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The 4 Monkeys

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The 201 rungs monkey bar hangs on the mid rung to keep balanced horizontally. Four monkeys are leveled ..(see illustration)
the 3 lbs stands and the 1 lbs hangs on the left end while the 2 lbs stands and another 3 lbs hangs at 20 rungs away from right end.
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If they all move (swing or walk) to the far ends at uniform rate of 1 rung/sec (they can pass each other),one of them shall move
at different uniform speed to keep the monkey bar from tilting. What then is the speed of the (fast or slow) monkey??
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3 answers to this question

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The center of gravity (c.g.)of both 3# monkeys is 90 rungs from left of the bar: (100+80)/2=90 or 10 rungs from the center hanger. While the c.g. of 1&2# monkeys is 60 rungs from the right of the bar:180/3=60 or 20 rungs from the center hanger.Thus, a torque balance at 10x6# = 20X3#. If both 3# monkeys move at same speed, their c.g. remains at the same point..so to balance the bar or keep the c.g. of 1 &2# monkeys at the same point, only the 2# monkey must walk half rung/sec. or else only the 1# monkey must swing two rungs/sec.

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Posted (edited) · Report post

you posted the answer too!

realized now that ts an old problem :P

Edited by m00li
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