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I am a physics teacher, but have found myself teaching engineering. I have come across the following pulley question, and am starting to doubt my own understanding of pulleys !

Until now the rule that mechanical advantage equals 2*no. of moveable pulleys seems to have hold fast, but I am not sure that it works in this case. I am wondering if instead a better rule is that mechanical advantage equals no. of supporting ropes ?

For this particular question, would the no. of ropes supporting the load be three or four ?

I am not sure whether or not to count the far left section where the effort is applied.

Any thoughts on which is the most reliable rule, and how to apply it to this question, would be most gratefully received please. Thank you.

enter image description here

NMech
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Matt Klein
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  • Why not start by solving the force/distance/etc. for the first pulley, replace it with an equivalent weight applied to the next pulley, repeat until all pulleys accounted for? That is the most rigorous path to a solution as well as a mathematical proof. NMech's answer is certainly more straightforward and simple. – Carl Witthoft Apr 05 '21 at 14:16

1 Answers1

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The tension on the rope is everywhere the same and its equal to F.

So if you did a free body diagram on the following system by sectioning along the ropes:

enter image description here

what you get from the equilibrium is $4F = 48[N]$.

I hope that is sufficient as an explanation, I tend to find that problems with pulleys can have different configurations and as such it is always better to turn to the basics.

NMech
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  • Thank you NMech for the very clear answer. I'm not sure why I didn't think of that. I was placing too much emphasis on worrying about a rule. Anyway, it all makes sense now. I've ticked your answer. – Matt Klein Apr 03 '21 at 08:47
  • I've had my share of deadends and head scratching, so I am glad it benefitted someone. I can also understand the *lure* of a rule (play of words intended), however, when you teach this to people, its best to be able to adapt. Rules are good, but they invariably have exceptions. – NMech Apr 03 '21 at 09:14
  • I'm just a little sad in viewing your very interesting and clever answers affected by the plague of "units in square brackets". I humbly invite you to consider avoiding this mistake. This definitely improves the answer quality – carloc Apr 03 '21 at 12:34
  • @carloc thank you trying for trying to correct me on what you perceive a mistake. I like the styling with the square brackets, for me it serves a purpose and I am used to it (keep in mind that I am rather old and in some ways antiquated). I would nevertheless be open and genuinely interest to hear why you consider it a *plague*. – NMech Apr 03 '21 at 13:24
  • @NMech I thought that square brackets are for dimensions, e.g. mass or length, rather than units, e.g. kg or m. Also, there should be a space between the value and the unit (Wikipedia: SI Units](https://en.wikipedia.org/wiki/International_System_of_Units#Lexicographic_conventions). – Andrew Morton Apr 03 '21 at 18:31
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    It's not much about what I perceive as a mistake, I align to what at least NIST https://physics.nist.gov/cuu/pdf/sp811.pdf and BIPM https://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf give as guidelines. In short [*] is the "units of *" operator. It should be used in something like [L]=m for instance meaning that the quantity L is given in metres. Finally I apologize if you felt me rude in defining plague what you wrote, but again, IMHO your beautiful answers deserve this little final tune up – carloc Apr 04 '21 at 05:23
  • @carloc Had I felt you were rude, I would have flagged your comment and I would not bothered to reply. My interest is genuine. I felt however that you used (unnecessarily) a very strong wording, on a matter that I consider there is much leeway, and its up to a point a matter of preference. – NMech Apr 04 '21 at 06:21
  • To chip in, SI units are never italicised and this along with the space before units generally makes the distinction quite clear. $ P = VI = 12 \times 2 = 24 \ \text W $. See [Manual of style - variables](https://en.wikipedia.org/wiki/Wikipedia:Manual_of_Style/Mathematics#Variables). Regarding the space between the numbers and units, they're separate words so it's just the same as "35bananas" or "35 bananas". The second is correct. – Transistor Apr 04 '21 at 09:22
  • @Transistor, (just to make a point) what is the correct way to indicate multiplication $VI$ ,$V \times I$ or $V\cdot I$? If only one of them is correct, how would you feel, if I asked/insisted that you followed "the right way", even though it is pretty clear what you mean? I understand everybody's point, and I am quite happy to accept (without using judgemental words like *plague*) that everybody has different views and is entitled to how he is willing to express himself. – NMech Apr 04 '21 at 10:43
  • @NMech, I'd go for clarity and in those examples all are clear. $VI$ would be the simplest. "*... how would you feel ...*" I've had to modify my writing style as I learned the error of my ways! I now add a space between the numbers and units as per SI recommendation. I feel good, thanks! – Transistor Apr 04 '21 at 10:46
  • @Transistor My apologies if you felt I was being rude with *"... how would you feel..."*. It was never my intention. – NMech Apr 04 '21 at 11:22
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    No, I didn't take it as rude at all. I've read enough of your posts to know you a bit. My point in response to your polite quesiton was that I'm open to correction. – Transistor Apr 04 '21 at 11:32
  • @transistor I am glad about that, because I've also read a lot of your posts, I 've learned a lot from them and I have a high respect for you and your contributions, and I would not think of being intentionally rude to anyone, much less to someone I respect. – NMech Apr 04 '21 at 11:39
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    I am an old schooler too, but only use enclose the ending units in brackets, or parenthesis, for mixed numerical operation without carrying units in every step. For instance, M = 2 x 6 = 12 (N-m), or 12 [kips-ft]. – r13 Apr 04 '21 at 21:00