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Question on motion of a ship

  1. Jul 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A ship which is at rest,is connected to land by a chain.The ship is powered by 2 propellers.When the ship is moving ,the 2 propellers push the water backward with velocity 20ms-1,relative to the earth.density(dw)of seawater is 1000kgm-3.Total mass of the ship(M)is 2*107kg

    (1)The effective area swept by the blades of a propeller is 12m2,calculate the mass of water pushed in 1 second.
    (2) Calculate the force exerted on the 2 propellers by water
    (3) what is the initial acceleration (a)of the ship if the chain is released?
    (4) goods of weight 4*106kg are unloaded from the ship at a harbor.Cross sectional area of the ship base is 1000m2.Assuming the hull of the ship is vertical,cal.the height raised by the unloading
    (5) As the ship is raised,there's a danger of it falling to a side.how much volume of water must be taken in to submerge it to the previous level?
    (6) The ship is loaded back with new goods & the water taken in is removed,before it starts sailing.Height submerged in the water is 5m.Due to a blast,a small hole is formed at the base.find the initial speed of water entering the ship?
    (7) water enters the ship at the rate of 10m3s-1,and the height of the ship is 8m.calculate the time taken for the ship to submerge fully
    (8) once the ship is fully submerged it reaches the sea bottom.If the average density of the material of the ship is 2500kgm-3.Cal the min,force requied to bring the ship back to the surface

    2. Relevant equations

    3. The attempt at a solution
    I'm sorry the question's so long.
    It seems like a very easy question and I managed to do it upto (6).

    (1) Volume of water pushed out/s = Au
    massof water/s=dw*V

    (2) [tex]\stackrel{\rightarrow}{F}[/tex] = ma
    480*1000*20= -9.6*106N

    (3) [tex]\stackrel{\rightarrow}{F}[/tex] = ma
    -9.6*106/2*107 = a
    = -0.48ms-2

    (4) P = F/A
    hdg = 4*107/1000

    (5) V= A*8

    (6) P1 at the bottom of vessel = pi + 5dg
    P2=pi=atmosphric pressure
    (P1-P2)A = m(v-u)/t
    5*104*103= m(v-u)/t
    It's here that i'm stuck,i don't know how to proceed without m/t,I'd like to think the m/t here is the same as what i got in (1),but i don't think that's possible.

    Hope someone can help.

  2. jcsd
  3. Jul 6, 2009 #2


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    I'm wondering why the answer in 4 seems different from the answer in 5? If the cargo displaced some amount of water won't that same volume be needed to reload the boat back to the same level?

    In 6) don't they want to know the speed of the water? Won't it be related to the kinetic energy in the water? Won't the potential energy released at 5 meters become kinetic energy as in

    p*g*h = 1/2*p*v2

    7 is just a how long to fill isn't it? It is already 5 m submerged with only 3 meters to go before swimming with the fishes.

    In 8 won't you need to add just enough buoyancy to bring it up? The density is 2500 but it already displaces 1000, so it will need additional displacement of water to get it to the surface?
  4. Jul 6, 2009 #3
    Yes,I think I've made a mistake here,the answer for(5) should have been ,
    Area of base*4=1000*4=4000m3 , right?

    which would mean,
    2gh= v2
    =[tex]\sqrt{100}[/tex] = 10ms-1 ?

    Therefore t = (3*1000)/10
    = 300s ?

    All of the above I understand,thanks to your very helpful hints,sir.

    Now,it is this part that I'm having trouble understanding,
    What is this 1000 you're referring to,sir?
    Since the ship needs to sink by 3m to get fully submerged,would it be wrong to just take,
    = (3*1000)*1000*10
    =3*107N ?
  5. Jul 6, 2009 #4


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    The 1000 kg/m3 is the density of water right?
    And if something submerged has a density of 2500 kg/m3 doesn't that mean it already displaces 1000 of the 2500 kg/m3? Doesn't that mean that you only need to supply an additional 1500 kg/m3 to make it buoyancy neutral in water? That's what is required to make it be able to come to the surface. To get it above the surface you will need more of course.

