 'В Ft We have (LBOW (Welo - (910) (0) (910)+(LT) = 95.949 gad/sec

Taking moment about point B, Тв. 00 — (ft). Y() = Тg bg — - Ig. oo –Co. 039) он): p = Трор (o od313) (35. 2999) — (o-o381 X7:49 од = (а 043b ) 6.9639 — 0. 00 чо( 0 = 0.otzt wg - ор = ( 6.964 o-07313 оголо-соолук да, - ФB = 9, 29 - 0-329OP Now, VAaVB : ХАША • Трор оре ЗА ДА ав ЧЗГРА 7:49 * og - DJNo ФА putting this expression in egr

50-5y07 од = 9 - 9946 — 0:3190 он *О. 9o92 0A = 9. 1946 : LDA= 104.7453 radly — Toy-453x ( 6 ) • w = [000. Дуо тут But, WB = 0.5807 WA • ФB = 0:3301.(1000. 24yo) - Фg - S80 - 3ч 12 Үр т But from equation Co. ft. O.039JTDA : ft - o-o23) х (1oq: 742) Pts Ч-03193 Дь 1