Question

The light right-angle boom which supports the 460-kg cylinder is supported by three cables and a ball-and-socket joint at O attached to the vertical x-y surface. Determine the reactions at O and the cable tensions. 1.25 m ! -1.50 m 1.60 m E 1.60 m RE 1.95 m 0.75 m 0.75 m 460 kg Answers: Z Ox = i 1739.2812 TAc = i 3917.3 N Oyr i 31.27 N Ted = 2642.037 N02= i 4976.282 TBE = i 886.672 N Z

Answer 1

Sada Dorawing the FBD of the boom!. Oz 01 BTB TBD 45126 N Detine position Co-ordinates!. 0; (0,0,0) A:(0,0,1.95) B: (1.5, 0,1-95) C:(-1.25, 1.60,0) D:(1.50, 3.2, 0) G: (0.75, 0,1095) E: (ISO ,0) ů AC = TAC = (-1.25 i 11.00 h - 1.95 Ås Irach 1: 57.925 û ac =-0.444 î + 0.5683 Ĵ - 0.6926 R .

1: TAC = TAC (-orunu î +0.5683 Ĵ - 0-6926 K) € HSE-SŤ PoE = - TBE E û BD = 3.29 -1.95 ñ - 0.8539 ) - 0.5203k 3.7473 Top = Too (0.8539 Ĵ - 0-5203 Â) € w = -(4512-6 ġ)N MoĄ = 1.95k GOG = 0.75 û +1-95 ĥ TOB = 1-5 û +1-95 ĥ Applying moment equilibrium about 0'. { Mezo: (CA * TAC ) + (gloc, x W) (910g * Toe ) + (170B * TOOD=0

k1 0 195 0.75 0 1-95, t -0.444 0-5683 -0-6926 O -4512-6 0 C TBE + Tarl 1-5 0 0 0 1-95 / -1 15 0 1-95 0-8539 -0-5203 TAC (-1.108 î - 0-8658 ġ) + (8799.57 h - 3384.45 h) + Toe (1-5 ſ) + Too 61.665 û + 0-7804ſ +1-26808) as (-1.108 TAC - 1.665 TBD + 8799.57) î + 8658 TAC + 1-5 Ter + 0-780 +(1-2808 Teo - 3384-45) Â = 0 î . 5 f ŕ coeff: Equating -1108 TAC - 1.665 TBO + 8799-57 = 0 - -0-8658 TAC + 0-7804 TB + 1-5 TBE EO 1.2808 TBO. - 3384-45 = 0 -

- Salving eqr , & TAC = 3971.02 N TBD - २४५२-५७ N TBE - 917-29 N Applying force equilibrium! E Fx eo Ox -0.444 TAC = Ox = 1763-13 N EFy=0 Oy + 0.5683 Tac + 0.8539 Tan - 451-6 = 0 => Oy=0 EFZ-O', Oz - 0.6926 TAC - Tee - 0.520 3 TBD = 0 O2 = 5042-48 N