Archimedes's principle of float

[justaus3r]

Homework Statement

A swimming pool with 5 M length and 4 M width.A big layer of ice floats on the surface of the pool and above the layer of ice a rock with 40 KG mass and 5 R.D.,If the ice is melted and the rock fall, calculate the change in the height of water in the pool

final answer (1.6 * 10^-3 m)

Homework Equations

F_b (upthrust force) = ρ_water * volume_immersed * gravity

The Attempt at a Solution

i tried by getting the volume of the displaced liquid as it equals the upthrust force and then divide it by the area to get the height of the displaced liquid but the answer is wrong. i think we should include the ice as it melted but it didn't give any info. about it though

 

[paisiello2]

This problem has nothing to do with Archimedes' Principle.

But you are right the thickness of ice is assumed to be negligible.

 

[SammyS]

Homework Statement

A swimming pool with 5 M length and 4 M width.A big layer of ice floats on the surface of the pool and above the layer of ice a rock with 40 KG mass and 5 R.D.,If the ice is melted and the rock fall, calculate the change in the height of water in the pool

final answer (1.6 * 10^-3 m)

Homework Equations

F_b (upthrust force) = ρ_water * volume_immersed * gravity

The Attempt at a Solution

i tried by getting the volume of the displaced liquid as it equals the upthrust force and then divide it by the area to get the height of the displaced liquid but the answer is wrong. i think we should include the ice as it melted but it didn't give any info. about it though

Hello justaus3r. Welcome to PF !

What is R.D. ?

Relative Density ?


Does the height of the water increase, or does it decrease ?

 

[SammyS]

This problem has nothing to do with Archimedes' Principle.

But you are right the thickness of ice is assumed to be negligible.

This problem has everything to do with Archimedes' Principle .

 

[dauto]

Your calculation is almost correct. It correctly computes the level rise due to the rock on top of the ice. When the ice melts, the rock doesn't vanish in thin air. It goes to the bottom where it still affects the water level, but the amount of water level rise is different than before. Compute the difference.

 

[paisiello2]

This problem has everything to do with Archimedes' Principle .

Yes, I see you are right now.

 

[justaus3r]

Hello justaus3r. Welcome to PF !

What is R.D. ?

Relative Density ?


Does the height of the water increase, or does it decrease ?

yes R.D. is relative density = Density of material / density of water

it increases

 

[justaus3r]

This problem has nothing to do with Archimedes' Principle.

But you are right the thickness of ice is assumed to be negligible.

archimedes' principle includes cases where opposing force(s) has higher value than upthrust force which leads to the sinking of the body

 

[justaus3r]

Your calculation is almost correct. It correctly computes the level rise due to the rock on top of the ice. When the ice melts, the rock doesn't vanish in thin air. It goes to the bottom where it still affects the water level, but the amount of water level rise is different than before. Compute the difference.

thanks in advance for helping but unfortunately i don't have the height of water in the pool

 

[SammyS]

thanks in advance for helping but unfortunately i don't have the height of water in the pool .

You don't need the height of water in the pool. you just need to find out how much it changes.

 

[justaus3r]

You don't need the height of water in the pool. you just need to find out how much it changes.

Ok, as for the ice part should i include it or i can't bec. i don't have its area ?

 

[SammyS]

Ok, as for the ice part should i include it or i can't bec. i don't have its area ?

Consider an ice cube floating in a glass of water.

How much water does it displace?


When it melts, what is the weight of the (liquid) water that the melted ice becomes?

 

[justaus3r]

Consider an ice cube floating in a glass of water.

How much water does it displace?


When it melts, what is the weight of the (liquid) water that the melted ice becomes?

it displaces amount of water equals to the upthrust force acting on ice from surface of water

I think it remains the same (am not sure though )

 

[SammyS]

it displaces amount of water equals to the upthrust force acting on ice from surface of water

I think it remains the same (am not sure though )

The ice cube displaces an amount of water equal to the weight of the ice cube.

When the ice cube melts the weight of the water it yields is equal to the weight of the water it used to displace.

 

[justaus3r]

The ice cube displaces an amount of water equal to the weight of the ice cube.

When the ice cube melts the weight of the water it yields is equal to the weight of the water it used to displace.

their masses are equal but their volumes aren't equal, right ?

 

[justaus3r]

Area = 20 m^(2)
Before melting
I think i solved it but correct me if i am wrong
if the height increased due to floatation of that ice layer with the body equals 2 * 10 ^(-3) m
since the body is floating therefore the M_body*g = ρ_water * vol._immersed * g
vol_immersed = M_body / ρ_water = 40 / 1000 = 4 * 10 ^(-2) then V = A * h, h = V / A
equals 4 * 10 ^ -2 / 20, therefore h = 2 * 10 ^(-3) m

since the ice had melted and the body fall
weight of water displaced = upthrust force
M * g = volume of body * ρ_water *g
M * g = 8 * 10 ^(-3) * 1000 * g
M = 8 Kg
volume of water = 8 / 1000 = 8 * 10(^-3) m^3
height = 8 * 10 ^(-3) / 20 = 4 * 10 ^(-4) m


change in height = 2 * 10^(-3) - 4 * 10^(-4) = 1.6 * 10 ^(-3) m

 

[ehild]

Area = 20 m^(2)
Before melting
I think i solved it but correct me if i am wrong
if the height increased due to floatation of that ice layer with the body equals 2 * 10 ^(-3) m
since the body is floating therefore the M_body*g = ρ_water * vol._immersed * g
vol_immersed = M_body / ρ_water = 40 / 1000 = 4 * 10 ^(-2) then V = A * h, h = V / A
equals 4 * 10 ^ -2 / 20, therefore h = 2 * 10 ^(-3) m

since the ice had melted and the body fall
weight of water displaced = upthrust force
M * g = volume of body * ρ_water *g
M * g = 8 * 10 ^(-3) * 1000 * g
M = 8 Kg
volume of water = 8 / 1000 = 8 * 10(^-3) m^3
height = 8 * 10 ^(-3) / 20 = 4 * 10 ^(-4) m


change in height = 2 * 10^(-3) - 4 * 10^(-4) = 1.6 * 10 ^(-3) m

Is the final height lower or higher then the initial height?

The change of height is the final height - initial height. You subtracted the final height from the initial one, so got wrong sign for the change. Otherwise, it is correct.

ehild

 

[justaus3r]

Is the final height lower or higher then the initial height?

The change of height is the final height - initial height. You subtracted the final height from the initial one, so got wrong sign for the change. Otherwise, it is correct.

ehild

I supposed that the change in height = height_big - height_small