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Volume of a boat problem

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    The design of boats is based on Archimedes' Principle, which states that the buoyant force on an object in water is equal to the weight of the water displaced. Suppose you want to build a sailboat whose hull is parabolic with cross section y=ax2, where a is a constant. Your boat will have length L and its maximum draft (the maximum vertical depth of any point of the boat beneath the water line) will be H. See the figure below.


    [PLAIN]https://webwork.math.nau.edu/webwork2_course_files/JLevy_137/tmp/gif/08_Webwork-prob6-q35fig.gif [Broken]

    Every cubic meter of water weighs 10000 newtons. What is the maximum possible weight for your boat and cargo?

    I really just don't know what to do with this problem, although I set up what I thought was correct.

    2. Relevant equations

    [itex]
    Volume (V) = 2 L \int^{\sqrt(\frac{h}{a})}_0 [h - ax^2] dx

    [/itex]


    3. The attempt at a solution
    [itex]
    V = 2 L \int^{\sqrt(\frac{h}{a})}_0 [h - ax^2] dx

    [/itex]

    [itex]
    V = 2 L [ hx - \frac{ax^3}{3} ]^{\sqrt(\frac{h}{a})}_0
    [/itex]

    The boat can be theoretically infinite in size if there's a big enough ocean to put it in, right?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 24, 2011 #2

    Mark44

    Staff: Mentor

    I don't know why you would say this. The three numbers, a, H, and L, are constants.

    Your expression for the volume is the same as what I get, but you need to evaluate your antiderivative.

    If you know the volume, that's the volume of water that the boat could displace, at most. How much would that volume of water weigh? That will be the maximum weight of the boat and its cargo.

    BTW, that's a really oddball shape for a boat of any kind, especially a sailboat!
     
    Last edited by a moderator: May 5, 2017
  4. Sep 24, 2011 #3

    LCKurtz

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    You need to double that integral to get the volume.

    The way I understand this problem, L and H are specified, and the only thing you have control over is the constant a. When you finish calculating your volume you get V as a function of a (H and L are not free to change). You want to maximize the volume by choosing a properly to displace the maximum volume of water.
     
  5. Sep 24, 2011 #4

    Mark44

    Staff: Mentor

    Good catch. My drawing shows the whole parabola, but the OP's integral was the same as mine - we both forgot to double the cross-sectional area.
    I interpreted the problem statement to mean that a was a constant, just like L and H, but your interpretation might be the correct one.
     
  6. Sep 24, 2011 #5
    I don't see where L and H are specified. The only information that it gives you is that 1 m3 of water is 10000 newtons.

    L and H are constants, but they're not specified. As far as I can tell, H is where y = (a constant) and provide the upper limit of the the top of the boat, but is not specified.

    The body of the boat,
    [ltex]
    f(x) = ax^2
    [/ltex]

    could be any squared value times a, depending on what H is. If there's a large enough body of water, the boat could have any volume as long as it displaced enough water.

    I copied the problem down word-for-word.

    btw, thanks for catching my mistake, LCKurtz!

    ***Edit***
    Apparently the constants, L,H, and A are what the question was looking for. Also, they are case-sensitive.

    smiley-bangheadonwall.gif
     
    Last edited by a moderator: Apr 14, 2017
  7. Sep 24, 2011 #6

    LCKurtz

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    After looking at this again, I think the problem is misunderstood, has something missing, or is a "trick" question. Because if H and L are fixed, whatever they may be, you can make the volume as large as you want by taking a sufficiently small.
     
  8. Sep 25, 2011 #7
    So now I'm finishing the evaluation:

    Original:
    [itex]
    V = 2L [Hx - \frac{a}{3} x^3]^{\sqrt(\frac{H}{a})}_0
    [/itex]

    [itex]
    V = 2L [H{\sqrt(\frac{H}{a})} - \frac{a}{3} {\sqrt(\frac{H}{a})}^3]
    [/itex]

    [itex]
    V = 2L [H{\sqrt(\frac{H}{a})} - \frac{a}{3} \frac{H}{a} {\sqrt(\frac{H}{a})}]
    [/itex]

    [itex]
    V = 2L [H{\sqrt(\frac{H}{a})} - \frac{H}{3} {\sqrt(\frac{H}{a})}]
    [/itex]

    [itex]
    V = 2LH{\sqrt(\frac{H}{a})} [\frac{3}{3} - \frac{1}{3} ]
    [/itex]

    [itex]
    V = 2LH{\sqrt(\frac{H}{a})} \frac{2}{3}
    [/itex]


    [itex]
    V = \frac{4}{3}LH{\sqrt(\frac{H}{a})}
    [/itex]

    "the answer above is NOT correct."

    [itex]
    \frac{4}{3}LH{\sqrt(\frac{H}{a})} * 10000
    [/itex] Newtons

    Correct.

    Thanks for all your help, guys!
     
    Last edited: Sep 25, 2011
  9. Sep 25, 2011 #8

    LCKurtz

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    Yes, that is the correct formula for the volume in terms of L, H, and a, for what it's worth.
     
  10. Sep 25, 2011 #9
    lol, I forgot the units and the 10,000. Thanks a ton! You've been a great help.
     
  11. Sep 25, 2011 #10

    LCKurtz

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    No problem. I think Mark44's interpretation must have been correct in the first place. It isn't an optimization problem, just a straightforward calculation.
     
  12. Sep 25, 2011 #11
    Agreed. Tip of the hat to Mark44!
     
  13. Mar 13, 2013 #12
    I stumbled across this problem while trying to figure out my own similar problem. I decided to post it here instead of creating a new topic.

    "You are designing the hull of a sailboat with parabolic cross sections of y=ax2, where a is a constant. Your boat will have length of 20 feet and the maximum vertical depth will be 10 feet. What is the volume of the hull?"

    Will I just be plugging in the length and maximum vertical depth for L and H in the equation provided by Ocasta?

    [itex]V=
    \frac{4}{3}20*10{\sqrt(\frac{10}{a})}
    [/itex]
     
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