Question

Table 2. Composition of Assigned Forces. Masse (g) 150 200 100 0 30° 1350 180° F F2 F3 Fi (N) = Fix (N)= Fix (N)= F2 (N) = Fax (N)= F2Y (N)= F3 (N) = F3x (N) =

Answer 1

Solo: Fi = 150X10-3x9.8) " FI=1.4715N. fix=1.4715 Xcos (30) 214715 * sin 130). fiya .: fix= 1.274N, Fly=0.736N F2= 2008 103x9.81 F2=1-962N. ) 1.62X=1962 x cos(135) f2y = 1.962x Sin(135) fzx=-1-3870, fay = 1.387N. f3= 100x10-3 X9.8) f3=0.98IN. f3 x 30.981 Xcoscilo) fix=-0.981 Fay 50.981 xsin (180) f3y = 0. A. FRx = fix t fax tf3x. = 1.274 + (-1.387) 7(-0.980. FRx = -1.9040] Fry = fly & fayt fzy 0936 +387 Ho :. [Fry = 2.117 N FR = V FRx² - Fry² ' IFR 20.925N OR= tanyl Fry! IFRX LORE OR 48.03 . (from cvet-aris),

FR AFRY 18-03 FEZ FR = 0.925N. O G = 48.030 m tvex-ans. FRA 418.030 from FE