Question

Prob 2 and has no a vertical plane, has an angul ar velocity of 2 rads /s CW in the the force is applied. Determine the shown is moving in The acceleration but position shown just before the force in theAsupporting cable and the magnitude and direction m system of the acceleration force is applied. of the mass center immediately after the 2m m 2.59 3ONm

Answer 1

Fixed print p ( measy 3ON 14lm about red point P) ese p (Temperature 1 '화) 30 n 90 TP (6+235 CGa 2.5

Hae - 1SNE (t1) (s-4Sa) 3(2+1 2m 213 Co 30N So targont al accderativeGn at ( ST (2t 3(21 X (34NE) ie 3 (4R+4 t2+Va ) ava (2+112 C645A ay (61Dty applied agtei the forte

(3 Frx いY at Suportioy the on Force applial by the upward dive ction Cable in Cable in horzontal binge the on the Fx t Forte apphed y Dise Ctin eve =4M6Y and Fy 1+ fy Fy fy-10 (1041 (132) Fy= t0Cio+ 5tY Fy 1D

for F 4mat t30 4XGSat-30 =10 X 3 (6+5) 30 Gtaty 6t5-6y E= 30 i 30 (3 ) fr (3+ta) Force in te Seppoting Callk Fatf 14 For accelerationi at-af = at ta at ta1-a4

3 (6+53) at (Waty and ac ) ac = R Rt) (413)