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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 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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Feynman is using definite small quantities (inches) in place of infinitessimals $\delta x$ etc. Probably he wanted to avoid non-essential mathematical formality, in line with his casual, hand-waving persona. The Principle of Virtual Work requires the structure to undergo infinitessimal displacements (hence "virtual"). He could instead have used units of nanometres (or smaller) but that would also be too pernickety. Inches are appropriately small compared with feet.

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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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