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In his book Chapter 4 Conservation of Energy, on Gravitational potential energy the discussion goes...

Take now the somewhat more complicated example shown in Fig. 4-6. A rod or bar, 8 feet long, is supported at one end. In the middle of the bar is a weight of 60 pounds, and at a distance of two feet from the support there is a weight of 100 pounds. How hard do we have to lift the end of the bar in order to keep it balanced, disregarding the weight of the bar? Suppose we put a pulley at one end and hang a weight on the pulley. How big would the weight W have to be in order for it to balance? We imagine that the weight falls any arbitrary distance — to make it easy for ourselves suppose it goes down 4 inches—how high would-the two load weights rise? The center rises 2 inches, and the point a quarter of the way from the fixed end lifts 1 inch. Therefore, the principle that the sum of the heights times the weights does not change tells us that the weight W times 4 inches down, plus 60 pounds times 2 inches up, plus 100 pounds times 1 inch has to add up to nothing: ...

But how do we know that the end point of the rod that is connected to the rope goes up 4 inches when the weight $W$ goes 4 inches down. My argument according to him is that if the weight $W$ goes 4 inches downwards. The rod is lifted a little less than 4 inches because the point where the rod and the rope is connected goes in a circular path. And the length of the path is 4 inches achieved by the weight $W$ by going downwards. Yet the concept of virtual work is true.

So, the 60 pound weight does not move 2 inch upward and the 100 pound weight does not move 1 inch upward. From triangle similarity if the rod moves vertically upwards with 4 inches, the rest of the weight moves according to Feynman's argument. But my argument is that it will be lesser than the value given in the book. So, if someone could help me resolve this, it'll be a very enriching experience.

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    $\begingroup$ Welcome to "the art of approximation" where $\sin x \approx \tan x \approx x$. :-) $\endgroup$ – CuriousOne Jul 1 '16 at 6:01
  • $\begingroup$ Another word for a mild cheat to complete the sentence, right? $\endgroup$ – Jyotishraj Thoudam Jul 1 '16 at 6:27
  • $\begingroup$ Very mild. It's really just an engineering problem... you could make some kind of cam-mechanism that keeps the string straight and compensates. It wouldn't change the physics, at all, and annoy the heck out of everyone. Look at it this way... you caught Feynman's tiny slight of hand, which makes you the smartest person in the room. That counts for something, for quite a bit, actually. $\endgroup$ – CuriousOne Jul 1 '16 at 6:58
  • $\begingroup$ @CuriousOne would you care to explain the solution presented in this link physics.stackexchange.com/q/265664. I tried to use the argument given above but I don't know how it'll work. $\endgroup$ – Jyotishraj Thoudam Jul 1 '16 at 13:37
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He made an approximation and didn't make it clear why (to avoid complicating things).

He had to move the bar a small amount to compare the movement of the pulley weight to the other two weights. He took an "arbitrary" small distance of 4" to make the math easy. In reality the bar doesn't really move much at all, the weights are resisting movement in both directions. The distance is infinitesimally small as sammy gerbil said. In that case the Small Angle Approximation applies, and $$\cos\theta \approx 1 - \frac {\theta ^2}{2}$$

and in this situation we have an infinitesimally small angle, so this theorem definitely applies, it also makes theta infinitesimally small, so $$\theta \approx 0\\\cos\theta \approx 1 - \frac {0}{2} = 1$$ if cosine is 1, so is the ratio of the two lines $$\cos\theta = \frac{adjacent}{hypotenuse}\\1 \approx \frac{adjacent}{hypotenuse}\\hypotenuse \approx adjacent$$ which means the distance of the rod from where it started should be essentially no further than when it began. I'm sure there was a 50x more elegant way to show that; but it's what I came up with.

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Feynman is using definite small quantities (inches) in place of infinitesimals $\delta x$ etc. Probably he wanted to avoid non-essential mathematical formality which would happen if he talked about 'infinitesimals.' This is in line with his casual, hand-waving persona.

He is imagining that the rotation of the beam is extremely small, so that the change in the direction of the rope attaching the end of the beam to the pulley is negligible. He is imagining that the displacements are infinitesimally small, which means that they are practically nothing. He could express these displacements in nano-inches or femto-inches or something even smaller. However instead he is measuring them in whole inches because it is a convenient unit and it doesn't matter what units we are using - the units all cancel out in the end.

It isn't the absolute value of the displacements which matter. (We do not need to know the value of work done in Joules.) It is only their ratios which are significant.

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The other answers are good, I just want to add that there's an audio version of the lectures selling on audible here (this one contains the Chapter 4) that shows that Feynman did explicitly state that this was an approximation, but got lost in the translation to the Feynman Lectures.

My crude transcription of the relevant section is as follows

...and we use a very small motion, because if I try to use 400 feet(s) here, it would get a little bit confuse(ing) by having gone around so many times *laugh* so we use a very tiny motion so it's easy to figure out that smaller the more easier it is to figure out. Actually, it goes in a curve and it isn't exactly 2 inches and so on, but you take an infinitesimal distance and that's called the principle of virtual work.

PS the audio chapter and the FLP chapter do not line up, they're all jumbled up, a conversion table can be found here

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