Talk:Atwood machine

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Simple but effective picture (just to give some feedback). Maybe give the masses labels, to go with the text, but it might only add an unnecessary layer of complexity. I mean, how many parts are there to lose track of? Femto 14:14, 16 Oct 2004 (UTC)

Gravity[edit]

If m2 is heavier than m1, shouldn't it be being pulled down by gravity? Meaning there should be a negative sign before the part of the m2 formula containing the gravity constant?

Or overall, shouldn't the first two formulas for the forces affecting m1 and m2 be switched?

Well, absolutely. But you must understand that our choice of which object is m1 and which object is m2 is arbitrary. Additionally, which object is accelerating upward is arbitrary as well. I will try to explain this with the concept of a sign convention.Carultch (talk) 02:49, 24 May 2010 (UTC)[reply]

Demonstration[edit]

Ok, if this is a popular demonstration then it surely isn't right to describe it as involving inelastic strings and ideal pulleys. I'll fix this if I can. Theoh 19:11, 6 January 2007 (UTC)[reply]

It's not so much a demonstration as an ideal model. At least that's how I remember from my I took intro Physics. It's used in labs, too, actually, and the non-ideal conditions are taken into account as experimental error. --Ttownfeen 00:04, 10 January 2007 (UTC)[reply]

Can someone please expand the section for 'Equation for an Ideal Pulley'? We just went over the Atwood machine in a Lab and also just learned about angular mom./acc./vel. and moment of Inertia and I found this part to be a bit vague. It'd be nice to have how to account for the pulley in the experiment/machine demonstrated. The rest of the page is well done and makes sense. --Mike Benton (talk) 05:29, 26 October 2008 (UTC)[reply]

Sorry, but what about...[edit]

...a point of application, direction and sense for a vector "sigma F"? Tadeusz Malinowski (talk) 21:43, 28 January 2009 (UTC)[reply]

Joke[edit]

Has anyone heard the joke about the Atwood machine? A medical student asked his physics professor, "why do we have to learn how to solve the Atwood machine? It seems so pointless.". The physics professor replied "So those who can't solve Atwood machine won't become doctors". Does anyone know the history of this joke..? Couldn't find it online. Danski14^(talk) 02:23, 29 January 2012 (UTC)[reply]

error[edit]

This, from the section on elevators, is wrong, " it has to overcome only weight difference and inertia of the two masses." The basic principle of the Atwood Machine is that the inertia of both masses still has to be overcome. I would change it but have been topic banned in other areas and have found some editors to be rather nasty. I suggest you look into it.

Arydberg (talk) 13:55, 2 July 2012 (UTC)[reply]

undefined term "traction"[edit]

The section where pulley inertia and friction are included has no segue from the previous. Worse, the term "traction" is introduced without link or explanation of what it means. It cannot be expected that readers for whom the earlier section is new will have the background to make a connection here where the authors make none. Zoetropo (talk) 08:04, 5 September 2017 (UTC)[reply]

d[edit]

What is 'd' in the equation for deriving acceleration (and hence g) from timing of the movement? I presume it stands for 'distance' or 'difference', i.e. the distance moved by the weights in the observed time, but it should not be introduced without explanation. So far as I recall, the only standard meaning of 'd' in mathematics is 'differential' or 'derivative'. 'Distance' is usually expressed by 's', for 'space', as in v = s/t.2A00:23C8:7904:EE01:6CD9:BB5F:C966:A696 (talk) 16:51, 16 July 2020 (UTC)[reply]

Thanks for having noticed this. I have removed the claim ([1]), which wasn't even supported by the cited source, whiich wasn't even reliable. - DVdm (talk) 17:38, 16 July 2020 (UTC)[reply]

Old explanation[edit]

I found an explanation of the machine in an old textbook (Deschanel's 'Natural Philosophy') which I found intuitively clearer, if maybe less rigorous, than the one in the article. Suppose that initially there are equal weights, each with mass P, on both sides of the machine. They will balance each other so there will be no movement. Now add a small weight with mass p to one side, so that on one side the mass is P and on the other it is P + p. The extra weight p will set the machine in motion, with one side rising, and the other falling, with some velocity v (disregarding sign). The two weights P still balance each other, so the motion must be entirely attributed to the extra weight p. But that has to move the combined mass of (2P + p) instead of just its own mass p. So if the effect of gravity on p alone would be g, the effect on the combined mass will be g [p/(2P + p)] . The result in substance is the same as in the article, but to me at least the explanation seems clearer.2A00:23C8:7904:EE01:3D91:74B2:BAAB:1F06 (talk) 12:00, 17 July 2020 (UTC)[reply]

Pulley Inertia - edit required in explanation[edit]

The need to include the inertia of the pulley is currently described as follows: "For very small mass differences between m1 and m2, the rotational inertia I of the pulley of radius r cannot be neglected." This must be incorrect, as the inertia only appears in the part of equation that involves the sum of the masses.PhysicistQuery (talk) 01:35, 16 February 2021 (UTC)[reply]