Question

4 ft 6.2 ft 4 ft 7.8 ft

Boom AC has negligible weight and is in equilibrium. The boom is supported by cables at D and E and a ball-and-socket joint at A. The suspended crate weighs 114 lb.

What is the tension (lb) in cable BDC?

What is the tension (lb) in cable CE?
What is the x component of the reaction (lb) (exerted on the boom by the wall) at A?

What is the y component of the reaction (lb) at A?

What is the z component of the reaction (lb) A?

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Answer #1

Gioon datas weight we 119 lb Consider the the sustem as shown In the figur below find the position 11 alb vector co Co = (-3Too = Top 1-êt 12.02 Too Too. (-01234 ê t 0.92 ŷ 4.912 el en find the position vector co as CE = 6e-1118 9 7 6.0² ) At LLLLL2TOD IDD = h tar TOO agit 4 J (-317 (-411 + a2/ TBD = TAD lagê -a ↑ + 4 f) TBP = Tool -0.968 ê -0.625 ↑ +0.625 Ê ] Que ResolveResolve the foro alo 2 direction Στο Azt 0.312 Teo to.95 te to.625 To - 119 50 616 Calccelek the momond about a axis Ema - 0.Salve es 7 & tension in Cable AD = tension in Cable co (Pulley System) TAD - TCD So from eg o 2.5 TBD - 1345.2 + 5.31 TCE +3.soon as of Ат - о. 21 х , , , , , , , , 6 4 2 4 9 22 x 152, 15 — 0.4664, 43 - 0 у 27. бсъдь , , , , , Със с G) — Ач - 6 92 x

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