Solving Confusing Exercises: Understanding Water Displacement and Density

In summary, the conversation discusses two tasks related to the principles of buoyancy and displacement of water. The first task involves finding the height of the water level rise and the change in cubic volume when a body with a given weight is placed in a glass of water. The solution is to give the body the same density as water and use the principle that a floating object displaces its own weight of fluid. The second task involves finding the volume of a ship that is under water, given its total mass and the density of water. The solution is to use the principle of buoyancy and the weight of the displaced water to determine the volume of the ship. The conversation also discusses the importance of understanding the difference between weight and mass in these calculations.
  • #1
fawk3s
342
1
Hi

First of all I'd like to say that I don't want to put this under "Homework help", because its not really homework. But it is something I can't wrap my mind around. There are 2 tasks:

1) There's 200 cm3 water in a glass. How high does the water level rise / how much does the cubage change, if you put a body with 40g to FLOAT in it?

Now it is solved by simply giving the body the density of the water.. But how the heck can you do that? It is said that the body floats, not swims/hovers/whatever.

To clear things up, this is how I understand it. My english is not so good so i don't know which is which:
float - Fgravity<Fbouyancy
swim - Fgravity=Fbouyancy
sink - Fgravity>Fbouyancy

2) A ship's and its stuff's total mass is 4000t=4000000kg. What is the cubage of the part of the ship which is under water? (Assuming the density of the water is 1030kg/m3, like in the sea).

I think the answer is supposed to be ~3883 m3, which you can get simply by 4000000/1030, but why and how does this give you the answer?


These are pretty elementary questions, but I am stupid, I know. No need to tell me that.
I appreaciate any help.

Thanks in advance,
fawk3s
 
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  • #2
I think it's best to work on the principle that a floating object displaces it's own weight of fluid.
As long as you can sort out the units ( weight and mass) appropriately, that will give you the answer to all these types of problems.
 
  • #3
I dont.. follow. :D
 
  • #4
If you know the mass of the object then you will know the weight (from local g knowledge).
That will be balanced by an upthrust due to an equal weight of displaced fluid. That's good ol' Archimedes' principle. Knowing the weight can then give the volume displaced (via the Mass).
I'm being pernickety about the weight - mass thing so that you can be sure of getting the right answer.
 
  • #5
I still don't understand. Where do you get the weight of the displaced fluid? You don't have the density nor the volume of the body.
 
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  • #6
You are missing the point, the thing is you don't have to know the density of the object that is (swimming/floating) The point is the object (ship) is imparting its weight on the water moving away (displacing) that much amount (weight of water). You can arrive at the volume of the displaced water as you know the water's density. Voila! that's the volume of the ship that is immersed in water.
 
  • #7
You are taking the objects weight as the weight of the displaced water? This is what confuses me!

Logic tells me that if you put a 5m3 body in a certain amount of water, the displaced fluids volume would be 5m3.
So let's say you have a body of 5m3 (cubemeter) and its made of iron (7800kg/m3). So its mass is m=5*7800=39000kg, therefore, weight=9.8*39000=382200N.
So let's say the displaced waters weight is also 382200N, therefore mass=382200/9.8=39000. And therefore, the volume would be 39000/1000=39m3.

Please, enlighten me.
 
  • #8
You are looking at the principle of buyoncy, the initial condition is that the object is floating/swimming, density of object <= density of fluid.

In th other case (iron block) when it sinks, density > density of fluid.
 
  • #9
How do you get the weight of the displaced water...? This is what I asked. And this would tell me how to solve the problems.
 
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  • #10
if the object is floating (DensityObj <= DensityFluid):

Weight of displaced water = weight of the object

Else

Weight of displaced water = volume of the submerged portion of object * Density of water
 
  • #11
PhyGuru said:
if the object is floating (DensityObj <= DensityFluid):

Weight of displaced water = weight of the object

How can you say that?

if DensityObj <= DensityFluid
then it MUST be
weight of displaced water <= weight of the object.

But that does not give me anything! I can't calculate it!
If you can't tell me how to calculate the weight of the displaced water, or solve the problem, then there is no point in posting the density-crap I already know about.
 
  • #12
Without involving density...

Any object, wholly or partly immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

Greater the volume of the displaced fluid, greather is the force.

If this force is greater than the weight of the object, it takes it up to the surface until the two forces balance each other out.

That's where the
Weight of displaced water = weight of the object

comes from, a perfect cancellation of 2 forces.
 
  • #13
The point is that whatever the density of the object it will sink until it displaces enough water to make it float (assming it doesn't sink of course). An object made of heavy wood (just floating) will displace the same amount ( weight or volume) of water as inflatable boat of negligable mass carrying the same object.
The upthrust is just enough to make it float and is there because of the water that has been pushed out of the way. In both cases that will be the same amount but the lump of wood will be mostly above the water surface in the second case. OR a huge block of foam, weighing the same would be almost all out of the water but still displaces the same amount of water.
I know I'm just repeating myself in essence but I'm putting it in practical terms this time.
 
  • #14
Oh god what have I written last night..
If the object is floating/swimming, the weight is the same as the displaced waters ! And the volume is only the same with the body's when the object has sinked ! HOW DID I NOT SEE THAT?!

I just discovered I am a retard.
But big thanks guys, I get it now! And thanks for your patience!
 

1. What are "2 confusing exercises"?

"2 confusing exercises" refers to a set of two activities or tasks that are difficult to understand or complete. These exercises are designed to challenge the mind and require problem-solving skills.

2. How do I know if an exercise is considered "confusing"?

An exercise can be considered confusing if it involves complex concepts, requires multiple steps, or has unclear instructions. If you find yourself struggling to understand or complete an exercise, it may be considered confusing.

3. Why are "2 confusing exercises" important in science?

Confusing exercises are important in science because they help develop critical thinking and problem-solving skills. These exercises also mimic real-world scenarios where scientists often encounter complex problems that require innovative solutions.

4. How can I effectively approach "2 confusing exercises"?

To effectively approach confusing exercises, it is important to break them down into smaller, manageable tasks. Take your time and carefully read through the instructions. If needed, seek help from a mentor or colleague.

5. Are there any benefits to completing "2 confusing exercises"?

Yes, completing confusing exercises can have several benefits. It can improve your analytical and problem-solving skills, increase your confidence in tackling difficult tasks, and prepare you for real-world challenges in the scientific field.

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