# qz 32 - .QQO‘ 6 ﬂEQ—D‘ﬁv MU‘QC‘ ENGINEERING...

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Unformatted text preview: .QQO‘ 6 ﬂEQ—D‘ﬁv MU‘QC‘ ENGINEERING MECHANICS - STATICS, 2nd. Ed. F. RILEY AND L. D. STURGES 6-65* The crane and boom, shown in Fig. P6-65, weigh 12,000 lb and 600 lb, respectively. When the boom is in the position shown, determine (a) The maximum load that can be lifted by the crane. (b) The tension in the cable used to raise and lower the boom when the load being lifted is 3600 lb. (c) The pin reaction at boom support A when the load . being lifted is 3600 lb. 3 SOLUTION (a) From a free-body diagram for the crane: + T SE = N - wc - wB - wL Y = N - 12,000 - soo - wL = o N = 12,600 + wL lb + C EMC = wc(9) — "3‘12 cos 30° - 1) - "1‘24 cos 30° + 1 — 1) - Ndx 12,000(9) - 600(12 cos 30° - 1) - "1‘24 cos 30° + 1 - 1) - (12,600 + WL)dx 102,365 - 20.785 WL - (12,600 + "lex = 0 For impending tipping: dx = 0 102,365 WL(max) = 20.785 = 4925 lb Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-65 (Continued) (b) From- a. free-body diagram for the pulley at B: + -e :0( B - 3600 cos 10° = 0 3545 lb 2'! + T 2Fy By — 3600 - 3600 sin ¢ = 0 4225 lb From a free-body diagram for the boom: + C EMA = 3x(24 sin 30°) — By(24 cos 30°) — "3‘12 cos 30°) + 1(24 sin 20°) = o 3545(24 sin 30°) — 4225(24 cos 30°) — 600(12 cos 30°) + 1(24 sin 20°) = o 6275 lb g 6.28 kip (c) + —e EFX = A — Bx - T cos 10° X = A - 3545 - 6275 cos-10° = o 9725 lb X + 1 SE = A - B - T sin 10° - 600 V V V = Ay - 4225 - 6275 sin 10° - 600 = 0 Av 5915 lb /(Ax)2 + (Av)2 = /(9725)2 + (5915)2 = 11,333 1b a 11.38 kip —1 5915 o o 9725 - 31.31 3 31.3 - tan 9.73 i kip = 11.33 kip a 31.3° ...
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