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Discover the step-by-step process to solve equations and calculate forces and torques in crane operations. Learn how to analyze and balance the forces exerted on cranes, including the impact of angles and distances. Enhance your understanding of mechanical principles in crane engineering.
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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
