Sinking Boat: Calculations & Results

[charlie05]

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

 

[mfb]

You used the diameter of the hole as radius.

The ship will sink before water flowing through the hole filled the whole interior.

 

[charlie05]

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

 

[BvU]

How much water do you need in there before it sinks ?

 

[charlie05]

when F_g > F_B...m * g > V * ϱ * g ...?

 

[BvU]

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

In terms of variables that we have encountered and explained ( ) so far !

 

[charlie05]

F_g...gravity
F_B...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

 

[BvU]

So what follows from post #5 ?

 

[charlie05]

m>V * ϱ...m>V * ϱ = abc*ϱ = 50*10*5*1000 = 2 500 000 kg of water ?

 

[BvU]

Nope

 

[charlie05]

so I do not know :-(

 

[BvU]

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

 

[haruspex]

How much water do you need in there before it sinks ?

so I do not know :-(

Did you answer part a? That makes it easy to answer BvU's question.

 

[charlie05]

Unfortunately, I do not answer a/ , I can not do this proof :-(

 

[charlie05]

F_B = V * ϱ * g = a*b* ( c-h)* ϱ * g = 50*10*(5-3,5)*1000*9,81= 7 357,5 N...?

 

[haruspex]

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 h_i and the height of the water outside is h_o, what is the volume of water displaced?

 

[charlie05]

body floats : F_g = F_B
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*h_i
h_i = c-h_0....?

 

[BvU]

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)

 

[haruspex]

body floats : F_g = F_B
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*h_i
h_i = c-h_0....?

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,
h_i is the height of water within the ship,
h_o is the height of water outside the ship
what does Archimedes' law say, using these symbols?

 

[charlie05]

I'm sorry, I reply later, I have duty now

 

[charlie05]

before impact: the ship is submerged c-h₀ = 5-3,5 = 1,5m
ship floats, is no water in ship...so F_g = F_B
M*g = V * ϱ * g
M*g = A*( c-h₀) *g

 

[BvU]

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

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 ?

 

[charlie05]

ufffff, no AHA yet :-(
level in the boat will go up and the ship will sink more and more

 

[BvU]

Well, then we still have work to do

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 (). Calculate total mass, so total water displacement and tell us what happened to the 3.5 m (quantitatively, not just 'a bit more').

 

[charlie05]

the height of the upper edge of the water level will decrease and when it is zero, the ship sinks...

 

[charlie05]

F_B = 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 ...

 

[charlie05]

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

 

[BvU]

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 ?

what happened to the 3.5 m?

the height of the upper edge of the water level will decrease and when it is zero, the ship sinks...

What are you saying ?

 

[charlie05]

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?

 

[BvU]

we are getting somewhere. Why the question marks ?

 

[charlie05]

AHA :-) :-) :-)

now I need to calculate b / velocity
0,5 ρ v² = hρg ?

 

[haruspex]

AHA :-) :-) :-)

now I need to calculate b / velocity
0,5 ρ v² = hρg ?

Where h is...?

 

[charlie05]

that is the question :-)... column of water above the hole. ? h=1,5 m ?

 

[haruspex]

that is the question :-)... column of water above the hole. ? h=1,5 m ?

What force pushes the water in through the hole? What height is responsible for that?

 

[charlie05]

hydrostatic pressure - is generated by weight of water.

 

[haruspex]

hydrostatic pressure - is generated by weight of water.

Right, but there is water on both sides of the hole, so two hydrostatic pressures.

 

[charlie05]

act against each other? pressure of column 5m is reduced by the pressure of column 1,5m?

 

[haruspex]

act against each other? pressure of column 5m is reduced by the pressure of column 1,5m?

The only time there is a 5m height of water is when the boat is about to sink, the water outside being 5m above the hole. What column is 1.5m at that time? Did you mean 3.5m?

 

[charlie05]

so in fact the hydrostatic pressure makes only 2 m column?

 

[mfb]

Where do the 2 meters come from? It is wrong, and I don't see any way to get that with the given 5m and 3.5m.

Did you draw a sketch?

 

[charlie05]

p_htotal = h₁ρg - h₂ρg = ( 5-3,5) ρg

 

[haruspex]

p_htotal = h₁ρg - h₂ρg = ( 5-3,5) ρg

Right.

 

[charlie05]

aha :-) so:

p_htotal = 1/2 ρv²...v = √ 2g(h₁-h₂) = √ 2g*(5-3,5) = 5,425 m/s

 

[charlie05]

hole area...S = π * 0,01² = 0,000314 m²
in 1 s...V1 = v*S = 5,425*0,000314 = 0,0017 m³
T = V /V1 = 1750/0,0017 = 1 029 412 s = 285,95 hour

 

[haruspex]

hole area...S = π * 0,01² = 0,000314 m²
in 1 s...V1 = v*S = 5,425*0,000314 = 0,0017 m³
T = V /V1 = 1750/0,0017 = 1 029 412 s = 285,95 hour

Looks right.
I am a little uncertain about the applicability of v^2=2gΔh here. The water that flows in then has to be arrested by the water already present. That suggests the pressure in that stream increases a little after entering the boat, which in turn suggests the water will flow in faster than that equation says. It is different from the case of water pouring out of a tank into the open air. However, I am unable to find any discussion of this on the net.
@mfb, any comments?

 

[charlie05]

Thanks very much - can we please have a look at the task a/ ?

I think the proff is based on the the definitions boundancy force ?

 

[haruspex]

Thanks very much - can we please have a look at the task a/ ?

I think the proff is based on the the definitions boundancy force ?

I thought you had solved a) in order to write your post #29.
Yes, it is to do with Archimedes' principle, as I wrote earlier.
Try to answer my post #19.

 

[charlie05]

a/ Show that difference between the levels inside and outside the ship is not changed.

without hole...outside 1,5m, inside 0
Water flows in1 m...outside 2,5m...inside 1m
ship sinking...outside 5 m..inside 3,5m

...level difference is always 1,5m?

can I prove it generally?
Archimedes' principle without hole...M*g = A ( h₀)*ρ * g
Archimedes' principle with hole...M*g + mg = F_B = ?

 

[haruspex]

Archimedes' principle without hole...M*g = A ( h0)*ρ * g

In post #19 I wrote h_o, not h₀, as being the height of water outside the boat relative to the hole. Your equation is correct, but h_o is a variable. Ah_o is the volume displaced at any time, not just at the start.

Archimedes' principle with hole...M*g + mg = F_B = ?

Write m in terms of h_i, A and ρ. For F_B, use the volume displacement mentioned above.

 

[charlie05]

water weight..m= Vρ = A*h_i *ρ
F_B = V * ϱ * g = A* h₀* ϱ * g

M*g + m*g =A ( h0)*ρ * g
A*c*ρ_boat *g+ A*h_i*ρ_water*g = A *h_o *ρ*g...

c*ρ_boat = ρ_water (h₀-h_i)
density of water and boat is constant...