Question

Please answer both if possible, thank you!

18 kN/m 12 kN/m 6 kN/m Fn = 300 N 12.AN/TITTTTTT6kN/m 0.5 m Fc = 225N 3 m .-3 m . 6 m Fig. for Q.2 0.5 m Fig. for Q.1 1. Two forces are acting on Point A along the direction as shown. Convert each force into Cartesian vector form. Show the various steps of your calculation. Then determine the magnitude and coordinate direction angles of the resultant. 15 Points 2. Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point A. Draw all the supporting diagrams (step by step). 15+5=20 Points

Answer 1

step 6 Co-ordnate of point A (0.5,-1.5, 0) B (-1.5, -2.5,2) CC-lis, OʻS, 3-5) Fg = 300 N T 3.5 m 0.5 m > Fc = 225N cao 0.5 m 1.5 m 1.5 m ② - Fig. for Q.1 amit vector calculation re-YA (-1.5-0.Slî + (-2.5=(-1.5))j + (2-0) FBA V 627 2+ G12+12 QAB = -2 0 - 1 3 3 J + 2 / 35 CAC - RCA amit vector Ac - re-VA (-1.5-0.5) T+ Co-5-(-1.5)) ſ + (3-5-0 lâ ✓ 6272+ (2)2 + (3.512 ů Ac = - 4 î + 7 9 + 73 k ③ force vector FÅR = FAB CAB = 300[ { 0 - 1 + { j ] Pag = - 200 l - 100j +200K FA a fac. ure = 225 [- 4 2 + 2 + 2 k] Face = -100@tlooſ + 175 Â |

Resultant Fr FR = FC + FAB z - 200 -100j +200k - 100 i +100j +175K IFR = -300 i +375 (FRL - V63.00) +(375)2 | FR = 457.19 N Direction angle of resultant Le cose che in ) = cote ( 2993) 2. 131° B = cover for me) = Coy . 10) = B= 900 Y= cost faza) = cost 31, 2013) Y = 34.89°

T8 kN/m 12 kN/m 6kNm 1. o 3m 3m 6m Resultant force = 12X3+ 18 x 6 + 6x3 = 162 KN FR = 162 KN 108kW 6x3=18 136kN Trisme tom Moment about o, FRX X = 36x1.5+ (0886+ 18x8.5 855 162 x = 5- 278 m Promo distanee from Az 5-278-3 TXA = 2.278m 1