Problem Solving about Archimedes' Principle

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Homework Statement


A piece of metal weighs 50.0 N in air, 36.0 N in water and 41.0 N in oil. Find the densities of the metal and the oil.


Homework Equations


Density of Water is 1000 kg / m[itex]^{3}[/itex]
Density of Air is 0.00121 g / cm[itex]^{3}[/itex]


The Attempt at a Solution


I really don't have any idea on how to solve the problem but I gave it a try :

Buoyant Force = Weight in Air - Apparent Weight on Fluid

so, Buoyant Force in Water = 50.0 N - 36 N
Buoyant Force in Water = 14 N

Converting the Weight of Displaced Water(Buoyant Force) to Mass gives :
14 N [itex]\div[/itex] 9.8 m/s[itex]^{2}[/itex]
Mass of Displaced Water = 1.43 kg

To get the volume of Water Displaced (Which is the same as the Volume of the Metal.... PLEASE CORRECT ME IF I'M WRONG BUT THIS IS WHAT I UNDERSTOOD FROM READING THE BOOK!!!) we do:

Volume of Water= Mass / Density
Volume = 1.43 kg [itex]\div[/itex] 1000 kg / m[itex]^{3}[/itex]
Volume of Water = 1.43 x 10[itex]^{-3}[/itex] m[itex]^{3}[/itex]
----- Which is also the Volume of the Metal....

To get the Density of Metal we need the Mass and Volume of the Metal. From this part, how do I get the density of Metal? How do I get the Mass of the Metal ?? How do I get the Density of the oil?

Buoyant Force in Oil = Weight in Air - Weight in Oil
Buoyant Force in Oil = 50.0 N - 41.0 N
Buoyant Force in Oil = 9 N

Converting it to Mass gives 0.92 kg.

Please help?
/frozonecom
 
Last edited:

Answers and Replies

  • #2
42
0

Homework Statement


A piece of metal weighs 50.0 N in air, 36.0 N in water and 41.0 N in oil. Find the densities of the metal and the oil.


Homework Equations


Density of Water is 1000 kg / m[itex]^{3}[/itex]
Density of Air is 0.00121 g / cm[itex]^{3}[/itex]


The Attempt at a Solution


I really don't have any idea on how to solve the problem but I gave it a try :

Buoyant Force = Weight in Air - Apparent Weight on Fluid

so, Buoyant Force in Water = 50.0 N - 36 N
Buoyant Force in Water = 14 N

Converting the Weight of Displaced Water(Buoyant Force) to Mass gives :
14 N [itex]\div[/itex] 9.8 m/s[itex]^{2}[/itex]
Mass of Displaced Water = 1.43 kg

To get the volume of Water Displaced (Which is the same as the Volume of the Metal.... PLEASE CORRECT ME IF I'M WRONG BUT THIS IS WHAT I UNDERSTOOD FROM READING THE BOOK!!!) we do:

Volume of Water= Mass / Density
Volume = 1.43 kg [itex]\div[/itex] 1000 kg / m[itex]^{3}[/itex]
Volume of Water = 1.43 x 10[itex]^{-3}[/itex] m[itex]^{3}[/itex]
----- Which is also the Volume of the Metal....

To get the Density of Metal we need the Mass and Volume of the Metal. From this part, how do I get the density of Metal? How do I get the Mass of the Metal ?? How do I get the Density of the oil?

Buoyant Force in Oil = Weight in Air - Weight in Oil
Buoyant Force in Oil = 50.0 N - 41.0 N
Buoyant Force in Oil = 9 N

Converting it to Mass gives 0.92 kg.

Please help?
/frozonecom


You can actually get the density of the Metal using the Specific Gravity :

Sp. Gr. = Weight in Air / Bouyant Force or [W(in air) - W(in water)]

Then another form of Specific Gravity is:

Sp. Gr. = Density of Substance/ Density of Water

Thus :

Density of Substance = Sp.Gr / Density of Water

Then:

Using Sp.Gr of Metal

Sp.Gr (Metal) = [Sp.Gr(Fluid)*(Weight in Air of Metal)] / Bouyant Force
 
Last edited:
  • #3
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So,

Sp. G. = 50 N / (50 N - 36 N)
Sp. G. = 50 N / 14 N
Sp. G. = 3.57

Then,
Sp. G. x Density of Water = Density of Substance ----- Is what you wrote up there a typo? I'm guessing it is since You have to multiply both sides of the equation by the Density of Water to cancel out the Density of Water in the right side of the equation.

If so,
3.57 x 1000 kg / m^3 = 3570 kg / m^3 or 3.57 g / cm^3 is the density of the metal.

