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Crane and Boom Review 33. Looking at the Forces at X. Hori Vert sign –2000 chain –C∙cos26 C∙ sin26 boom B∙cos46 B∙sin46. –C∙ cos26 + B ∙ cos46 = 0. –2000 + C∙sin26 + B∙sin 46 = 0. Solve the Equations. –C∙cos26 + B∙ cos46 = 0
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Looking at the Forces at X Hori Vert sign –2000 chain –C∙cos26 C∙sin26 boom B∙cos46 B∙sin46 –C∙cos26 + B ∙cos46 = 0 –2000 + C∙sin26 + B∙sin 46 = 0
Solve the Equations • –C∙cos26 + B∙cos46 = 0 • B∙cos46 = C∙cos26 • B = C∙cos26 / cos46 = 1.294 C • –2000 + C∙sin26 + B ∙sin 46 = 0 • C∙sin26 + B ∙ sin 46 = 2000 • C∙ sin26 + 1.294C ∙ sin 46 = 2000 • C∙ .438 + 1.294 C ∙ .719 = 2000 • .438 C + .931 C = 2000 • 1.369 C = 2000 C = 2000/1.369 = 1461 B = 1.294∙C = 1890
The Torque About A is Zero T = F x s distance perpdicular to force cos 46 = AU / AX AU = AX cos 46 sin 72 = AY / AX AY = AX sin 72 T of sign = 2000 ∙ AX ∙ cos46 T of chain = C∙AY = C ∙ AX sin 72
Net Torque is Zero • Torqe • sign = − 2000 ∙ AX ∙ cos46 (clockwise) • chain = C ∙ AX sin 72 (counter clockwise) • They total zero • 2000 ∙ AX ∙ cos46 = C ∙ AX sin 72 • 2000 ∙ cos46 = C ∙ sin 72 • 2000 ∙ .695 = C ∙ 0.951 • C = 2000 ∙ .695 / 0.951 = 1460.8