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  1. #1
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    Re: An Hourglass Puzzle

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">I'd say it's akin to (literally) jumping on the bathroom scales and registering a higher weight than just standing on them. In the case of the balance pan, it is providing a reaction force to the static weight of the hourglass, plus a force equivalent to the rate of momentum change of the sand as it comes to rest. It therefore "feels" extra weight compared to the static-weight-only felt by the other pan.</span hide>

    ... I think.

    Alan

  2. #2
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    Re: An Hourglass Puzzle

    <img src=/S/hmmn.gif border=0 alt=hmmn width=15 height=15> I am going to come from a different angle to Alan and will make the following assuptions.

    1) It is in a vacuum
    2) There is no friction on the fucrum and balance

    In diagram A the centre of gravity for the identical hourglasses is exactly the same place ie the bottom bulb and we can epect that the balance board is exactly horizontal

    However, if in diagram B the balance board is perfectly horizontal then the centres of gravity should pass through the samer point, but if there is a slight deviation in the board angle the centre of gravity will move out to the left of the balance board by dx (where x is the distance of the hourglass from the fulcrum)

    This dx will be an increase from the centre and will cause the left hand to move down, similarly if the board tips to the right it will not balance
    Jerry

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    An Hourglass Puzzle

    Two identical hourglasses with all sand in the bottom half are placed on a balance scale. As the hourglasses are identical and all of the sand is in the bottom half, the scale is, in fact, balanced. If one of the hourglasses is flipped over, putting the sand in the top half, does the scale stay "balanced" or will it only move back to a balanced position after all the sand in the hourglass has fallen? The hourglasses are sealed, "air-tight" containers.

    <img src=/S/bubbles.gif border=0 alt=bubbles width=31 height=17>
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    Re: An Hourglass Puzzle

    HG_var = hourgass with sand in top; HG_control = hourglass with sand in the bottom

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold"> when the sand is first released in HG_var, HG_var actually becomes lighter than HG_control as there is sand falling through the air, not being weighed. So HG_var is lighter so the scale tips toward HG_control. When the freefalling sand begins to land on the bottom, the force of the impact all but negates the loss of mass from the freefalling sand so the scale is very close to balanced. When the freefalling sand is "running-out" (as it were) there is less sand in freefall so the force of the impacting sand causes the scale to tip towards HG_var. Of course, when everything is again in a resting state the scale is balanced.</span hide>
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  5. #5
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    Re: An Hourglass Puzzle

    The plot thickens...

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Thinking (hopefully) more clearly about this one, I think the two would always be in balance. Regardless of the "internals" of the closed system (hourglass) the mass is always the same. It's probably going to end up that the reduced "resting" mass resulting from the airborn sand is going to be exactly compensated by its momentum change contribution, when it comes to a stop. Regardless, this is a very interesting combination of phenomena.</span hide>

    Alan

  6. #6
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    Re: An Hourglass Puzzle

    I'm not saying you are incorrect...

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Using that rationale, think how funny it would be if I were shipping you a case of hourglasses. I walk into the post office and set the case of hourglasses [upside down] on the precision postal scale. I say to the clerk, please check the weight quickly <img src=/S/smile.gif border=0 alt=smile width=15 height=15>, under the assumption that the case weighs less while the sand is falling. Cheaper postage! </span hide>

    <img src=/S/hmmn.gif border=0 alt=hmmn width=15 height=15>
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  7. #7
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    Re: An Hourglass Puzzle

    Well, I don't think that would happen as the total weight of the system is unchanged.
    Also I did a search and found this. <img src=/S/shrug.gif border=0 alt=shrug width=39 height=15>
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  8. #8
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    Re: An Hourglass Puzzle

    Wow! I had no idea that Galileo had already tackled this issue and made it a "famous" puzzle.

    I have two daughters in two different phases of physics and I thought of this "problem" while having discussions at the dinner table. It created several tracks of thought and made dinner lasts a bit longer than usual. We each came to different conclusions with one daughter promising to deliver the question to her physics teacher...

    If there is or was a "correct" answer to the question, I wasn't sure of it. The link you've provided shot down my theory that an hourglass weighs the same regardless of it's positioning. Your choccy bar should arrive shortly; enjoy. <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

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  9. #9
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    Re: An Hourglass Puzzle

    Thank you. <img src=/S/cool.gif border=0 alt=cool width=15 height=15> It will come in handy for a snack a little later. <img src=/S/thumbup.gif border=0 alt=thumbup width=15 height=15>
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