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Hole in a boat

  1. Mar 21, 2017 #1
    1. The problem statement, all variables and given/known data
    The bottom of the boat has dimensions a = 50 m, b = 10 m, the height of the ship is c = 5m. Impact created in the bottom a circular hole with a diameter d = 20 mm. The initial height of the upper edge of the ship above the water level is h = 3.5 m. The ship is empty and open at the top.
    a/ Show that difference between the levels inside and outside the ship is not changed
    b/ Determine the velocity of water in the hole.
    c/ Calculate how long the boat sinks.

    2. Relevant equations
    a/ ......
    b/ 0,5 ρ v2 = hρg v = √ 2gh

    c/ volume of boat......V = abc
    flat of hole......S = π r2

    3. The attempt at a solution
    b/ v = √2*9,9-81*3,5 = 8,28 m/s
    c/ V = abc = 50*10*5= 2500 m3
    S = π * 0,022 = 0,00126 m2
    in 1 s.......V1 = v*S = 8,28*0,00126 = 0,01 m3
    T = V /V1 = 2500/0,01 = 250 000 s = 69,44 hour
     
  2. jcsd
  3. Mar 21, 2017 #2

    mfb

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    You used the diameter of the hole as radius.

    The ship will sink before water flowing through the hole filled the whole interior.
     
  4. Mar 21, 2017 #3
    yes :-(

    S = π * 0,012 = 0,000314 m2
    in 1 s.......V1 = v*S = 8,28*0,000314 = 0,0026 m3
    T = V /V1 = 2500/0,0026 = 961 538,5 s = 267 hour
     
  5. Mar 21, 2017 #4

    BvU

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    How much water do you need in there before it sinks ?
     
  6. Mar 21, 2017 #5
    when Fg > FB.......m * g > V * ϱ * g ....?
     
  7. Mar 21, 2017 #6

    BvU

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    You want us to guess what you mean with these symbols ?

    In terms of variables that we have encountered and explained ( :rolleyes: ) so far !
     
  8. Mar 21, 2017 #7
    Fg...gravity
    FB...buoyancy force
    m......water weight...not in whole ship, only flooded part
    V.....the volume of whole ship
    ϱ.....density of water
    g....9,81 m.s2
     
  9. Mar 21, 2017 #8

    BvU

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    So what follows from post #5 ?
     
  10. Mar 21, 2017 #9
    m>V * ϱ.......m>V * ϱ = abc*ϱ = 50*10*5*1000 = 2 500 000 kg of water ?
     
  11. Mar 21, 2017 #10

    BvU

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    Nope
     
  12. Mar 21, 2017 #11
    so I do not know :-(
     
  13. Mar 21, 2017 #12

    BvU

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  14. Mar 22, 2017 #13

    haruspex

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    Did you answer part a? That makes it easy to answer BvU's question.
     
  15. Mar 22, 2017 #14
    Unfortunately, I do not answer a/ , I can not do this proof :-(
     
  16. Mar 22, 2017 #15
    FB = V * ϱ * g = a*b* ( c-h)* ϱ * g = 50*10*(5-3,5)*1000*9,81= 7 357,5 N.....?
     
  17. Mar 22, 2017 #16

    haruspex

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    What does Archimedes' principle say? The 'boat' is a rectangular box. If its base area is A, the height of the water inside is hi and the height of the water outside is ho, what is the volume of water displaced?
     
  18. Mar 22, 2017 #17
    body floats : Fg = FB
    m * g = V * ϱ * g
    m......water weight...not in whole ship, only flooded part m = ϱ*V = ϱ* a*b*h0
    V.....the volume of whole ship in water .......V = a*b*hi
    hi = c-h0....?
     
  19. Mar 22, 2017 #18

    BvU

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    Ah, I mistakenly interpreted this ... as "I have answered this and did not have any problem there". (This is a hint for you for a possible next time)
     
  20. Mar 22, 2017 #19

    haruspex

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    I cannot make sense of your equations. You seem to reuse symbolic names with different meanings.
    Nowhere do I see a reference to the mass of the ship itself.

    If
    M is the mass of the ship,
    H is the height of the ship,
    A is the base area
    ρ Is the density of water,
    hi is the height of water within the ship,
    ho is the height of water outside the ship
    what does Archimedes' law say, using these symbols?
     
  21. Mar 22, 2017 #20
    I'm sorry, I reply later, I have duty now
     
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