    The only thing you are missing is what is the volume of the 2500 kg/m3. But that it looks to me like you can figure from the original mass of the boat and the additional mass of the cargo added. Then that volume times 1500 kg/m3 should be all that's needed to bring it off the bottom to just at the surface.
  6. Jul 6, 2009 #5
    I'm still finding it difficult to understand this.
    so interms of density,we have to supply the additional density (for any submerged object),to balance its density?

    Does this include the density of the load as well?
    I'm just so confused.
  7. Jul 7, 2009 #6


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    The statement says the average, so that would be the cargo too supposedly.

    As to density ... you are not going to supply density, you need to supply volume ...

    Total mass / average density = Volume of submerged ship and cargo.

    That volume times the difference in average density and water ... 2500 kg/m3 - 1000 kg/m3 or 1500 kg/m3 is the buoyant force you need.
  8. Jul 7, 2009 #7
    Total mass / average density = Volume of submerged ship and cargo.

    To find the total mass of the ship+ new cargo,
    Assuming all the cargo was removed in part(4),
    to find the mass of the ship only,
    =20*106 - 4*106

    To find the mass(x) of the new cargo added in part(6),
    P=F/A = xg/1000
    =xg/1000 = h*1000*g
    =x= 5*106kg

    so total mass= 5*106 + 16*106kg

    Total mass / average density = Volume of submerged ship and cargo
    21*10^6/2500 = V

    V*1500 *10 = force needed
    (21*10^6/2500)*15000 = F

    Is this correct?

    Though I'm still a bit unclear about the density part.

  9. Jul 7, 2009 #8


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    It may be. But my reading of the problem suggests that the mass of the ship alone may be 2*107 kg. So either you need to consider that the ship + cargo is 25*106 or 21*106 as you found. I think the problem is ambiguous on this point.

    The important thing is you grasp the concept that buoyant force is given by the amount of water displaced, and that for a submerged object it already has a buoyant force on it given by the volume of water it displaces. Meaning that the weight (downward force) of a submerged object is really only the weight given by the difference in density between it and water multiplied by its volume.
  10. Jul 7, 2009 #9
    I'm sorry,I still find it difficult to understand this.
    Are we talking about the Archimedes' principle,here?
    i.e.,if an object floats totally or partially immeresed in a fluid at rest,then the weight of the body is equal to the weight of the fluid displaced by the body?(this is what you mean here,ryt?)
    so the upthrust or bouyant force acting on this body = volume of the body*density of fluid immersed in*g?

    are you referring to the apparent weight of the body in the liquid,here?
    & do we always have to find the difference in densities to find the upthrust of an object immersed in a fluid?
    I've never substracted the densities before to find the upthrust acting on a body,so this is very new to me.
    Do we always have to find the difference or is it only under certain conditons?
    I'm just so thoroughly confused and I would really like to understand.

    thank you
  11. Jul 7, 2009 #10


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    Whatever volume a submerged object occupies times the density of water is the buoyant force acting on it. If it is heavier than water, then the "apparent" weight, say if you put a scale under the water, would be lessened by that amount.

    Mass = volume * density

    Weight = mass * gravity

    Weight under water = mass * gravity - buoyant force

    Buoyant force = density water * volume * gravity

    Weight under water = density object * volume * gravity - density water * volume * gravity

    Weight under water = volume * gravity * (density of object - density water)
    Well you won't find "upthrust" that way, subtracting densities is used to determine the excess mass that needs to be supported for an object submerged. The "up" force is the buoyant force, which is how much water is displaced.
    Usually problems involving buoyancy are about things floating on the water. But when an object doesn't displace enough water to float the force is still there at least to the extent that it is still displacing water and off setting some of it's mass (weight) in the downward direction.
    Last edited: Jul 7, 2009
  12. Jul 8, 2009 #11
    Wow.this is perfect.
    this is just the explanation i needed,and i feel like i finally understand it all,but i guess the only way to really findout is to attempt more questions,so if i have any problem,i will post.

    Thank you
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