Answer to the Friday Puzzle…..

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coverOn Friday I set this puzzle….

A boat is at anchor.  There is a rope ladder hanging over the side of the boat, and the rungs of the ladder are a foot apart.  The sea is rising at a rate of 15 inches per hour.  After 6 hours, how much of the rope ladder will remain above the water, assuming that 10 feet of the ladder was above the water when the tide began to rise?
If you have not tried to solve it, have a go now.  For everyone else the answer is after the break….
Since the boat is afloat, the water level in  relation to the ship stays the same. Therefore, 10 feet of the ladder always shows above the water!  Did you solve it?
I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.

38 comments on “Answer to the Friday Puzzle…..

  1. Stan says:

    yes, but then I went into the meta business of wondering why you asked it.

  2. Heather wiseman says:

    Ech…. Bit easy!

  3. Anne Elk says:

    As I said on Friday, a nice bit of puzzle nostalgia. I haven’t seen that one since I was at school. (Now let’s wait for the debates!)

  4. Ike Newborough says:

    Could a terrible pedant argue that because the boat is that bit further away from the centre of the Earth, the gravitational attraction between the two objects is fractionally weaker and so the boat is fractionally higher in the water? I wouldn’t argue that, of course; my answer was the same as Richard’s.

    • Dougal McTavish says:

      An even more terrible pedant could point out that the boat will sink until it’s displaced its own weight in water. The surrounding water is also fractionally further way from the centre of the earth, so the boat will displace just as much

    • Ike Newborough says:

      That’s true, Dougal, and I confess that I hadn’t thought that far ahead in my pedantic (but not quite pedantic enough) analysis. That said, because the centre of gravity of the boat is slightly further away from the centre of the Earth than the centre of gravity of the water, and the gravitational relationships are proportional to the square of the distances, the reduction in the pull of the boat is disproportionately higher. I think. I could be wrong and would be happy to be put right. This reply is an attempt to save some face. This all puts me in mind of the old puzzle about how many boats can safely cross an aqueduct.

    • Steve Jones says:

      A much better argument is that the rising of boat relative to the seabed means that more of the anchor chain will have to be lifted from the sea floor. This means that the boat will have to displace more water to counterbalance the extra weight (assuming, as is reasonable, that the anchor chain is denser than seawater). For this reason, at high tide, the ladder will be slightly closer to the waterline than at low tide.

      Only if the boat is tied off to a mooring buoy, or the like, would this be untrue.

    • You’re right, Ike. A terrible pedant will always argue things like that. It doesn’t make them right though…

    • Steve Jones says:

      nb. need to add to my post (before anyboy else points it out), is that even if the anchor chain is fully raised off the seabed at low tide, the rising tide will raise the boat and therefore put more tension on the anchor chain which would also lower cause the boat to float lower than at low tide.

    • Ike Newborough says:

      Withering way you put it aside, greatproductmanagement, you make a fair point. But was it to my original post or to my follow-up? I’d still be interested to hear from any Newtonian and/or Archimedian experts (or even amateurs) generous enough to elaborate on whether the boat’s distance from the centre of the Earth has any microscopic bearing on its position relative to the surface of the water. I know the formulae (or I could at least manipulate them if I looked them up) and the principles at stake (when I pause for thought). But I don’t have the time to go through the formulae nor do I have quite the intuitive feel that some people have as to whether the effects of Archimedes’ principles entirely or only partially balance things out when the sea level changes. Thinking about it with all my concentration before clicking “post comment” for a third and hopefully final time, I think I now appreciate that my second comment still misses the point and that all that’s relevant is the centre of gravity of the water that the boat displaces, not the general centre of gravity of the body of water that it’s in. Am I right?

