Sinking Boat: Calculations & Results

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Homework Help Overview

The discussion revolves around a physics problem involving a sinking boat with specific dimensions and a hole in its bottom. Participants explore the implications of water entering the boat through the hole and the resulting effects on buoyancy and water levels inside and outside the boat.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to calculate the velocity of water entering through the hole and the time it takes for the boat to sink. They also discuss the relationship between the water levels inside and outside the boat and question the assumptions made regarding buoyancy and hydrostatic pressure.

Discussion Status

The discussion is ongoing, with participants providing various calculations and questioning each other's reasoning. Some participants express uncertainty about their calculations, while others suggest reconsidering certain assumptions, particularly regarding the hydrostatic pressure and the implications of water levels on buoyancy.

Contextual Notes

There are constraints regarding the initial conditions of the boat, including its dimensions and the height of water above the hole. Participants also note the need to clarify the definitions of variables used in their calculations, particularly in relation to buoyancy and the mass of the water inside the boat.

charlie05
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Homework Statement


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.

Homework Equations


a/ ...
b/ 0,5 ρ v2 = hρg v = √ 2gh

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

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
 
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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.
 
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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
 
How much water do you need in there before it sinks ?
 
when Fg > FB...m * g > V * ϱ * g ...?
 
You want us to guess what you mean with these symbols ?

In terms of variables that we have encountered and explained ( :rolleyes: ) so far !
 
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
 
So what follows from post #5 ?
 
m>V * ϱ...m>V * ϱ = abc*ϱ = 50*10*5*1000 = 2 500 000 kg of water ?
 
  • #10
Nope
 
  • #11
so I do not know :-(
 
  • #12
charlie05 said:
the height of the ship is c = 5m. ...
The initial height of the upper edge of the ship above the water level is h = 3.5 m
 
  • #13
BvU said:
How much water do you need in there before it sinks ?
charlie05 said:
so I do not know :-(
Did you answer part a? That makes it easy to answer BvU's question.
 
  • #14
Unfortunately, I do not answer a/ , I can not do this proof :-(
 
  • #15
FB = V * ϱ * g = a*b* ( c-h)* ϱ * g = 50*10*(5-3,5)*1000*9,81= 7 357,5 N...?
 
  • #16
charlie05 said:
Unfortunately, I do not answer a/ , I can not do this proof :-(
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?
 
  • #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....?
 
  • #18
charlie05 said:
Unfortunately, I do not answer a/ , I can not do this proof :-(
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)
 
  • #19
charlie05 said:
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....?
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?
 
  • #20
I'm sorry, I reply later, I have duty now
 
  • #21
before impact: the ship is submerged c-h0 = 5-3,5 = 1,5m
ship floats, is no water in ship...so Fg = FB
M*g = V * ϱ * g
M*g = A*( c-h0) *g
 
  • #22
Yes, so the 'ship' itself isn't 2.5 Mt but only 0.75. (Check your work in post #15!)

Which is good, because otherwise it would sink straightaway :smile:

But I don't see a big AHA emerging on your part ?

Suppose A * 1 m3 water has poured in through the little hole. What about the levels inside and outside the ship ? So how long until it sinks ?
 
  • #23
ufffff, no AHA yet :-(
level in the boat will go up and the ship will sink more and more
 
  • #24
Well, then we still have work to do
BvU said:
Suppose A * 1 m3 water has poured in through the little hole. What about the levels inside and outside the ship ?
Maybe you missed that question (:rolleyes:). Calculate total mass, so total water displacement and tell us what happened to the 3.5 m (quantitatively, not just 'a bit more').
 
  • #25
the height of the upper edge of the water level will decrease and when it is zero, the ship sinks...
 
  • #26
FB = V * ϱ * g = a*b* ( c-h)* ϱ * g = 50*10*(5-3,5)*1000*9,81= 7 357,5 N...

submerged portion will be greater and the buoyant force will increase ...
 
  • #27
when is in boat m kg water ( boat without water is 1,5 m submerged), M is the mass of the ship,

M * g = V * ϱ * g
(M+m)*g = V´ * ϱ * g
(M+m) = V´ * ϱ
 
  • #28
charlie05 said:
boat without water is 1,5 m submerged
Correct (given). Now:

how far is it submerged when the water level inside has risen from 0 above the ship floor to 1 m ?

BvU said:
what happened to the 3.5 m?

charlie05 said:
the height of the upper edge of the water level will decrease and when it is zero, the ship sinks...
What are you saying ?
 
  • #29
when the water level inside has risen from 0 above the ship floor to 1 m , ship is it submerged 1,5 + 1 = 2,5m?the ships sinks, in that case...h = 0...There is nothing above the surface...is submerged entire height c = 5m
must to drain height of 3.5 meters of water?
V = 50*10*3,5 = 1 750 m3 of water is necessary for sinking ship?
 
Last edited:
  • #30
we are getting somewhere. Why the question marks ?
 

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