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Bouyancy Question

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A large number of people are on a sinking ship, and decide to use a large crate with an open top as a lifeboat.
    A physicist in the crowd has been tasked with finding how many people can board the craft before it would sink. She assumes the average person has a mass of 70kg.
    The crate measures 5metres long by 3metres wide by 2metres high and has a mass of 19 tonnes.
    How many average people can the crate safely hold before the top will drop below the surface, and water comes comes pouring in?

    2. Relevant equations
    I'm assuming that
    are going to be used somehow.

    3. The attempt at a solution
    Well, I've had a lot of problems with this question. None of my classmates, nor myself, have been taught how to do a bouyancy question, and the question above appeared in our homework. I've looked up how to do bouyancy questions on the Internet, with some success. I don't really understand what I'm doing, but this is what I have done so far, but I'm not sure how/where to continue.

    Volume of the crate = 30m^3, (as 5*3*2).
    Weight of the crate = 186200N, (as 19000kg[19tonnes]*9.8)
    Pressure on the bottom of the crate = 12413.33... (as 186200N/15m^2)

    This is as far as I've got, and I don't know if I am doing this correctly, and/or where to continue.
  2. jcsd
  3. Oct 26, 2014 #2


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    Do you remember the Principle of Archimedes? If you do, state it right now here...
  4. Oct 26, 2014 #3
    As I've said, my class hasn't been taught how to do bouyancy questions yet. I looked over my Notebook incase I completely forgot about doing it, but there is absoltuely nothing when it comes to bouyancy. When I searched up the Principal on the Internet, it looked to me as if it can only be used for an object that is fully submerged? My question is asking how many people can fit in the crate, before it submerges.
  5. Oct 26, 2014 #4


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    The principle of Archimedes is applicable to everything that is submerged, either completely or in part, in a fluid. Quote it...
  6. Oct 26, 2014 #5
    Archemides Principle states that the upward bouyancy force is exerted on a body immersed in liquid, is equal to the weight of the fluid that the body displaces.

    So, does that mean I need to calculate the weight of the displaces liquid? Do I use the equation:

    V= (Normal Weight in air - Weight when in the fluid)/ gp?

    If so, that gives me the bouyancy of the object? How do I then go about answering the question?
  7. Oct 26, 2014 #6


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    Yes, the upward bouyancy force is exerted on a body immersed in liquid, is equal to the weight of the fluid that the body displaces.

    Now imagine that you place the empty crate on he water. It will sink somewhat. Exactly enough to balance its weight

    ¿What's the volume of water that weighs the same as the empty crate? ¿How much will the empty crate sink when put on the water..?
  8. Oct 26, 2014 #7
    Okay, I think I get that part... So, because the bouyancy force is equal to the weight of the fluid, I'd use

    Fb=gpV, which is 9.8*1.025(Salt water)*30=301.25, which means that the weight of the displaced liquid is 301.25N? Continuing from that, does that make the volume of the water 301.25L?
  9. Oct 26, 2014 #8


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    The empty crate weighs186200 N.

    That's the weight (on the Earth) of a mass of 186200/9,8 = 19000kg.

    It's the same as the mass of a volume of 19 m3 of pure water.

    But since salt water has a higher density, we would need displacing less volume of liquid in order to float the empty crate: 19 * (1,0000/1,0250) = 18,54 m3

    Now, we know the volume that the crate has to displace to stay afloat

    ¿How much has that crate to sink in order to dispace exactly that volume?
  10. Oct 26, 2014 #9
    Well, I assuming that all you do for that is 18.54m3/15m2=1.236m. So, it has sunk 1.236 metres, meaning that 0.764 metres of the crate isn't submerged?
  11. Oct 26, 2014 #10


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    Those 0,764 meters are an indicator of the 'reserve buoyancy' that you still have. Not its value, but an indicator...
  12. Oct 26, 2014 #11
    Right... So I'm guessing that you would need to find how many people can be added into the crate? Like, for 1 person, it would be 19*(10070/10250)? =Ans. Then, Ans/15? Then, just continue going until that value goes to 0 or less?
  13. Oct 26, 2014 #12


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    Once the crate is floating empty, you have 0,764 m left above the water level. If submerged to the rim, the extra displacement will be:

    0,764 * 5 * 3 = 11,46 m3. Now you can calculate how much does that volume weighs, in Newtons of course... That will be the 'maximum loading capacity'...

    You want to put on board a number of 'average people' with a mass of 70 kg each... You know that means an individual weight of 70 * 9,8 = 686 N

    Little is left to find the solution...
  14. Oct 26, 2014 #13
    Okay, I think I understand it now (hopefully!)

    So, 11,46m3 is equivalent to 11460kg. Then, to find the Weight of that, I'd times it by 9.8, giving me 112308N. Then, i'd just divide that by 686?
  15. Oct 26, 2014 #14


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