2.173 Given: In the configuration shown, the monkey on the right weighs twice that of the monkey on the left. The monkey on the right travels, relative to the rope, down 15 ft in 2 s. Required: The distance traveled by the monkey on the right, if it clings tightly to the rope. Solution: If we calculate the moments about the center of the pulley we can write HFrM&=×=∑∑. (1) ;irrA−=;jFMgA−=;irrB=;2jFMgB−=kMrMg−=∑Substituting this into equation (1) and integrating: ∑×==−vrHkmrMg;irrA−=;jvAAy&=;irrB=;jvBAy&=so BAyrmyrmrMgt&&2+−=−i.e. BAyygt&&2+−=−Integrating this equation w.r.t. t gives BAyygt∆+∆−=−2)2/1(2(2) If Monkey A moves up ∆yArelative to the ground in those two seconds, Monkey B moves down ∆yBplus the additional 15 ft relative to the ground. In other words ftyyAB15−∆−=∆
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This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.