I'm just wondering, what Law or Principle says this? Is it stated in Archimedes' Principle? Because I might have passed through it while reading the book. :)
 
  • #4
42
0
So,

Sp. G. = 50 N / (50 N - 36 N)
Sp. G. = 50 N / 14 N
Sp. G. = 3.57

Then,
Sp. G. x Density of Water = Density of Substance ----- Is what you wrote up there a typo? I'm guessing it is since You have to multiply both sides of the equation by the Density of Water to cancel out the Density of Water in the right side of the equation.

If so,
3.57 x 1000 kg / m^3 = 3570 kg / m^3 or 3.57 g / cm^3 is the density of the metal.

I'm just wondering, what Law or Principle says this? Is it stated in Archimedes' Principle? Because I might have passed through it while reading the book. :)

Yes, It is included in Archimedes Principle. But these formula are not present in all of the books that has his Principle.

You got the density correct now how do you find the Density of the Oil? Read my previous post I edited it.
 
  • #5
63
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I can't quite understand the equation for the Specific Gravity of the Metal.

Can you be specific? I got lost in identifying which data I should use. What is the value I should use for the Specific Gravity of the fluid? and for the buoyant force. I got dizzy. Thank you for helping me.


Edit : Tried.....

Sp. G. Metal = ((3.57) (50 N)) / (50 N - 36 N)

Which gives 12.75 as the Specific Gravity.
 
Last edited:
  • #6
42
0
I can't quite understand the equation for the Specific Gravity of the Metal.

Can you be specific? I got lost in identifying which data I should use. What is the value I should use for the Specific Gravity of the fluid? and for the buoyant force. I got dizzy. Thank you for helping me.


Edit : Tried.....

Sp. G. Metal = ((3.57) (50 N)) / (50 N - 36 N)

Which gives 12.75 as the Specific Gravity.

Remember, You have the Specific Gravity of your Metal which is "3.57"

Now the formula states that:

Sp.Gr (Metal) = [Sp.Gr(Fluid)*(Weight in Air of Metal)] / Bouyant Force

Thus:

3.57 = [Sp.Gr(Fluid)*(Weight in Air of Metal)] / Bouyant Force

By Fluid I meant your unknown fluid which is : Oil

and by bouyant force I meant : ( Weight in Air of Metal - Weight in Oil of Metal )
 
  • #7
63
0
Oh...

3.57 = (Sp. G. Fluid)(50 N) / (50 N - 41 N)
3.57 = (Sp. G Fluid)(50 N) / 9 N
3.57 x 9 N = Sp. G Fluid x 50 N
(3.57 x 9 N) / 50 N = Sp. G. Fluid
Sp. G of Oil = 0.64

And then, Sp. G of oil = Density of Oil / Density of Water ???

Then 0.64 x (1 g / cm^3) = D of Oil
D of oil = .64 g / cm^3 ??
 
  • #8
42
0
Oh...

3.57 = (Sp. G. Fluid)(50 N) / (50 N - 41 N)
3.57 = (Sp. G Fluid)(50 N) / 9 N
3.57 x 9 N = Sp. G Fluid x 50 N
(3.57 x 9 N) / 50 N = Sp. G. Fluid
Sp. G of Oil = 0.64

And then, Sp. G of oil = Density of Oil / Density of Water ???

Then 0.64 x (1 g / cm^3) = D of Oil
D of oil = .64 g / cm^3 ??

Yep, Thats about it ;) Glad you understood the formulas.

Remember to make your life easier:

Sp.Gr.(Object) = Sp.Gr.(Fluid)*Weight of Object in Air / Bouyant Force
 
  • #9
63
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Yep, Thats about it ;) Glad you understood the formulas.

Remember to make your life easier:

Sp.Gr.(Object) = Sp.Gr.(Fluid)*Weight of Object in Air / Bouyant Force

Thank you very much for the help. :)

If you have any link or sites that you think can help me understand it more please, can you give it to me? I can't seem to see it in my book, which is really frustrating since something really new to me. I tried searching but still no luck.
 
  • #10
42
0
Thank you very much for the help. :)

If you have any link or sites that you think can help me understand it more please, can you give it to me? I can't seem to see it in my book, which is really frustrating since something really new to me. I tried searching but still no luck.

Actually, I just got the formula from my book. It's a custom made book of my Professor in our University.

Glad I can help.
 
  • #11
63
0
Actually, I just got the formula from my book. It's a custom made book of my Professor in our University.

Glad I can help.

Wow. You don't know how lucky you are! :)
Okay... Going kinda off topic now.

Thanks again! :)
 

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