Floating in Sea Water vs. Floating in Fresh Water

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SUMMARY

This discussion focuses on the buoyancy of a ship floating in fresh water versus sea water, specifically addressing how much additional weight a ship can carry when transitioning from fresh water (specific gravity 1.0) to sea water (specific gravity 1.03). The key equations discussed include the relationship between mass, volume, and density, as well as the concept of specific gravity as it relates to buoyancy. The conclusion is that a ship can carry an additional 1,500 metric tons when floating in sea water due to its higher density compared to fresh water.

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Homework Statement


Really need help with this one. I don't quite know where to begin.

If a ship loaded when floating in fresh water to a weight of 50,000 metric tons were floated in sea water of specific gravity 1.03, how much additional weight could she take on board without being immersed to a greater extent than in the fresh water?
 
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You need some relevant equations. How about something to tell us what "specific gravity" means?
 
Ok, specific gravity means the ratio of the density of a substance to the density of a given reference material. I believe fresh water has a specific gravity of 1.0 so it appears that the sea water is denser than the fresh water because its specific gravity is >1.0.

So, the sea water ship will be able to hold more cargo b/c it's denser and the buyoancy will be greater.

i've seen these two formulas before but don't know which to use:
sp gr = weight of the substance/weight of an equal volume of water
sp gr = density of the substance/density of water

i'm guessing you might use the first one and set fresh water and salt water equal to each other, solving for weight of substance B?

weight of freshwater/weight of equal volume of water = weight of saltwater/weight of equal volumeof water.

is this correct? if so, how do we find the wieghts of the salt and fresh water?

thanks for any help
 
yes that is all you need. you just need to find the density. a 50K metric ton boat displaces 50K metric tons of water. So you need to find the weight of salt water that occupies the same space as the weight of fresh water. then you have your answer
 
I'm still not sure I understand where specific gravity comes into play. if we set the equation up like this would it be:

1.0 = 50,000/weight of an equal volume of water?
 
been working on this but not sure where to go from here. to use this equation:

weight of freshwater/weight of equal volume of water = weight of saltwater/weight of equal volume of water

wouldn't make sense b/c it would be 50k/50k = weight of saltwater/weight of equal volume of water

which is un-solveable.

would you use specific gravity instead: 50k/1.0 = weight of saltwater/1.3?
That doesn't seem to make sense either. Density must come into play somewhere but I can't figure out where.
 
Specific gravity is just a different way of indicating what the density of a substance is. If sea water has a specific gravity of 1.03, that just means its density \rho_{sw} is equal to \rho_{sw} = 1.03\rho_{pure water} = 1.03~\mbox{g}/\mbox{cm}^3.
 
so would the equation be:
50,000/1.0 = equal volume of salt water

?
do we need to conver 50,000 to Newtons?
and would we multiply the answer by 9.81 to get the weight?
 
I suggest you draw the free-body diagrams of the two cases and apply F=ma. From there, derive a relationship between the two cases rather than just trying to guess what ratio, if any, will work.
 
  • #10
i'm sorry - I'm completely lost here. We can only do F=ma for fresh water b/c that's the only one we're given a mass for: F=(50,000)(9.81) = 490,500. I don't understand how this relates to specific gravities, weights or displacements. Please explain - thank you!
 
  • #11
kriegera said:
i'm sorry - I'm completely lost here. We can only do F=ma for fresh water b/c that's the only one we're given a mass for: F=(50,000)(9.81) = 490,500. I don't understand how this relates to specific gravities, weights or displacements. Please explain - thank you!

F is not important here. The ship would float to the same level regardless of g.

The level at which the ship is floating relates to the volume of water displaced.

The volume of water displaced has a mass of 50,000 metric tons, for fresh water.

In sea water, the SAME volume will have a greater mass, right? That will tell you how much more you can load onto the ship when it is in the ocean.

Given that a certain volume of fresh water weighs 50,000 metric tons, how much more will the same volume of sea water weigh?

Cheers -- sylas
 
  • #12
Would it be
weight of same volume of sea weater = 50,000/1.03 - 48543

i get that the sea water is denser and can therefore hold more weight i just don't understand the relationship between weight, volume and specific gravity.
 
  • #13
kriegera said:
Would it be
weight of same volume of sea weater = 50,000/1.03 - 48543

i get that the sea water is denser and can therefore hold more weight i just don't understand the relationship between weight, volume and specific gravity.

The weight of a certain volume of fresh water is 50,000 metric tons, right?

The question is... what is the weight of the same volume of sea water. Give a number. It will be more than 50,000; because sea water is more dense.

The basic relationships are:

Mass = Volume * Density
Density = SpecificGravity * DensityFreshWater

specific gravity is just density relative to some standard reference; in this case relative to fresh water.

Cheers -- sylas
 
  • #14
kriegera said:
i'm sorry - I'm completely lost here. We can only do F=ma for fresh water b/c that's the only one we're given a mass for: F=(50,000)(9.81) = 490,500. I don't understand how this relates to specific gravities, weights or displacements. Please explain - thank you!
That's only its weight Fg. For the ship not to sink, there has to be another force to counter its weight. That's the bouyancy force Fb, which depends on the density of the fluid \rho, g, and the volume of fluid displaced V. Since the ship is in equilibrium, you have

\Sigma F = F_g-F_b = mg-\rho gV=0
 
  • #15
sylas said:
The weight of a certain volume of fresh water is 50,000 metric tons, right?

The question is... what is the weight of the same volume of sea water. Give a number. It will be more than 50,000; because sea water is more dense.

The basic relationships are:

Mass = Volume * Density
Density = SpecificGravity * DensityFreshWater

So if Density = SpecificGravity*DensityFreshWater then
Sea Water Density/Specific Gravity = 1.03*1.0 = 1.03
And
Sea Water Mass = 50,000 * 1.03 = 51,500 metric tons

Would final weight be: 51,500*9.81 = 505,215. is this correct?
 
  • #16
A metric ton is a unit of weight, not mass.
 
  • #17
i must have gone one step too far/wrong then - so it would be 51,500 metric tons then?
 
  • #18
kriegera said:
i must have gone one step too far/wrong then - so it would be 51,500 metric tons then?

Right. And now from that you can see how much additional weight can be loaded.
 
  • #19
1,500 metric tons!
 
  • #20
kriegera said:
1,500 metric tons!

:approve:
 
  • #21
Thank you! :)
 

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