Pulley rotational speed in block and tackle

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KentVibEngineer
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Homework Statement


For the pictured block and tackle system, formulate an equation to solve for the rotational speed of each sheave wheel for a given line pull speed. (ignore friction, slippage, line stretch) (Mass, force, efficiency, mechanical advantage are not the focus of this. Pulley speeds are desired for bearing fault detection by frequency spectrum analysis)
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Homework Equations


All pulley sheave diameters are same.

The Attempt at a Solution


Numbering sheave from draw line toward fixed rope anchor, I know 1 rotates fastest and 4 rotates slowest, but I can't determine the reduction % at each wheel.
 
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BvU said:
Hi Kent, :welcome:

PF rules dictate that you make an attempt at solution. If four pulleys is confusing, try two to begin with.

My actual application has 5 sets of pulleys above and below. But its not the number that is the issue, it is the complete lack of equations applicable to this topic. Sure there a tons of equations about the FORCE of pulley sets, but 2 weeks of scouring every bit of written material google can access got me nowhere.
[Mentor's note - edited to remove some discussion of the HW forum rules, which might be better in the forum feedback section]
 
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Forget the pulleys and focus on the amount of rope passing each pulley, starting at the right. The load rises a distance X. What happens to the rightmost vertical length of rope, and what length of rope passes the rightmost bottom pulley? That gives you its speed.

Now whatever length passes that rightmost bottom pulley must also pass all the other pulleys... But there's an additional length that must pass the second pulley in the chain (the rightmost top one) because the distance between that pulley and the first (bottom right) pulley has also decreased. Keep working from right to left and the general formula will become apparent pretty quickly.

(And a historical note: I first saw this as an extra-credit problem in a high-school math class long ago. I thought it was a fun problem then, and seeing it again, think it is still is).