1 a) B 3m F2 6m C TE 1 TB Te Fx 6 m Fy 30' 400 sin30 b) Coordinates: 400 cos 30 400 N A= (6, 0, 0) 1 AB = (-6,-3,2) B = (0,-3,2) C = (0, 0, 2) > AC=(-6,0,2) From B, TAB = TAB TAB = (-61 -3) +2K) X PAB From C, TAC = TAC PC = (-61 + 2K) Y PAC From A, A = 400 sin30 }' - 400 cos 30 F From O, o = Oxi + Oy j + O2 K EFx: Ox - 6x - 6Y = 0 - (1) EFy: Oy + 400sin 30 - 3X = 0 - (2) E F2 : Oz - 400 Los 30 + 2Y + 2X = 0 -- (3) c) From moments E (Mo). = 0 = (POR X TAB) + (POR X TAC) + (SOA XA) + (-600} + 720R) i k + 1 6 8 2%) + -6x -3X 2x / 1-64 27/ i j K 6 0 O ) 0 400 sin30 - 400Cos30 1 = j(12x)+ k(-18x) -j(12y) +j(6.400cos30) + k(6.400sin30) + = j (12x -12Y +6.400cos30-600) + k (-18x + 6.400 sin 30 +720) E My =- 12x-12Y +6.400 cos30-600=0 - (4) EM= = - 18x + 6.400 sin30 +720 = 0 (5 )
2. Wc = 80 lb P = 180 lb Q = 30 16 FA = 190 lb Q cos 50 € 50° 0 sin50 3 O A 1 >P FA NA FB No a) EF2: Qsin 50 - FB +P - NA = 0 E Fy: Q cos50 + NB - WC - FA = 0 b) moment about point O, = 0 E (Mo). = 0 (30)+ (3FB) +(3FA) - P=0 P-3Q-3FB-3FA =0 EFx: 30 sin 50 - 190 + 180 - FB = 0 E Fy : 30 COS50 + N8 - 80 - FA = 0 E (Mo): 180 - 3(30) - 3FB - 3FA = O : FB = 12.98 : FA = 17.02 : NB= 77.7 d) FARMANA = 0.2 ( 190) = 38 N FB = MB NB = 0.2 (77.7) = 15.54 N : No, not in impeding motion. A and B are because FA < Frictional force of A. V We