What is the weight of the weight in water?

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The discussion revolves around calculating the weight of an object submerged in water, applying Archimedes' principle. Participants explore the relationship between the object's density and buoyancy, noting that the object, with a density of 0.9 g/cm³, will float. The calculations involve determining the upward force exerted by water and how it affects the object's apparent weight. Confusion arises regarding the distinction between weight and mass, with clarification that while the object has mass, its apparent weight in water can be considered zero when in equilibrium. Ultimately, the conversation emphasizes the importance of understanding buoyancy and the forces at play when an object is submerged.
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Homework Statement


A weight is 20 centimeters in length, 20 centimetres in width, 5 centimetres in height. Its mass is 1.5 kg. Find the weight's weight in water.

m = 1.8 kg
h = 5 centimeters = 0.05m
l = 20 centimeters = 0.2 m
width = length

Homework Equations



FA = pliquidgV
P = mg

The Attempt at a Solution


There is a drawing in the attachment. I know i should of put measurements there.

Now my problem. Should i attempt to put the weight in equilibrium so i could:
FA - P = 0
FA = P
FA = mg
pliquidgV = mg
pliquidghlw = mg

And then what?
 

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Avathacis said:
FA - P = 0

No.

Have you heard of Archimedes principle?
 
Borek said:
No.

Have you heard of Archimedes principle?

If it's the force, then yes. Otherwise no.
 
Yes, it is about force and displacement.
 
Borek said:
Yes, it is about force and displacement.

So where am i wrong? If i put the weight in equilibrium while underwater (meaning it stays where we put it) the answer should be 0, no? Isn't that the first Newton's law?
 
That would be true for floating object, you are not told object is floating.

You can calculate upward force from Archimedes principle.
 
Borek said:
That would be true for floating object, you are not told object is floating.

You can calculate upward force from Archimedes principle.

I can do that. But what does it give me? I need an answer in letters before i can start calculating anyway.
pliquidghlw = 1960 N
 
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one cubic centimeter of water has a mass of one gram

how many cubic centimeters is the object
 
granpa said:
one cubic centimeter of water has a mass of one gram

how many cubic centimeters is the object

2000

I'm confused. Where are you guys turning to?
 
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  • #10
https://www.youtube.com/watch?v=VDSYXmvjg6M
 
  • #11
Ok what I've figured out.

I guess i can get the density of the weight:

p = 1800 g / 2000 cm^3 = 0.9 g/cm^3.

Now i know that it will float. Now what? I'm still confused.

What i figured out just now.

px = 1 g/cm^3 - 0.9 g/cm^3 = 0.1 g/cm^3

mx = 0.1 g/cm^3 * 2000 cm^3 = 200g = 0.2 kg

P = 0.2 kg * 9.8 m/s^2 = 1.96 N.

Feels a little short on the weight. Or it's 0.9 g/cm^3 * 2000 cm^3 = 1800 g = 1.8 kg
P = 1.8 kg * 9.8 m/s^2 = 17.64 N
 
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  • #12
Sorry, I was wrong about one thing - if its density is lower than that of water, it will float. But you can't start calculations assuming it as you did, that's what you can find after doing at least part of the calculations.

Now, if it floats - does it weight anything?
 
  • #13
Borek said:
Sorry, I was wrong about one thing - if its density is lower than that of water, it will float. But you can't start calculations assuming it as you did, that's what you can find after doing at least part of the calculations.

Now, if it floats - does it weight anything?

It does. An object doesn't weight anything only if it doesn't have mass (probably impossible) or is not affected by gravity.

I kinda need this solved in a few hours >.>...
 
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  • #14
Imagine you have rope that u can use to lift the weight while it's under water.

Water helps you lifting the block as if another person was pushing the block from bottom-up.

Find the upward force of water, then take it away from the "out if water" weight,
 
  • #15
Quinzio said:
Imagine you have rope that u can use to lift the weight while it's under water.

Water helps you lifting the block as if another person was pushing the block from bottom-up.

Find the upward force of water, then take it away from the "out if water" weight,

Pw = FA - P

Pw = pliquidghlw - mg

Like this?

No wait. The object actually becomes heavier then.
 
  • #16
Avathacis said:
Pw = FA - P
Like this?

No wait. The object actually becomes heavier then.

Take it away from the out-of-water weight, not the other way round.

Pw = mg - pliquidghlw
 
  • #17
Quinzio said:
Take it away from the out-of-water weight, not the other way round.

Pw = mg-pliquidghlw

It means the answer is supposed to be negative? I guess this is because the upward force is stronger than the downward force?
 
  • #18
Yeah, ok, the result is negative, so the object will ...... ?
 
  • #19
Quinzio said:
Yeah, ok, the result is negative, so the object will ...... ?

Float?
 
  • #20
Ok. !
 
  • #21
And when it floats - does it weight anything?

Imagine you put it on the balance. Does it measure anything?
 
  • #22
Borek said:
And when it floats - does it weight anything?

Imagine you put it on the balance. Does it measure anything?

I guess it does weight something, but the weight is a lot lower. It's easier to lift things underwater anyway.
 
  • #23
Think a little bit more - if it would weight anything there would be a force pulling the object down. But your object is not going down - it floats. So there is no force pulling it down - which means it doesn't weight anything.

You may be confusing weight with a mass - your object still has the mass, it just doesn't weight anything.
 
  • #24
Borek said:
Think a little bit more - if it would weight anything there would be a force pulling the object down. But your object is not going down - it floats. So there is no force pulling it down - which means it doesn't weight anything.

You may be confusing weight with a mass - your object still has the mass, it just doesn't weight anything.

So the answer is 0? I asked my teacher, he said the answer can't be negative. So it's either 0 or he was wrong :O.

Or i am not getting something.
 
  • #25
When you push the object under the water surface, net force will be directed up - in a way that's equivalent to the weight being negative. However, when the system is at equilibrium, part of the object sticks above the water, net force acting on the object is zero - so its weight is zero.

This is a little bit tricky, perhaps it would be better to replace all occurrences of weight in this thread with apparent weight.
 
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  • #26
Borek said:
When you push the object under the water surface, net force will be directed up - in a way that's equivalent to the weight being negative. However, when the system is at equilibrium, part of the object sticks above the water, net force acting on the object is zero - so its apparent weight is zero.

Err, i was assuming i could put it in equilibrium, but since the density is 0.9 g/cm^3 it can't be in equilibrium, right?

This can't be that hard to solve. Borek, can you write equations on what you based your assumptions and your answer? :3
 
  • #27
bump, still haven't figured this out.
 
  • #28
It is partially in water, partially above - and in equilibrium.
 
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