    • Ike Newborough says:

      My word, has this been nagging away at the back of my mind when I should have been concentrating on my boring job? Yes is the definitive and undeniable answer to that. Embarrassed by my first blunder, I’ve been trying to modify Dougal McTavish’s eloquent reply, which I know does undermine the general point in my first post. But is there a valid refinement to his comment? I admitted defeat in my third post but now I’ve reconsidered my second post. I’m imagining a flat and solid cuboid of ultra-light polystyrene. If that lay on the surface of a body of water, I can intuitively feel that its centre of gravity is higher than that of the water it’s displaced. Am I going mad or is that right? (I might be going mad for other reasons, of course.) Or does surface tension explain that away? No! Thinking about it, anything that’s solid and regular in shape that floats must by definition have a higher centre of gravity than the water it displaces. So my point about squares of distances might have some substance. Unless I’m missing something eureka-like about the nature of the water that’s displaced. By the way, people, I’m well aware that I’m basically conversing with myself now. Heat, work stress and impending fatherhood is not a conducive combination for retaining sanity.

    • Ike Newborough says:

      No, I think I’m talking *******s. But I still can’t rid these nagging doubts. Thank goodness I used a rubbish pseudonym.

  5. mgm75 says:

    I didn’t see it Friday but I got it before I even read to the end. A great trick question!

  6. Baker's Dozen says:

    Not if the boat is tightly anchored.

    #FAIL

  7. Stan says:

    The first word of the puzzle is “A boat is at anchor”. Suppose the anchor weighs as much as the boat (an unusual case but useful for thinking). As the tide wants to raise the boat the anchor will resist. So doesn’t it all depend on the weight of the anchor and shape of the boat.

  8. Hugh Jarse says:

    Good grief! There are clearly too many people with too much time on their hands

  9. Ralf says:

    Nice trick question. But I refuse to even think about puzzles that have “foot” and “inches” in it.
    We are in the year 2013.

  10. JeffJo says:

    Speaking of pendants: The cycle of the tides is approximately twelve hours. We can loosely approximate the sea’s height, relative to its average, by H=H0*sin(t*pi/12), where H0 is half of the total range of the tides. That means it changes at a rate of H0*pi/12*cos(t*pi/12)

    In the Bay of Fundy, H0 has been measured at 335 inches. That means it is rising at 15 inches/hour approximately 40 minutes before high tide. Five hours, 20 minutes after that, the boat may be resting on the sea floor, with a dry (but muddy) ladder.

    It only takes H0~=7 feet for the boat to be lower after six hours.

    • JeffJo says:

      Oh, dear, I think I left a factor of two out of that. Too early in the morning (obviously) to fix it, though.

    • Dave says:

      I don’t think anyone was speaking of pendants.

      A little pedantry for you there.

  11. Steve says:

    I worked out an answer quite quickly and then straight after realised my ‘mistake’. Very interesting comments about this from nearly everyone.

  12. de N'Hug says:

    Yes, nostalgia. I remember that when I was boy I explained this one to a gypsy. The happy old days.

  13. Gus Snarp says:

    Damn it, I fell for it.

  14. Henry says:

    Richard conveniently left out the fact that in 6 hours all the of guests for my onboard dinner party will have arrived and their weight will lower the relative height of the boat. How many rungs lower will we be? Without knowing the weights we can’t know. Do we care? No, as long as the beer stays dry.

  15. JeffJo says:

    So much for giving hints. Twice now, with no notice.

    If the tide is rising now, in six hours it will (A) be receding, and (B) be lower than it is now. The ladder is either just as high above the water level, or higher if it reaches the sea floor.

  16. My wife and I correctly agreed immediately. Brilliant!

  17. john jahanpour-burke says:

    I Loved This Puzzle. Me And My Sister Were The Only Ones Of My Family Who Got This In About 10 Seconds.

  18. issahamati says:

    I correctly agreed

  19. Miss Chili says:

    Despite having gotten this puzzle solved in as much time as it took to read it, I still feel like bonking my desk with my head repeatedly.

  20. Evan Cohrs says:

    Regarding the boat riddle; If the boat was at anchor, (depending on the size of the boat,) wouldn’t it stay the same level as the water rose?

    – Evan Cohrs

    • Steve says:

      Hi Evans. It depends I suppose on whether the boat was so heavy it was stuck on the bottom or perhaps it had a hole in it and so was technically sunk. Perhaps the clue to the answer is that the “boat is at anchor”. Unless you are fairly knowledgeable on boats etc. that statement might be a bit ambiguous.

    • Steve says:

      Hi Jim88. Some more words please. Was it the wrong answer or the right answer that was easy ?

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